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Example 5 Let $m, n, k$ be positive integers, $n \geqslant m+2, k$ be an odd number greater than 1, and $p=k \times 2^{n}+1$ be a prime number, $p \mid 2^{2^{m}}+1$. Prove: $k^{2^{n-1}} \equiv 1(\bmod p)$.
|
Given that $2^{2^{m}} \equiv-1(\bmod p)$, and $n \geqslant m+2$, hence $2^{m+1}$ is a factor of $n \cdot 2^{n-1}$, so, $2^{n \cdot 2^{n-1}} \equiv(-1)^{2 t}=1(\bmod p)$ (here $\left.t=n \cdot 2^{n m-2}\right)$.
Now, from $k \cdot 2^{n} \equiv-1(\bmod p)$, we know $k^{2^{n-1}} \cdot 2^{n \cdot 2^{n-1}} \equiv\left(-13^{2^{n-1}}=1(\bmod p)\right.$, combining the conclusion above, we can get $k^{2^{n-1}} \equiv 1(\bmod p)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Let $m$ be a positive integer, prove: there exist integers $a, b, k$, such that $a, b$ are both odd, and $k \geqslant 0$, and
$$2 m=a^{20}+b^{11}+k \cdot 2^{2011}.$$
|
To prove that (1) is equivalent to (under the condition that the left side is not less than the right side)
$$2 m \equiv a^{20}+b^{11}\left(\bmod 2^{2011}\right)$$
We first prove: there exist odd numbers $a, b$ satisfying (2).
Notice that, for any odd numbers $x, y$, we have
$$x^{11}-y^{11}=(x-y)\left(x^{10}+x^{9} y+\cdots+y^{10}\right),$$
The right side of the above equation, $x^{10}+x^{9} y+\cdots+y^{10}$, is the sum of 11 odd numbers, which must be odd. Therefore, $x^{11}-y^{11} \equiv 0\left(\bmod 2^{2011}\right) \Leftrightarrow x \equiv y\left(\bmod 2^{2011}\right)$. This indicates that, modulo $2^{2011}$, the numbers $1^{2011}$, $3^{2011}, \cdots, \left(2^{2011}-1\right)^{11}$ are a permutation of the numbers $1, 3, 5, \cdots, 2^{2011}-1$. Thus, there exists an odd number $b_{0}$ such that $b_{0}^{11} \equiv 2 m-1\left(\bmod 2^{2011}\right)$.
Now, take a sufficiently small negative odd number $b$ such that $b \equiv b_{0}\left(\bmod 2^{2011}\right)$ and $2 m-1-b^{11} \geqslant 0$, then $2 m-1-b^{11} \equiv 2 m-1-b_{0}^{11} \equiv 0\left(\bmod 2^{2011}\right)$. Therefore, let $(a, b, k) = \left(1, b, \frac{2 m-1-b^{11}}{2^{2011}}\right)$, which satisfies (1).
Thus, there exist $a, b, k$ that satisfy the conditions.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Proof: In decimal notation, among any 39 consecutive positive integers, there must be one number whose digit sum is a multiple of 11.
|
Proof: Since among any 10 consecutive positive integers, there must be one that is a multiple of 10, among any 39 consecutive positive integers, there must be 3 numbers that are multiples of 10. Among these 3 numbers, there must be one whose tens digit is no greater than 8, and this number must have at least 19 numbers following it within the 39 consecutive integers. Let this number be \(a\), and let its digit sum be \(S(a)\). Now consider the numbers
$$a, a+1, \cdots, a+9, a+19$$
These 11 numbers are all within the 39 consecutive integers, and by the choice of \(a\), their digit sums are \(S(a), S(a)+1, \cdots, S(a)+10\), forming 11 consecutive positive integers, among which there must be one that is a multiple of 11. The proposition is thus proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let $n$ be a positive odd number. Prove: Among the numbers
$$2-1,2^{2}-1, \cdots, 2^{n-1}-1$$
there must be a number that is a multiple of $n$.
|
Prove that when $n=1$, the proposition is obviously true.
Consider the case when $n>1$, at this point, in the numbers
$$1,2, \cdots, 2^{n-1}$$
there is no number that is a multiple of $n$, so the remainders when these numbers are divided by $n$ can only be $1,2, \cdots, n-1$. Therefore, there must be two numbers among these $n$ numbers that are congruent modulo $n$, i.e., there exist $0 \leqslant i<j \leqslant n-1$, such that $2^{i} \equiv 2^{j}(\bmod n)$. Since $n$ is odd, we have $\left(2^{i}, n\right)=1$, so $2^{j-i} \equiv 1(\bmod n)$, i.e., $n \mid 2^{j-i}-1$. The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let $m$ and $n$ be positive integers, $m$ is odd, and $\left(m, 2^{n}-1\right)=1$. Prove: The number $1^{n}+$ $2^{n}+\cdots+m^{n}$ is a multiple of $m$.
|
Prove that since $m$ is odd, and $1,2, \cdots, m$ is a complete residue system modulo $m$, then $2 \times 1, 2 \times 2, \cdots, 2 \times m$ is also a complete residue system modulo $m$, so,
$$1^{n}+2^{n}+\cdots+m^{n} \equiv (2 \times 1)^{n}+(2 \times 2)^{n}+\cdots+(2 \times m)^{n} \pmod{m}.$$
Thus, $m \mid \left(2^{n}-1\right)\left(1^{n}+2^{n}+\cdots+m^{n}\right)$, and combining with $\left(m, 2^{n}-1\right)=1$, the proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 (1) Prove: There exist infinitely many sets of integers $(x, a, b, c)$ such that
$$x^{2}+a^{2}=(x+1)^{2}+b^{2}=(x+2)^{2}+c^{2}$$
(2) Question: Does there exist an integer set $(x, a, b, c, d)$ such that
$$x^{2}+a^{2}=(x+1)^{2}+b^{2}=(x+2)^{2}+c^{2}=(x+3)^{2}+d^{2} ?$$
|
(1) For positive integers $k$ greater than 1, let $x=4 k^{3}-1, a=2 k^{2}+2 k, b=2 k^{2}+$ $1, c=2 k^{2}-2 k$, it can be known that the integer tuple $(x, a, b, c)$ meets the requirements.
The construction idea for $(x, a, b, c)$ is as follows:
From the requirements of the problem, we know $a^{2}-b^{2}=2 x+1, b^{2}-c^{2}=2 x+3$, thus, set $b=c+$ $n, a=b+m=c+n+m$, there should be
$$\left\{\begin{array}{l}
2 c n+n^{2}=2 x+3 \\
2 c m+2 m n+m^{2}=2 x+1
\end{array}\right.$$
This requires $m$ and $n$ to both be odd numbers. Subtracting the two equations, we get $c=\frac{1+n m}{n-m}-\frac{n+m}{2}$, to make it an integer, take $n=m+2$, then $c=\frac{1+m(m+2)}{2}-\frac{2 m+2}{2}=\frac{m^{2}-1}{2}$, let $m=2 k+1$, and we get our construction.
(2) There does not exist such an integer tuple.
In fact, for any integer $y$, we have
$$y^{2} \equiv\left\{\begin{array}{l}
0(\bmod 8), \text { if } y \equiv 0(\bmod 4) ； \\
1(\bmod 8), \text { if } y \equiv 1 \text { or } 3(\bmod 4) ； \\
4(\bmod 8), \text { if } y \equiv 2(\bmod 4) .
\end{array}\right.$$
Therefore, for integers $y$ and $z$, we have
$$y^{2}+z^{2} \equiv\left\{\begin{array}{l}
0,1 \text { or } 4(\bmod 8), \text { if } y \equiv 0(\bmod 4) ； \\
1,2 \text { or } 5(\bmod 8) ， \text { if } y \equiv 1 \text { or } 3(\bmod 4) ； \\
0,4 \text { or } 5(\bmod 8) ， \text { if } y \equiv 2(\bmod 4) .
\end{array}\right.$$
If there exists an integer tuple $(x, a, b, c, d)$ that meets the requirements, let $T=x^{2}+a^{2}$, since $x$, $x+1, x+2, x+3$ form a complete residue system modulo 4, without loss of generality, assume $x \equiv 0(\bmod 4)$, then $x+1 \equiv$ $1(\bmod 4), x+2 \equiv 2(\bmod 4)$, so, there should be
$$T(\bmod 8) \in\{0,1,4\} \cap\{1,2,5\} \cap\{0,4,5\}=\varnothing$$
This is a contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Let $n$ be a positive integer. Prove: there exists a positive integer whose digits are all odd, which is a multiple of $5^{n}$.
The text is translated while preserving the original line breaks and format.
|
Prove that we can construct a qualified $n$-digit positive integer using the recursive method.
When $n=1$, take $a_{1}=5$.
Assume that when $n=m$, there exists an $m$-digit positive integer $a_{m}$ with all digits being odd, such that $5^{m} \mid a_{m}$. Let $a_{m}=5^{m} \times q$, where $q \equiv r(\bmod 5)$, and $r=0,1,2,3$ or 4. Now consider the numbers
$$10^{m}, 3 \times 10^{m}, 5 \times 10^{m}, 7 \times 10^{m}, 9 \times 10^{m}$$
When divided by $5^{m}$, the quotients are
$$2^{m}, 3 \times 2^{m}, 5 \times 2^{m}, 7 \times 2^{m}, 9 \times 2^{m},$$
where the difference between any two numbers is not a multiple of 5, forming a complete residue system modulo 5. Therefore, there must be one number $\equiv 5-r(\bmod 5)$. Let $a \times 2^{m} \equiv 5-r(\bmod 5)$, where $a$ is one of the numbers $1, 3, 5, 7, 9$. Let $a_{m+1}=a \times 10^{m}+a_{m}$, then
$$5^{m} \mid a_{m+1},$$
and $\quad \frac{a_{m+1}}{5^{m}} \equiv a \times 2^{m}+r \equiv 5-r+r \equiv 0(\bmod 5)$,
thus
$$5^{m+1} \mid a_{m+1}$$
Therefore, there exists an $(m+1)$-digit positive integer $a_{m+1}$, with all digits being odd, and $5^{m+1} \mid a_{m+1}$. The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1
Define "Fermat number": $F_{n}=2^{2 n}+1$. When $n>1$, prove: if prime $p$ is a factor of $F_{n}$, then $p \equiv 1\left(\bmod 2^{n+1}\right)$
|
None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". Here is the formatted output as requested:
None
|
not found
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3
Define "Fermat number": $F_{n}=2^{2^{n}}+1$. When $n>1$, prove: if a prime $p$ is a factor of $F_{n}$, then $p \equiv 1\left(\bmod 2^{n+2}\right)$.
|
None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". Here is the formatted output as requested:
None
|
not found
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Proof: $1 \underset{200 \uparrow}{10 \cdots} 1$ is divisible by 1001.
|
$$\begin{aligned}
\underbrace{0 \cdots 01}_{200 \uparrow 0} & =10^{201}+1=\left(10^{3}\right)^{67}+1 \\
& =\left(10^{3}+1\right)\left[\left(10^{3}\right)^{66}-\left(10^{3}\right)^{65}+\cdots-10^{3}+1\right],
\end{aligned}$$
Therefore, $10^{3}+1(=1001)$ divides $1 \underbrace{0 \cdot \cdots 0}_{200 \uparrow 0} 1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
211 girls and $n$ boys went to pick mushrooms. All these children collected a total of $n^{2}+9 n-2$ mushrooms, and each child collected the same number of mushrooms. Determine whether there are more girls or boys among the children picking mushrooms.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
2. Since each child collected the same number of mushrooms, the total number of children $n+11$ must divide the total number of mushrooms
$$n^{2}+9 n-2=(n+11)(n-2)+20,$$
thus $n+11$ divides 20. Since $n+11>11$, $n$ can only be 9. Therefore, the number of girls is greater than the number of boys.
| null |
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let $n>1,2 \nmid n$, then for any integer $m>0$, we have $n \nmid\left(m^{n-1}+1\right)$.
|
Proof: Suppose there exists an odd number $n > 1$ such that $n \mid (m^{n-1} + 1)$, then $(m, n) = 1$. Let $p$ be any prime divisor of $n$, and $r$ be the order of $m$ modulo $p$ (note that $p \nmid m$). Assume $n-1 = 2^k t$, where $k \geq 1$ and $2 \nmid t$. Then we have
$$m^{2^k t} \equiv -1 \pmod{p}$$
Thus, $m^{2^{k+1} t} \equiv 1 \pmod{p}$, so $r \mid 2^{k+1} t$.
The key point is to prove that $2^{k+1} \mid r$. Suppose this conclusion is not true, then $r = 2^s r_1$, where $0 \leq s \leq k$ and $r_1 \mid t$. From $m^r \equiv 1 \pmod{p}$, we get $m^{2^k t} \equiv 1 \pmod{p}$, combining with (1) we get $p = 2$, which is a contradiction! Therefore, $2^{k+1} \mid r$.
Now, from $(p, m) = 1$, we get $m^{p-1} \equiv 1 \pmod{p}$, so $r \mid (p-1)$, hence $2^{k+1} \mid (p-1)$, i.e., $p \equiv 1 \pmod{2^{k+1}}$. Since $p$ is any prime divisor of $n$, decomposing $n$ into its standard form, we know $n \equiv 1 \pmod{2^{k+1}}$, i.e., $2^{k+1} \mid (n-1)$, but this contradicts the assumption that $2^k \parallel (n-1)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Let $p$ be an odd prime. Prove: any positive divisor of $\frac{p^{2 p}+1}{p^{2}+1}$ is $\equiv 1(\bmod 4 p)$.
|
To prove that any prime divisor $q$ of $\frac{p^{2 p}+1}{p^{2}+1}$ satisfies $q \equiv 1(\bmod 4 p)$. First, note that
$$\frac{p^{2 p}+1}{p^{2}+1}=p^{2(p-1)}-p^{2(p-2)}+\cdots-p^{2}+1$$
Thus, $q \neq p$. Let $r$ be the order of $p$ modulo $q$. Since
$$p^{2 p} \equiv-1(\bmod q)$$
it follows that $p^{4 p} \equiv 1(\bmod q)$, so $r \mid 4 p$. Therefore, $r=1,2,4, p, 2 p$ or $4 p$.
If $r=1,2, p, 2 p$, this would imply $p^{2 p} \equiv 1(\bmod q)$, which, combined with (2), leads to $q=2$, which is impossible; if $r=4$, then since $q$ is a prime, we conclude that $q \mid\left(p^{2}-1\right)$ or $q \mid\left(p^{2}+1\right)$. The former has been proven impossible. If the latter holds, i.e., $p^{2} \equiv-1(\bmod q)$. We take (1) modulo $q$, the left side is of course 0 modulo $q$, and the right side $\equiv(-1)^{p-1}-(-1)^{p-2}+\cdots-(-1)+1 \equiv p(\bmod q)$. Therefore, $p = q$, which is impossible, so $r \neq 4$. Therefore, it must be that $r=4 p$.
Finally, since $(p, q)=1$, by Fermat's Little Theorem, we have $p^{q-1} \equiv 1(\bmod q)$, so $r \mid(q-1)$, i.e., $4 p \mid(q-1)$, thus $q \equiv 1(\bmod 4 p)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 (1) Let $p$ be an odd prime, $a \neq \pm 1, p \nmid a$. Let $r$ be the order of $a$ modulo $p$, and $k_{0}$ satisfy $p^{k_{0}} \|\left(a^{r}-1\right)$. Denote $r_{k}$ as the order of $a$ modulo $p^{k}$, then we have
$$r_{k}=\left\{\begin{array}{l}
r, \text { if } k=1, \cdots, k_{0}, \\
r p^{k-k_{0}}, \\
\text { if } k>k_{0} .
\end{array}\right.$$
(2) Let $a$ be an odd number, $a \equiv 1(\bmod 4), a \neq 1, k_{0}$ satisfy $2^{k_{0}} \|(a-1)$. Denote $l_{k}$ as the order of $a$ modulo $2^{k}$, then we have
$$l_{k}=\left\{\begin{array}{l}
1, \text { if } k=1, \cdots, k_{0}, \\
2^{k-k_{0}}, \text { if } k>k_{0} .
\end{array}\right.$$
(3) Let $a$ be an odd number, $a \equiv-1(\bmod 4), a \neq-1, k_{0}$ satisfy $2^{k_{0}} \|(a+1)$. Denote $l_{k}$ as the order of $a$ modulo $2^{k}$, then we have
$$l_{k}=\left\{\begin{array}{l}
1, \text { if } k=1, \\
2, \text { if } k=2, \cdots, k_{0}+1, \\
2^{k-k_{0}}, \text { if } k>k_{0}+1 .
\end{array}\right.$$
|
Prove (1) When $1 \leqslant k \leqslant k_{0}$, from $a^{r_{k}} \equiv 1\left(\bmod p^{k}\right)$ we can deduce $a^{r_{k}} \equiv 1(\bmod p)$, hence by the definition of $r$ we have $r \mid r_{k}$. On the other hand, from $a^{r} \equiv 1\left(\bmod p^{k_{0}}\right)$ we can get $a^{r} \equiv 1\left(\bmod p^{k}\right)$, hence by the definition of $r_{k}$ we have $r_{k} \mid r$, thus $r_{k}=r\left(k=1, \cdots, k_{0}\right)$.
Now let $k>k_{0}$. We first prove that for each $i=0,1, \cdots$, we have $p^{k_{0}+i} \|\left(a^{\not p^{i}}-1\right)$, i.e.,
$$a^{r p^{i}}=1+p^{k_{0}+i} u_{i},\left(u_{i}, p\right)=1$$
This can be proven by induction: When $i=0$, by the definition of $k_{0}$, (1) holds. Suppose (1) holds for $i \geqslant 0$, then by the binomial theorem it is easy to see
$$\begin{aligned}
a^{a^{i+1}} & =\left(1+p^{k_{0}+i} u_{i}\right)^{p}=1+p^{k_{0}+i+1} u_{i}+C_{p}^{2} p^{2 k_{0}+2 i} u_{i}^{2}+\cdots \\
& =1+p^{k_{0}+i+1}\left(u_{i}+C_{p}^{2} p^{k_{0}+i-1} u_{i}^{2}+\cdots\right) \\
& =1+p^{k_{0}+i+1} u_{i+1}
\end{aligned}$$
It is easy to see that $p \nmid u_{i+1}$ (note that we need $p \geqslant 3$), thus (1) holds for all $i \geqslant 0$.
Using (1), we prove by induction on $k \geqslant k_{0}$ that $r_{k}=r p^{k-k_{0}}$. When $k=k_{0}$, the conclusion has been proven. If $k>k_{0}$, suppose we have $r_{k-1}=r p^{k-k_{0}-1}$. On one hand, taking $i=k-k_{0}$ in (1) we know $a^{p^{p-k_{0}}} \equiv 1\left(\bmod p^{k}\right)$, hence $r_{k} \mid r p^{k-k_{0}}$. On the other hand, from $a^{r_{k}} \equiv 1\left(\bmod p^{k}\right)$ we can deduce $a^{r_{k}} \equiv 1\left(\bmod p^{k-1}\right)$, hence $r_{k-1} \mid r_{k}$, thus $r_{k}=r p^{k-k_{0}}$ or $r p^{k-k_{0}-1}$. But taking $i=k-k_{0}-1$ in (1), we know $a^{r p^{k-k_{0}-1}} \neq 1\left(\bmod p^{k}\right)$, hence we must have $r_{k}=r p^{k-k_{0}}$.
(2) When $1 \leqslant k \leqslant k_{0}$, the conclusion is obviously true. When $k>k_{0}$, note that $a \equiv 1(\bmod 4)$, $a \neq 1$ implies $k_{0} \geqslant 2$, from which it is easy to prove by induction on $i=0,1, \cdots$ that
$$a^{2^{i}}=1+2^{k_{0}+i} u_{i}, 2 \nmid u_{i}$$
From (2), it is not difficult to deduce with the same argument as in (1) that $l_{k}=2^{k-k_{0}}\left(k \geqslant k_{0}\right)$.
(3) From $a \equiv-1(\bmod 4)$, it is easy to prove the conclusion for $k=1,2, \cdots, k_{0}+1$. Also, by induction it is not difficult to see that for $i=1,2, \cdots$, we have
$$a^{2^{i}}=1+2^{k_{0}+i} u_{i}, 2 \nmid u_{i}^{*}$$
From this, we can deduce with the same argument as in (1) that $l_{k}=2^{k-k_{0}}$ (for $k \geqslant k_{0}+1$).
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Let $a$ and $n$ be integers, neither of which is $\pm 1$, and $(a, n)=1$. Prove: there are at most finitely many $k$ such that $n^{k} \mid\left(a^{k}-1\right)$.
|
To prove that since $n \neq \pm 1$, $n$ has a prime factor. First, assume $n$ has an odd prime factor $p$, then $p \nmid a$. Let the order of $a$ modulo $p$ be $r$. Since $a \neq \pm 1$, there exists a positive integer $k_{0}$ such that $p^{k_{0}} \|\left(a^{r}-1\right)$.
If there are infinitely many $k$ such that $n^{k} \mid\left(a^{k}-1\right)$, then there are infinitely many $k > k_{0}$ satisfying
$$a^{k} \equiv 1\left(\bmod p^{k}\right)$$
But by Example 5, the order of $a$ modulo $p^{k}$ is $r p^{k-k_{0}}$, so by (1) we have $r p^{k-k_{0}} \mid k$, thus $k \geqslant r p^{k-k_{0}} \geqslant 3^{k-k_{0}}$, which clearly has only finitely many such $k$, leading to a contradiction.
If $n$ has no odd prime factors, then $n$ is a power of 2. First, note that if an odd $k$ makes $n^{k} \mid\left(a^{k}-1\right)$, then
$$a^{k}-1=(a-1)\left(a^{k-1}+\cdots+a+1\right)$$
is divisible by $2^{k}$. But the second factor in (2) is the sum of an odd number of odd numbers, hence it is odd, so $2^{k} \mid (a-1)$. Since $a \neq 1$, there are at most finitely many such $k$.
If there are infinitely many even $k=2 l$ such that $n^{k} \mid\left(a^{k}-1\right)$, then
$$\left(a^{2}\right)^{l} \equiv 1\left(\bmod 2^{l}\right)$$
Define $k_{0}$ such that $2^{k_{0}} \|\left(a^{2}-1\right)$, then $k_{0} \geqslant 3$. By Example 5, when $l > k_{0}$, the order of $a^{2}$ modulo $2^{l}$ is $2^{l-k_{0}}$, so for $l > k_{0}$, (3) implies $2^{l-k_{0}} \mid l$, thus $l \geqslant 2^{l-k_{0}}$, but there are at most finitely many such $l$, leading to a contradiction!
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
1 Prove: Any divisor of the Fermat number $F_{k}=2^{2^{k}}+1(k \geqslant 0)$ is $\equiv 1\left(\bmod 2^{k+1}\right)$.
|
1. It suffices to prove that any prime factor $p$ of $F_{k}$ satisfies $p \equiv 1\left(\bmod 2^{k+1}\right)$. Clearly, $p \neq 2$. Let the order of 2 modulo $p$ be $r$. From $p \mid F_{k}$, we get
$$2^{2^{k}} \equiv-1(\bmod p)$$
Thus, $2^{2^{k+1}} \equiv 1(\bmod p)$, so $r \mid 2^{k+1}$, meaning $r$ is a power of 2. Let $r=2^{l}$, where $0 \leqslant l \leqslant k+1$. If $l \leqslant k$, then from $2^{2^{l}} \equiv 1(\bmod p)$, repeated squaring can lead to $2^{2^{k}} \equiv 1(\bmod p)$, combining with (1) we get $p=2$, which is impossible. Therefore, it must be that $l=k+1$. Also, $2^{p-1} \equiv 1(\bmod p)$, so $r \mid (p-1)$, hence $2^{k+1} \mid (p-1)$, i.e., $p \equiv 1\left(\bmod 2^{k+1}\right)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
3 Prove that for any integer $k>0$, there exists a positive integer $n$, such that $2^{k} \mid\left(3^{n}+5\right)$.
|
3. Using induction. For $k=1,2$, the conclusion is obviously true. Suppose for $k \geqslant 3$ there exists $n_{0}$ such that $2^{k} \mid \left(3^{n_{0}}+5\right)$, and let $3^{n_{0}}=2^{k} u-5$. If $u$ is even, then $2^{k+1} \mid \left(3^{n_{0}}+5\right)$. Now assume $u$ is odd.
The key to the argument is to note that for $k \geqslant 3$,
$$3^{2^{2^{-2}}}=1+2^{k} v, \text{ where } v \text{ is odd.}$$
(See (3) in Example 5 of this unit.) Now we have
$$\begin{aligned}
3^{n_{0}+2^{2^{-2}}} & =3^{n_{0}} \cdot 3^{2^{2-2}}=\left(-5+2^{k} u\right) \left(1+2^{k} v\right) \\
& =-5+\left(u-5 v+2^{k} u v\right) \cdot 2^{k} .
\end{aligned}$$
The number in the parentheses is even, so $2^{k+1}$ divides $3^{n_{0}+2^{k-2}}+5$. This completes the inductive proof.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 If $n \equiv 4(\bmod 9)$, prove that the indeterminate equation
$$x^{3}+y^{3}+z^{3}=n$$
has no integer solutions $(x, y, z)$.
|
Prove that if equation (1) has integer solutions, then (1) modulo 9 also has integer solutions. It is well known that a perfect cube modulo 9 is congruent to $0, 1, -1$ (or equivalently $0, 1, 8$), hence
$$x^{3}+y^{3}+z^{3} \equiv 0,1,2,3,6,7,8(\bmod 9)$$
But $n \equiv 4(\bmod 9)$, so (1) modulo 9 has no solutions, which contradicts the previous statement, hence equation (1) has no integer solutions. $\quad$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
3 Let the decimal representation of the positive integer $n$ be $n=\overline{a_{k} \cdots a_{1} a_{0}}\left(0 \leqslant a_{i} \leqslant 9, a_{k} \neq 0\right)$, and let $T(n)=a_{0}-a_{1}+\cdots+(-1)^{k} a_{k}$ (the alternating sum of the digits of $n$ starting from the units digit). Prove that $n-T(n)$ is divisible by 11. From this, derive the digital characteristic for divisibility by 11: 11 divides $n$ if and only if 11 divides $T(n)$.
|
3. We have
$$n-T(n)=\left(a_{0}-a_{0}\right)+\left(10 a_{1}+a_{1}\right)+\cdots+\left(a_{k} \times 10^{k}-(-1)^{k} a_{k}\right) .$$
It is easy to see that for $i=0,1, \cdots, k$, the number $a_{i} \times 10^{i}-(-1)^{i} a_{i}$ is divisible by 11 (using decomposition (5) for even $i$ and decomposition (6) for odd $i$). Therefore, $n-T(n)$ is divisible by 11, so both conclusions in the problem are valid.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Proof: The following numbers cannot be expressed as the sum of cubes of several consecutive integers.
(1) $385^{97}$;
(2) $366^{17}$.
|
Prove that using
$$1^{3}+2^{3}+\cdots+k^{3}=\left(\frac{k(k+1)}{2}\right)^{2}$$
it is easy to see that the sum of cubes of several consecutive integers can be expressed in the form
$$\left(\frac{m(m+1)}{2}\right)^{2}-\left(\frac{n(n+1)}{2}\right)^{2},$$
where \( m \) and \( n \) are integers. We need to prove that for the integers in (1) and (2), there do not exist \( m \) and \( n \) such that they can be expressed in the form (1). While factorization can also solve the problem, it is quite cumbersome; using congruence arguments, however, is much more straightforward.
First, classify the integer \( x \) modulo 9 and check each case. It is not difficult to see that \(\left(\frac{x(x+1)}{2}\right)^{2}\) is congruent to 0 or -1 modulo 9. Therefore, numbers of the form (1) modulo 9 can only be \( 0, 1, -1 \). On the other hand, by Euler's theorem,
$$385^{97} \equiv 385 \times\left(385^{16}\right)^{6} \equiv 385 \equiv 7(\bmod 9)$$
This proves that \( 385^{97} \) cannot be expressed in the form (1).
However, since \( 366^{17} \equiv 0(\bmod 9) \), the modulo 9 approach cannot resolve the problem for the number \( 366^{17} \).
This time, we use modulo 7. It is easy to verify that for any integer \( x \), the number \(\left(\frac{x(x+1)}{2}\right)^{2}\) is congruent to 0, 1, or -1 modulo 7. Hence, numbers of the form (1) modulo 7 can only be \( 0, \pm 1, \pm 2 \). But
$$366^{17} \equiv 2^{17} \equiv 2 \times 2^{4} \equiv 4(\bmod 7)$$
Therefore, our assertion holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Proof, the indeterminate equation
$$(x+2)^{2 m}=x^{n}+2$$
has no positive integer solutions.
|
To prove, for the subsequent argument, we first derive some simple conclusions from equation (1).
Clearly, $n>1$. Moreover, $x$ must be odd; otherwise, taking (1) modulo 4 leads to a contradiction. Furthermore, $n$ is also odd, because if $2 \mid n$, then $x^{n}$ is the square of an odd number, making the right side of (1) $\equiv 1+2=3(\bmod 4)$, while the left side $\equiv 1(\bmod 4)$, which is impossible. Hence, $2 \nmid n$.
Let $x+1=2^{\alpha} x_{1}$, where $x_{1}$ is odd and $\alpha>0$ (since $x$ is odd). Rewrite equation (1) as
$$(x+2)^{2 m}-1=x^{n}+1$$
The left side of (2) has a factor $(x+2)^{2}-1=\left(2^{\alpha} x_{1}+1\right)^{2}-1=2^{a+1}\left(2^{\sigma-1} x_{1}^{2}+x_{1}\right)$, so $2^{a+1}$ divides the left side of (2). On the other hand, since $n-1>0$ is even, using the binomial theorem, we easily get
$$x^{n}+1=x\left(2^{a} x_{1}-1\right)^{n-1}+1 \equiv x \cdot 1+1=2^{a} x_{1}\left(\bmod 2^{a+1}\right)$$
Since $2 \nmid x_{1}$, the right side of (2) $x^{n}+1 \not \equiv 0\left(\bmod 2^{a+1}\right)$, leading to a contradiction!
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 7 Proof: The indeterminate equation
$$8^{x}+15^{y}=17^{x}$$
has the only positive integer solution $x=y=2$.
|
First, we use congruences to prove that $y$ and $z$ are both even.
Equation (1) modulo 4 gives
$$(-1)^{y} \equiv 1(\bmod 4)$$
Thus, $y$ is even. Modulo 16 of equation (1) gives
$$8^{x}+(-1)^{y} \equiv 1(\bmod 16)$$
which simplifies to $8^{x} \equiv 0(\bmod 16)$, hence $x \geqslant 2$.
Note that $17^{2} \equiv 1,15^{2} \equiv 1(\bmod 32)$. If $z$ is odd, then from (1) with $2 \mid y$ and $x \geqslant 2$, we get
$$1 \equiv 17(\bmod 32),$$
which is impossible. Therefore, $z$ must be even.
Let $y=2 y_{1}, z=2 z_{1}$, then equation (1) can be factored as
$$\left(17^{z_{1}}-15^{y_{1}}\right)\left(17^{z_{1}}+15^{y_{1}}\right)=8^{x} .$$
It is easy to see that the greatest common divisor of the two factors on the left side of (2) is 2, and the right side of (2) is a power of 2, so we must have
$$\left\{\begin{array}{l}
17^{z_{1}}-15^{y_{1}}=2 \\
17^{z_{1}}+15^{y_{1}}=2^{3 x_{n}-1}
\end{array}\right.$$
Taking (3) modulo 32, we see that $z_{1}$ and $y_{1}$ must both be odd (otherwise, the left side of (3) $\left.\equiv 0,-14,16(\bmod 32)\right)$. Adding (3) and (4), we get
$$17^{z_{1}}=1+2^{3-2}$$
If $x \geqslant 3$, then the right side of (5) $\equiv 1(\bmod 32)$; and since $z_{1}$ is odd, the left side $\equiv 17(\bmod 32)$, which is impossible. Therefore, we must have $x=2$. From this and (5), we get $z_{1}=1$, i.e., $z=2$, and it follows easily that $y_{1}=1$, i.e., $y=2$. Thus, $x=y=z=2$.
|
x=y=2
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
I prove that the indeterminate equation
$$x^{2}+3 x y-2 y^{2}=122$$
has no integer solutions.
|
1. Transform the equation into
$$(2 x+3 y)^{2}=17 y^{2}+4 \times 122$$
Taking modulo 17, we get $(2 x+3 y)^{2} \equiv 12(\bmod 17)$. However, it is easy to verify that the square of an integer modulo 17 can only be one of $0, 1, 2, 4, 8, 9, 13, 15, 16$, and cannot be 12. Therefore, the original equation has no integer solutions.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4 Let $n$ integers have the following property: the product of any $n-1$ of them minus the remaining number is divisible by $n$. Prove: the sum of the squares of these $n$ numbers is also divisible by $n$.
保留源文本的换行和格式，直接输出翻译结果。
|
4. Let $a_{1}, \cdots, a_{n}$ be integers with the given property, and let $A$ be their product. For $1 \leqslant i \leqslant n$, the number $n$ divides $\frac{A}{a_{i}}-a_{i}$, and thus it also divides
$$a_{i}\left(\frac{A}{a_{i}}-a_{i}\right)=A-a_{i}^{2}$$
Therefore, $n$ divides the sum $\left(A-a_{1}^{2}\right)+\cdots+\left(A-a_{n}^{2}\right)=n A-\left(a_{1}^{2}+\cdots+a_{n}^{2}\right)$. Hence, $n$ divides $a_{1}^{2}+\cdots+a_{n}^{2}$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4 Prove: The indeterminate equation
$$5^{x}-3^{y}=2$$
has only the positive integer solution $x=y=1$.
|
4. The equation clearly has the solution $x=y=1$. Taking the equation modulo 4, it is easy to see that $y$ is odd. If $y>1$, taking the equation modulo 9 yields
$$5^{x} \equiv 2(\bmod 9)$$
It is not difficult to find that for $x=1,2, \cdots, 5^{x}$ modulo 9 cycles as $5,7,8,4,2,1$. Hence, by (1), $x$ must be of the form $6 k+5$. Taking the original equation modulo 7, it is easy to verify that for odd $y$,
$$3^{y} \equiv 3,5,6(\bmod 7)$$
When $x=6 k+5$, by Fermat's Little Theorem, $5^{6} \equiv 1(\bmod 7)$, so
$$5^{x}=5^{6 k+5} \equiv 5^{5} \equiv 3(\bmod 7)$$
Thus, the two sides of the original equation are not congruent modulo 7, so it has no solutions for $y>1$. Therefore, the only positive integer solution is $y=1, x=1$
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5 Prove that $x^{3}+y^{4}=7$ has no integer solutions.
|
5. It is easy to verify that $x^{3} \equiv 0,1,5,8,12(\bmod 13) ; y^{4} \equiv 0,1,3,9(\bmod 13)$. From these, it is clear that $x^{3}+y^{4} \not \equiv 7(\bmod 13)$, hence the equation has no integer solutions.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Let $u$ be a given positive integer, prove that the equation
$$n!=u^{x}-u^{y}$$
has at most finitely many positive integer solutions $(n, x, y)$.
|
Prove that we can assume $u>1$. The conclusion is equivalent to proving that the equation
$$n!=u^{r}\left(u^{s}-1\right)$$
has at most finitely many positive integer solutions $(n, r, s)$.
First, note that for a given $n$, equation (1) clearly has at most finitely many solutions $(r, s)$. We will prove that when $n$ is sufficiently large, equation (1) has no solutions, which will prove the above conclusion.
Choose a prime $p \nmid u$. We can assume (1) has a solution $n>p$ (otherwise, there is nothing to prove), and let $p^{\alpha} \| n!$, then
$$\alpha=\sum_{l=1}^{\infty}\left[\frac{n}{p^{l}}\right] \geqslant\left[\frac{n}{p}\right]>a n,$$
where $a$ is a (positive) constant that depends only on $p$.
Let the order of $u$ modulo $p$ be $d$ and $p^{k_{0}} \|\left(u^{d}-1\right)$, then by Example 5 of Unit 8, when $\alpha>k_{0}$, the order of $u$ modulo $p^{\alpha}$ is $d p^{\alpha-k_{0}}$. Since $u$ and $p$ are fixed numbers, $k_{0}$ and $d$ are also fixed numbers. If (1) has solutions for sufficiently large $n$, then by (2) we know $\alpha>k_{0}$. From (1) we get
$$u^{s} \equiv 1\left(\bmod p^{\alpha}\right),$$
hence by the property of the order, we have $d p^{\alpha-k_{0}} \mid s$; in particular, $s \geqslant d p^{\alpha-k_{0}}$. Therefore,
$$u^{s}-1 \geqslant u^{d p^{\alpha-k_{0}}}-1>u^{d p^{\alpha-k_{0}}-k_{0}}-1.$$
But when $n$ is sufficiently large, it is easy to see that the right-hand side of the above inequality $\geqslant n^{n}-1$. Hence by (3) we get $u^{s}-1>n!$, and thus $u^{r}\left(u^{s}-1\right)>n!$, so when $n$ is sufficiently large, (1) has no positive integer solutions $(r, s)$. This completes the proof.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Proof: For each $n>1$, the equation
$$\frac{x^{n}}{n!}+\frac{x^{n-1}}{(n-1)!}+\cdots+\frac{x^{2}}{2!}+\frac{x}{1!}+1=0$$
has no rational roots.
|
Proof Let $a$ be a rational root of the given equation, then it is easy to see that
$$a^{n}+\frac{n!}{(n-1)!} a^{n-1}+\cdots+\frac{n!}{k!} a^{k}+\cdots+\frac{n!}{1!} a+n!=0,$$
Thus, $a$ is a rational root of an integral polynomial with leading coefficient 1, so $a$ must be an integer (see Exercise 2, Question 4).
Since $n>1$, $n$ has a prime factor $p$ (a basic fact that has been used multiple times). Because $n \left\lvert\, \frac{n!}{k!}(k=\right.$ $0,1, \cdots, n-1$ ), it follows from (1) that $p \mid a^{n}$, hence the prime $p$ divides $a$. Now compare the powers of $p$ in the terms on the left side of (1). Since the number of times $p$ appears in $k!$ is
$$\sum_{l=1}^{\infty}\left[\frac{k}{p^{l}}\right]<\sum_{l=1}^{\infty} \frac{k}{p^{l}}<k$$
it follows that $p^{k} \nmid k!(k \geqslant 1)$. Let $p^{r} \| n!$. Then, since $p^{k} \mid a^{k}$ and $p^{k} \nmid k!$, we have $p^{r+1} \left\lvert\, \frac{n!}{k!} a^{k}(k=\right.$ $1,2, \cdots, n)$, and thus from (1) we get $p^{r+1} \mid n!$, which contradicts the definition of $r$.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Let $n>1, x_{1}, \cdots, x_{n}$ be $n$ real numbers, and their product be denoted as $A$. If for $i=1, \cdots, n$, the number $A-x_{i}$ is an odd integer. Prove: Each $x_{i}$ is an irrational number.
|
Proof by contradiction: If there exists an $i$ such that $x_{i}$ is a rational number, then because $A-x_{i}$ is an odd integer, $A$ must be a rational number. Let $A-x_{i}=a_{i} (i=1, \cdots, n)$. Then from $x_{1} \cdots x_{n}=A$, we get
$$\left(A-a_{1}\right) \cdots\left(A-a_{n}\right)=A$$
Since $a_{i}$ are all (odd) integers, $A$ satisfies a monic polynomial equation with integer coefficients, so the rational number $A$ must be an integer. However, on the other hand, whether $A$ is odd or even, it is easy to see that the parity of both sides of equation (1) is different, so (1) cannot hold, leading to a contradiction! Therefore, each $x_{i}$ must be an irrational number.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Let $a, b, c$ be integers, $f(x)=x^{3}+a x^{2}+b x+c$. Prove: there are infinitely many positive integers $n$, such that $f(n)$ is not a perfect square.
|
We prove that for any positive integer $n \equiv 1(\bmod 4)$, among the four integers $f(n), f(n+1), f(n+2), f(n+3)$, at least one is not a perfect square, thereby proving the conclusion of the problem.
It is easy to see that
$$\begin{array}{l}
f(n) \equiv 1+a+b+c(\bmod 4) \\
f(n+1) \equiv 2 b+c(\bmod 4) \\
f(n+2) \equiv-1+a-b+c(\bmod 4) \\
f(n+3) \equiv c(\bmod 4)
\end{array}$$
Eliminating $a$ and $c$, we get
$$f(n+1)-f(n+3) \equiv 2 b, \quad f(n)-f(n+2) \equiv 2 b+2(\bmod 4)$$
Therefore, either $f(n+1)-f(n+3) \equiv 2(\bmod 4)$, or $f(n)-f(n+2) \equiv 2(\bmod 4)$. Since a perfect square is congruent to 0 or 1 modulo 4, either $f(n+1)$ and $f(n+3)$ include at least one non-square number, or $f(n)$ and $f(n+2)$ include at least one non-square number. Thus, among $f(n), f(n+1), f(n+2), f(n+3)$, at least one is not a perfect square.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Let $p(x)$ be a polynomial with integer coefficients, and for any $n \geqslant 1$ we have $p(n)>n$. Define $x_{1}=$ $1, x_{2}=p\left(x_{1}\right), \cdots, x_{n}=p\left(x_{n-1}\right)(n \geqslant 2)$. If for any positive integer $N$, the sequence $\left\{x_{n}\right\}$ $(n \geqslant 1)$ contains a term divisible by $N$. Prove that $p(x)=x+1$.
|
Proof We proceed in two steps. First, we prove that for any fixed $m>1$, the sequence $\left\{x_{n}\right\}$ modulo $x_{m}-1$ is a periodic sequence. Clearly, $x_{m} \equiv 1=x_{1}\left(\bmod x_{m}-1\right)$. Since $p(x)$ is a polynomial with integer coefficients, for any integers $u, v (u \neq v)$, we have $(u-v) \mid (p(u)-p(v))$, i.e.,
$$p(u) \equiv p(v)(\bmod u-v)$$
Taking $u=x_{m}, v=x_{1}=1$ in the above, we get $x_{m+1} \equiv x_{2}\left(\bmod x_{m}-1\right)$. Proceeding similarly, $x_{m+2} \equiv x_{3}, x_{m+3} \equiv x_{4}, \cdots\left(\bmod x_{m}-1\right)$, which shows that $\left\{x_{n}\right\}$ modulo $x_{m}-1$ is the periodic sequence $x_{1}, \cdots, x_{m-1}$, $x_{1}, \cdots, x_{n-1}, \cdots$
In the second step, we prove
$$x_{m}-1=x_{m-1}$$
By the given condition, for the number $N=x_{m}-1$, there exists $x_{k}$ such that $\left(x_{m}-1\right) \mid x_{k}$. From the conclusion of the previous part, we can assume $1 \leqslant k \leqslant m-1$. Moreover, $p\left(x_{m-1}\right)>x_{m-1}$, so $x_{m}-1 \geqslant x_{m-1}$. Therefore, $k$ must be $m-1$, i.e., $\left(x_{m}-1\right) \mid x_{m-1}$. Thus, $x_{m-1} \geqslant x_{m}-1$, combining this with the previous inequality, we conclude that (1) holds.
Since (1) is $p\left(x_{m-1}\right)-1=x_{m-1}$. Given that $m$ is any integer greater than 1, this implies that $p(x)=x+1$ has infinitely many distinct roots, so $p(x)$ must be identically equal to the polynomial $x+1$. This completes the proof of the conclusion.
|
p(x)=x+1
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 7 Let $f(x)$ be a quadratic polynomial with real coefficients. If for all positive integers $n$, $f(n)$ is the square of an integer. Prove that $f(x)$ is the square of a linear polynomial with integer coefficients.
|
Proving this problem is not easy, but there are several completely different solutions. The method introduced here is based on the knowledge of the limit of sequences, which is relatively simple.
Let $f(x)=a x^{2}+b x+c, a_{n}=f(n)(n \geqslant 1)$, then it is easy to know
$$\begin{aligned}
\sqrt{a_{n}}-\sqrt{a_{n-1}} & =\frac{a_{n}-a_{n-1}}{\sqrt{a_{n}}+\sqrt{a_{n-1}}} \\
& =\frac{2 a n-a+b}{\sqrt{a n^{2}+b n+c}+\sqrt{a n^{2}+(-2 a+b) n+a-b+c}} \\
& =\frac{2 a+\frac{b-a}{n}}{\sqrt{a+\frac{b}{n}+\frac{c}{n^{2}}}+\sqrt{a+\frac{b-2 a}{n}+\frac{a-b+c}{n^{2}}}}
\end{aligned}$$
Therefore, when $n \rightarrow \infty$, $\sqrt{a_{n}}-\sqrt{a_{n-1}}$ has a limit, and the limit value is $\frac{2 a}{\sqrt{a}+\sqrt{a}}=\sqrt{a}$. But it is known that $\sqrt{a_{n}}$ are all integers, so $\left\{\sqrt{a_{n}}-\sqrt{a_{n-1}}\right\}(n \geqslant 2)$ is an integer sequence, thus its limit value $\sqrt{a}$ must be an integer, and for sufficiently large $n$, all terms $\sqrt{a_{n}}-\sqrt{a_{n-1}}$ are equal to the limit $\sqrt{a}$, i.e., there is a (fixed) positive integer $k$, such that
$$\sqrt{a_{n}}-\sqrt{a_{n-1}}=\sqrt{a} \text {, for } n \geqslant k+1 \text {. }$$
Now let $m$ be any integer greater than $k$, summing the above equation for $n=k+1, \cdots, m$, we get $\sqrt{a_{m}}=$ $\sqrt{a_{k}}+(m-k) \sqrt{a}$, i.e.,
$$a_{m}=\left(m \sqrt{a}+\sqrt{a_{k}}-k \sqrt{a}\right)^{2} .$$
Let $\alpha=\sqrt{a}, \beta=\sqrt{a_{k}}-k \sqrt{a}$, then $\alpha, \beta$ are fixed integers independent of $m$, thus, (1) indicates that all integers $m$ greater than $k$ are roots of the polynomial
$$f(x)-(\alpha x+\beta)^{2}$$
Therefore, this polynomial must be the zero polynomial, i.e., $f(x)=(\alpha x+\beta)^{2}$.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5 Let integers $a, b, c, d$ satisfy $ad - bc > 1$, prove: at least one of $a, b, c, d$ is not divisible by $ad - bc$.
保留源文本的换行和格式，直接输出翻译结果。
|
5. If $a, b, c, d$ are all divisible by $ad-bc$, then $(ad-bc)^2$ divides $ad$ and $bc$, hence divides $ad-bc$, from which it follows that $|ad-bc|=1$, which contradicts the given $ad-bc>1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 8 Let $n>1, n$ positive integers sum to $2 n$. Prove that, among them, we can always select some numbers whose sum equals $n$, unless the given numbers satisfy one of the following conditions:
(1) There is one number that is $n+1$, and the rest are 1;
(2) When $n$ is odd, all numbers are equal to 2.
|
Proof - Let the given positive integer be $01$, then we have
$$\ddot{a_{2}}+\cdots+a_{k}+a_{n} \equiv 0(\bmod n)$$
The left side of the above equation is clearly a positive integer less than $a_{1}+\cdots+a_{n}=2 n$, hence
$$a_{2}+\cdots+a_{k}+a_{n}=n$$
(iv) Suppose $a_{1}-a_{n} \equiv 0(\bmod n)$. We have already proven that $a_{n} \leqslant n+1$ (see (2)). If $a_{n}=$ $n+1$, then the sum of the $n-1$ positive integers $a_{1}, \cdots, a_{n-1}$ equals $2 n-a_{n}=n-1$, thus they all equal 1, which is precisely the case excluded in the problem (1).
Assume $a_{n} \leqslant n$, then $0 \leqslant a_{n}-a_{1} \leqslant n-1$, combining with $a_{n}-a_{1} \equiv 0(\bmod n)$, we deduce $a_{n}=\cdots=a_{2}=a_{1}=2$. When $n$ is odd, this is the case excluded in the problem (2). If $n$ is even, then the sum of any $\frac{n}{2}$ $a_{i}$'s equals $n$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 9 Let $p$ be a prime, and given $p+1$ distinct positive integers. Prove that one can select a pair of these numbers such that when the larger of the two is divided by their greatest common divisor, the quotient is at least $p+1$.
|
Prove that dividing the given $p+1$ numbers by their greatest common divisor obviously does not affect the conclusion of this problem, so we can assume that these $p+1$ numbers are coprime. In particular, there must be a number that is not divisible by $p$. Let these $p+1$ numbers be
$$x_{1}, \cdots, x_{k}, x_{k+1}=p^{l_{k+1}} y_{k+1}, \cdots, x_{p+1}=p^{l_{p+1}} y_{p+1}$$
Here, $\dot{x}_{1}, \cdots, x_{k}$ are distinct and coprime with $p$ $(k \geqslant 1)$, $l_{k+1}, \cdots, l_{p+1}$ are positive integers, and $y_{k+1}, \cdots, y_{p+1}$ are positive integers not divisible by $p$.
Among the $p+1$ numbers
$$x_{1}, \cdots, x_{k}, y_{k+1}, \cdots, y_{p+1}$$
there must be two that are congruent modulo $p$. We discuss three cases.
(1) At least three numbers in (1) are equal. In this case, the conclusion is easy to prove. If $y_{r}=y_{s}=y_{t}$, then $p^{l_{r}}, p^{l_{s}}, p^{l_{t}}$ are distinct, and the largest number is at least $p^{2}$ times the smallest. Without loss of generality, assume $p^{l_{r}} \geqslant p^{2} \cdot p^{l_{t}}$, then $x_{r}$ and $x_{t}$ satisfy the condition; if $y_{r}=y_{s}=x_{t}(1 \leqslant t \leqslant k)$, without loss of generality, assume $l_{r}>l_{s}$, then $l_{r} \geqslant 2$, so $x_{r}$ and $x_{t}$ satisfy the condition.
(2) Two pairs of numbers in (1) are equal. If $y_{i}=y_{j}, y_{r}=y_{s}$, then when $\left|l_{i}-l_{j}\right| \geqslant 2$ or $\left|l_{r}-l_{s}\right| \geqslant 2$, the conclusion holds as above; when $\left|l_{i}-l_{j}\right| \leqslant 1$ and $\left|l_{r}-l_{s}\right| \leqslant 1$, we can rename $x_{i}, x_{j}, x_{r}, x_{s}$ as $a, a p, b, b p$, and $ap$
Thus, the integer $\frac{b p}{(a, b p)} \geqslant p+1$.
If $x_{i}=y_{r}, x_{j}=y_{s}(1 \leqslant i, j \leqslant k)$, the conclusion can also be proven similarly.
(3) Exactly two numbers in (1) are equal. This can only be $y_{r}=y_{s}$, or $x_{i}=y_{r}(1 \leqslant i \leqslant k)$. In this case, we can remove $y_{r}$ from (1), leaving $p$ numbers that are distinct but still have two that are congruent modulo $p$. Now there are three possibilities:
(i) Suppose $y_{r} \equiv y_{s}(\bmod p)$. Without loss of generality, assume $y_{r}>y_{s}$. If $l_{r}>l_{s}$, the conclusion is obvious. If $l_{r} \leqslant l_{s}$, let $y_{r}=y_{s}+n$, then $n>0$, and $p \mid n$. Let $\left(y_{r}, y_{s}\right)=d$, then $p \nmid d$, so $\left(x_{r}, x_{s}\right)=p^{l} d$. We have (note $d|n, p| n$, and $p \nmid d$)
$$\frac{x_{r}}{\left(x_{r}, x_{s}\right)}=\frac{y_{r}}{d}=\frac{y_{s}}{d}+\frac{n}{d} \geqslant 1+p$$
Thus, the larger of $x_{r}$ and $x_{s}$, when divided by their greatest common divisor, yields a quotient of at least $p+1$.
(ii) Suppose $x_{r} \equiv x_{s}(\bmod p)(1 \leqslant r < s \leqslant k)$. Without loss of generality, assume $x_{r}>x_{s}$. If $y_{s} < y_{r}$, the conclusion is obvious. If $y_{s} > y_{r}$, let $y_{s}=y_{r}+n$, then $n>0$, and $p \mid n$. Let $\left(x_{r}, y_{s}\right)=d$, then $p \nmid d$, so $\left(x_{r}, x_{s}\right)=\left(x_{r}, p^{l_{s}} y_{s}\right)=d$, thus
$$\frac{x_{r}}{\left(x_{r}, x_{s}\right)}=\frac{y_{s}}{d}+\frac{n}{d} \geqslant 1+p$$
Thus, the larger of $x_{r}$ and $x_{s}$, when divided by their greatest common divisor, yields a quotient of at least $p+1$. This completes the proof of the problem.
We note that if the $p+1$ integers in Example 9 are replaced by $p$ integers, the conclusion may not hold. For example, the $p$ numbers $1,2, \cdots, p$ clearly do not have two numbers that satisfy the condition.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 10 Let $S$ be a subset of $\left\{1,2, \cdots, 2^{m} n\right\}$, and the number of elements in $S$ is $|S| \geqslant\left(2^{m}-\right.$ 1) $n+1$. Prove that there are $m+1$ distinct numbers $a_{0}, \cdots, a_{m}$ in $S$, such that $a_{i-1} \mid a_{i}(i=1, \cdots, m)$.
|
Prove that each positive integer $a$ can be uniquely represented in the form $2 u k$, where $u \geqslant 0$ and $k$ is odd. We call $k$ the odd part of $a$, and if the odd part of $a$ does not exceed $n$, then $n$ is called a good number. This proof is based on a lower bound estimate of the number of good numbers in $S$. For this, we first count how many good numbers are in the interval $\left(n, 2^{m} n\right]$.
Let the interval $[1, n]$ contain $t$ odd numbers (which is actually equal to $\left[\frac{n+1}{2}\right]$, but we do not need this). Let $k$ be any such odd number, then the integers $u$ satisfying $n<2^{u} k \leqslant 2^{m} n$ are exactly $m$ in number. This is because, if the integer $v$ satisfies $2^{v-1} \leqslant \frac{n}{k}<2^{v}$, then $2^{v} k, 2^{v+1} k, \cdots, 2^{v-1+m} k$ are all the numbers that meet the requirement, i.e., there are $m$ numbers in the interval $\left(n, 2^{m} n\right]$ whose odd part is $k$, so there are exactly $m t$ good numbers in this interval. Therefore, the number of non-good numbers in this interval is $2^{m} n-n-m t$, and thus the number of good numbers in $S$ is $\geqslant|S|-\left(2^{m} n-n-m t\right)=m t+1$.
Let $k_{1}, \cdots, k_{t}$ be all the odd numbers in $[1, n]$, and let $S$ contain exactly $x_{i}$ numbers with $k_{i}$ as their odd part $(k=1, \cdots, t)$, then by the conclusion of the previous paragraph, the number of good numbers in $S$ is
$$x_{1}+\cdots+x_{t} \geqslant m t+1$$
Thus, there must be an $x_{i}(1 \leqslant i \leqslant t)$ such that $x_{i} \geqslant m+1$, i.e., $S$ contains at least $m+1$ integers with the same odd part $k_{i}$. These numbers, arranged from smallest to largest, are $a_{0}, a_{1}, \cdots, a_{m}$, which are the $m$ numbers that meet the requirement, thus completing the proof.
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 11 Let $A$ be an $n$-element set of positive integers $(n \geqslant 2)$. Prove that $A$ has a subset $B$, satisfying $|B|>\frac{n}{3}$, and for any $x, y \in B$, $x+y \notin B$.
|
Let the numbers in $A$ be $a_{1}, \cdots, a_{n}$. From Exercise 3, part 2, we know that there are infinitely many primes that are $-1$ modulo 3, so we can take such a prime $p > a_{i} (1 \leqslant i \leqslant n)$, and set $p = 3k - 1$. Consider the following ($p$ rows and $n$ columns) $pn$ numbers:
$$\begin{array}{l}
a_{1}, a_{2}, \cdots, a_{n} \\
2a_{1}, 2a_{2}, \cdots, 2a_{n} \\
\cdots \cdots \\
pa_{1}, pa_{2}, \cdots, pa_{n}
\end{array}$$
Since $p > a_{i}$, we have $\left(p, a_{i}\right) = 1$. Therefore, each column in (1) forms a complete residue system modulo $p$ (see Unit 6, (10)), and thus for each $j (0 \leqslant j \frac{n}{3},$
i.e., there is an $l (1 \leqslant l \leqslant p)$ such that among $l a_{1}, l a_{2}, \cdots, l a_{n}$, the number of elements that are congruent to $k, k+1, \cdots, 2k-1$ modulo $p$ is greater than $\frac{n}{3}$. We take
$$B = \{a \in A \mid l a \text{ is congruent to } k, k+1, \cdots, 2k-1 \text{ modulo } p\},$$
then $B$ meets the requirement: because for any $x, y \in B$, it is easy to see that $l(x+y) (= l x + l y)$ modulo $p$ is either $\geqslant 2k$ or $\leqslant k-1$, thus $x+y \notin B$.
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 12 Given $n \geqslant 2$. Prove that there exist $n$ distinct positive integers with the following properties:
(1) These numbers are pairwise coprime;
(2) The sum of any $k$ of these numbers $(2 \leqslant k \leqslant n)$ is a composite number.
|
Proof. When $n=2$, the conclusion is obviously true. Suppose we already have $n$ positive integers $a_{1}, \cdots, a_{n}$ that meet the requirements. We will construct $n+1$ numbers based on this.
Since there are infinitely many primes, we can take $2^{n}-1$ distinct primes $p_{i}\left(1 \leqslant i \leqslant 2^{n}-1\right)$ that are all coprime with $a_{1} a_{2} \cdots a_{n}$. Let the $2^{n}-1$ sums formed by any $k$ of $a_{1}, \cdots, a_{n}$ (where $1 \leqslant k \leqslant n$) be denoted as $S_{j}\left(1 \leqslant j \leqslant 2^{n}-1\right)$, where the sum for $k=1$ is the number $a_{i}(1 \leqslant i \leqslant n)$.
Since $\left(p_{i}, a_{1} \cdots a_{n}\right)=1$, there exists $b_{i}$ such that $a_{1} \cdots a_{n} \cdot b_{i} \equiv 1\left(\bmod p_{i}\right)\left(1 \leqslant i \leqslant 2^{n}-1\right)$. By the Chinese Remainder Theorem, the system of congruences
$$x \equiv -b_{i}-b_{i} S_{i}\left(\bmod p_{i}\right), 1 \leqslant i \leqslant 2^{n}-1$$
has infinitely many positive integer solutions $x$. We choose a solution $x_{0}>p_{i}\left(1 \leqslant i \leqslant 2^{n}-1\right)$, and multiply both sides of (1) by $a_{1} \cdots a_{n}$, obtaining
$$a_{1} \cdots a_{n} x_{0}+1+S_{i} \equiv 0\left(\bmod p_{i}\right), 1 \leqslant i \leqslant 2^{n}-1$$
Let $a_{n+1}=a_{1} \cdots a_{n} x_{0}+1$, then $a_{1}, \cdots, a_{n}, a_{n+1}$ these $n+1$ numbers meet the requirements: since $x_{0}>p_{i}$, we have $a_{n+1}+S_{i}>p_{i}$; and (2) implies that $a_{n+1}+S_{i}$ has a divisor $p_{i}$, so for any $i, a_{n+1}+S_{i}$ is composite. By the construction of $a_{n+1}$, it is certainly coprime with each $a_{i}$ $(1 \leqslant i \leqslant n)$. This completes the inductive construction.
The essence of the above solution is that if we already have $a_{1}, \cdots, a_{n}$, we hope to take a value of the parameter $x$ such that the number $a_{1} \cdots a_{n} x+1$ can serve as $a_{n+1}$. The main benefit of constructing a number of this form is that the required $\left(a_{n+1}, a_{i}\right)=1(1 \leqslant i \leqslant n)$ is automatically satisfied.
Things that meet the requirements of the problem often exist in more than one form, and we can choose certain things with special properties to try, i.e., to satisfy appropriate sufficient conditions to ensure that they meet part of the requirements of the problem. This technique of taking a step back to move forward and choosing less to achieve more is widely used in constructive proofs.
This problem can also be solved using the following (more direct) construction: take $a_{i}=i \cdot n!+1$, then $a_{1}, \cdots, a_{n}$ meet the requirements. This is because:
First, for $i \neq j$ we have $\left(a_{i}, a_{j}\right)=1$. This is because if $\left(a_{i}, a_{j}\right)=d$, then $j a_{i}-i a_{j}$ is a multiple of $d$, i.e., $d \mid(i-j)$. But $1 \leqslant|i-j|<n$, so it follows that $d \mid n!$, and thus by $d \mid a_{i}$ we have $d=1$.
Furthermore, the sum of any $k$ of $a_{i}$ (where $2 \leqslant k \leqslant n$) has the form $m \cdot n!+k$ (where $m$ is some integer), which clearly has a proper divisor $k$, and thus is not a prime.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
1 Prove that for any integer $a \geqslant 3$, there are infinitely many positive integers $n$ such that $a^{n}-1$ is divisible by $n$. (Please compare Example 2 of Unit 8.)
|
1. Since $a \geqslant 3$, it follows that $a-1$ has a prime factor $p$. By Fermat's Little Theorem, we know that $a^{p} \equiv a \equiv 1(\bmod p)$. Using induction, it is easy to prove that $n=p^{k}(k=1,2, \cdots)$ all meet the requirements.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2 Let $n_{1}, \cdots, n_{k}$ be positive integers, with the following property:
$$n_{1}\left|\left(2^{n_{2}}-1\right), n_{2}\right|\left(2^{n_{3}}-1\right), \cdots, n_{k} \mid\left(2^{n_{1}}-1\right) .$$
Prove: $n_{1}=\cdots=n_{k}=1$.
|
2. The given conditions can be restated as
$$2^{n_{2}} \equiv 1\left(\bmod n_{1}\right), 2^{n_{3}} \equiv 1\left(\bmod n_{2}\right), \cdots, 2^{n_{1}} \equiv 1\left(\bmod n_{k}\right)$$
Let $D=\left[n_{1}, \cdots, n_{k}\right]$. Then from the above equations, we have
$$2^{D} \equiv 1\left(\bmod n_{i}\right)(i=1, \cdots, k)$$
Thus, $2^{D} \equiv 1(\bmod D)$, and by Example 2 in Unit 8, we know $D=1$, so $n_{1}=n_{2}=\cdots=n_{k}=1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4 Prove: The indeterminate equation $x^{n}+1=y^{n+1}$ has no positive integer solutions $(x, y, n)$, where $(x, n+1)=1$, $n>1$.
|
4. Clearly $y>1$. The original equation can be factored as
$$(y-1)\left(y^{n}+y^{n-1}+\cdots+y+1\right)=x^{n}$$
The key is to prove that $y-1$ and $y^{n}+y^{n-1}+\cdots+y+1$ are coprime. If their greatest common divisor $d>1$, then $d$ has a prime factor $p$. From $y \equiv 1(\bmod p)$, we know that $y^{i} \equiv 1(\bmod p)$, thus
$$y^{n}+y^{n-1}+\cdots+y+1 \equiv n+1(\bmod p)$$
Therefore, $p \mid(n+1)$; but from (1) it also follows that $p \mid x^{n}$, hence the prime $p \mid x$, which contradicts $(x, n+1)=1$, so $d=1$. Now from (1) it follows that there exist positive integers $a, b$, such that
$$y-1=a^{n}, y^{n}+y^{n-1}+\cdots+y+1=b^{n}$$
But $y^{n}<y^{n}+y^{n-1}+\cdots+y+1<(y+1)^{n}$, i.e., $y^{n}+y^{n-1}+\cdots+y+1$ lies between two consecutive $n$-th powers, so it cannot be an $n$-th power of an integer, which contradicts the already proven (2).
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5 (1) Prove: For any given positive integer $n$, there exist positive rational numbers $a, b, a \neq b$, that are not integers, such that $a-b, a^{2}-b^{2}, \cdots, a^{n}-b^{n}$ are all integers.
(2) Let $a, b$ be positive rational numbers, $a \neq b$. If there are infinitely many positive integers $n$ such that $a^{n}-b^{n}$ is an integer, then $a, b$ are both integers.
|
5. (1) For example, take $a=2^{n}+\frac{1}{2}, b=\frac{1}{2}$, then for $k=1, \cdots, n$,
$$\begin{aligned}
a^{k}-b^{k} & =(a-b)\left(a^{k-1}+a^{k-2} b+\cdots+a b^{k-2}+b^{k-1}\right) \\
& =2^{n} \cdot a^{k-1}+2^{n} \cdot a^{k-2} b+\cdots+2^{n} \cdot a b^{k-2}+2^{n} \cdot b^{k-1}
\end{aligned}$$
Since $k \leqslant n$, it is easy to see that each term on the right-hand side of the above equation is an integer, hence their sum is an integer.
(2) Let $a=\frac{x}{z}, b=\frac{y}{z}, x, y, z$ be positive integers, and $(x, y, z)=1$. Then $a^{n}-b^{n}$ being an integer is equivalent to
$$x^{n} \equiv y^{n}\left(\bmod z^{n}\right)$$
We need to prove $z=1$, which implies that $a, b$ are both integers.
Assume $z>1$, then $z$ has a prime factor. If $z$ has an odd prime factor $p$, we set $r$ to be the smallest positive integer such that $x^{r} \equiv$ $y^{r}(\bmod p)$. By (1), we know $x^{n} \equiv y^{n}(\bmod p)$, hence $r \mid n$ (refer to Note 3 in Example 5 of Unit 8). Let $p^{a}\left\|n, p^{\beta}\right\|\left(x^{r}-y^{r}\right)$ (note, since $a \neq b$, hence $x \neq y$), then by (1) in Example 5 of Unit 8, we have $p^{\alpha+\beta} \|\left(x^{n}-y^{n}\right)$, but (1) implies $p^{n} \mid\left(x^{n}-y^{n}\right)$, thus $p^{n} \leqslant$ $p^{\alpha+\beta}$, so $n \leqslant \alpha+\beta$, and $p^{\alpha} \leqslant n$, hence $\alpha \leqslant \log _{p} n$, thus
$$n \leqslant \log _{p} n+\beta$$
This cannot hold for sufficiently large $n$ (note that $\beta$ is a fixed number), therefore (1) cannot hold for infinitely many $n$, leading to a contradiction.
If $z$ has no odd prime factors, then $z$ is a power of 2. By (1) and $(x, y, z)=1$, we know that $x, y$ are both odd. When $n$ is odd, by
$$x^{n}-y^{n}=(x-y)\left(x^{n-1}+x^{n-2} y+\cdots+x y^{n-2}+y^{n-1}\right),$$
and noting that the second factor on the right-hand side of the above equation is odd, thus $2^{n} \mid\left(x^{n}-y^{n}\right)$ implies $2^{n} \mid(x-y)$, since $x \neq y$, such $n$ can only be finite. When $n$ is even, let $2^{s} \|\left(x^{2}-y^{2}\right)$, by Note 3 (2) in Example 5 of Unit 8, if $2^{a} \| n$, then $2^{a+s-1} \|\left(x^{n}-y^{n}\right)$. Combining with (1), we get $n \leqslant \alpha+s-1$. Since $\alpha \leqslant \log _{2} n$, hence
$$n \leqslant \log _{2} n+s-1$$
This cannot hold for sufficiently large even $n$, leading to a contradiction.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
6 Let $n \geqslant 4$ be an integer, and $a_{1}, \cdots, a_{n}$ be distinct positive integers less than $2 n$. Prove: from these numbers, one can select several such that their sum is divisible by $2 n$.
untranslated text remains the same as requested. However, if you need any further assistance or a different format, feel free to let me know!
|
6. If each $a_{i}$ is not equal to $n$, the conclusion is easy to prove. Because the $2n$ numbers
$$a_{1}, a_{2}, \cdots, a_{n}, 2n-a_{1}, 2n-a_{2}, \cdots, 2n-a_{n}$$
are all positive integers and less than $2n$, hence there must be two that are equal, i.e., there exist $i, j$ such that $a_{i}=2n-a_{j}$. If $i=j$, it implies $a_{i}=n$, which contradicts the assumption, so $i \neq j$, thus $a_{i}+a_{j}=2n$, which is divisible by $2n$.
Now, without loss of generality, assume $a_{n}=n$. Consider the $n-1 (\geqslant 3)$ integers $a_{1}, a_{2}, \cdots, a_{n-1}$, among which there must be two numbers whose difference is not divisible by $n$, because if all $\mathrm{C}_{n-1}^{2}$ differences of two numbers are divisible by $n$, then due to $\mathrm{C}_{n-1}^{2} \geqslant 3$, there are three numbers $a_{i}<a_{j}<a_{k}$, such that $n\left|\left(a_{j}-a_{i}\right), n\right|\left(a_{k}-a_{j}\right)$, thus $a_{k}-a_{i}=\left(a_{k}-a_{j}\right)+\left(a_{j}-a_{i}\right) \geqslant 2n$, which is impossible.
Without loss of generality, assume $a_{1}-a_{2}$ is not divisible by $n$. Consider the following $n$ numbers
$$a_{1}, a_{2}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \cdots, a_{1}+a_{2}+\cdots+a_{n-1}.$$
If they are all incongruent modulo $n$, then one of them is divisible by $n$; if two of the numbers in (1) are congruent modulo $n$, then the difference of these two numbers is divisible by $n$, thus it must produce a sum of some of the numbers $a_{1}, \cdots, a_{n-1}$ that is divisible by $n$ (since $a_{1}-a_{2}$ is not divisible by $n$), denote this sum as $kn$. If $k$ is even, the conclusion is already established; if $k$ is odd, adding $a_{n}=n$ to the said sum, the result is obtained.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 For any integer $n$, prove that the fraction $\frac{21 n+4}{14 n+3}$ is in its simplest form.
|
To prove that $21n+4$ and $14n+3$ are coprime, it is easy to see that these two numbers satisfy Bézout's identity:
$$3(14n+3)-2(21n+4)=1$$
Therefore, the conclusion holds.
In general, it is not easy to derive Bézout's identity for coprime integers $a$ and $b$. Therefore, we often use the following workaround: create an auxiliary equation similar to Bézout's identity:
$$ax + by = r$$
where $r$ is an appropriate integer. If we set $(a, b) = d$, then from the above equation, we know $d \mid r$. The appropriate $r$ means: from $d \mid r$, we can further deduce $d=1$, or $r$ has fewer divisors, which can lead to the conclusion by elimination.
Moreover, the above auxiliary equation is equivalent to $a \mid (by - r)$ or $b \mid (ax - r)$, and sometimes, these divisibility conditions are easier to derive.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let $n$ be a positive integer, prove that $(n!+1,(n+1)!+1)=1$.
The text is translated while preserving the original line breaks and format.
|
Prove we have the equation
$$(n!+1)(n+1)-((n+1)!+1)=n$$
Let $d=(n!+1,(n+1)!+1)$, then by (1) we know $d \mid n$.
Furthermore, since $d \mid n$, hence $d \mid n!$, combining with $d \mid(n!+1)$ we know $d \mid 1$, so $d=1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let $F_{k}=2^{2^{k}}+1, k \geqslant 0$. Prove: If $m \neq n$, then $\left(F_{m}, F_{n}\right)=1$.
---
The translation maintains the original text's line breaks and format.
|
Assume without loss of generality that $m>n$. The key to the argument is to use $F_{n} \mid\left(F_{m}-2\right)$ (see Example 2 in Unit 1), which means there is an integer $x$ such that
$$F_{m}+x F_{n}=2 \text {. }$$
Let $d=\left(F_{m}, F_{n}\right)$, then from the above equation, we deduce $d \mid 2$, so $d=1$ or 2. But $F_{n}$ is clearly odd, hence $d$ must be 1.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Let $a>1, m, n>0$, prove:
$$\left(a^{m}-1, a^{n}-1\right)=a^{(m, n)}-1$$
|
Let $D=\left(a^{m}-1, a^{n}-1\right)$. We prove that $\left(a^{(m, n)}-1\right) \mid D$ and $D \mid \left(a^{(m, n)}-1\right)$ to derive $D=a^{(m, n)}-1$, which is a common method in number theory to prove the equality of two numbers.
Since $(m, n)|m, (m, n)| n$, by the factorization formula (5) in the first unit, we know that $\left(a^{(m, n)}-1\right) \mid \left(a^{m}-1\right)$ and $\left(a^{(m, n)}-1\right) \mid \left(a^{n}-1\right)$. Therefore, by property (3) of this unit, $a^{(m, n)}-1$ divides $\left(a^{m}-1, a^{n}-1\right)$, i.e., $\left(a^{(m, n)}-1\right) \mid D$.
To prove $D \mid \left(a^{(m, n)}-1\right)$, let $d=(m, n)$. Since $m, n>0$, we can choose $u$, $v>0$ such that (see the note in (1) of this unit)
$$m u-n v=d$$
Since $D \mid \left(a^{m}-1\right)$, it follows that $D \mid \left(a^{m u}-1\right)$. Similarly, $D \mid \left(a^{m w}-1\right)$. Therefore, $D \mid \left(a^{m u}-a^{n v}\right)$, and from (1), we get
$$D \mid a^{m v}\left(a^{d}-1\right)$$
Moreover, since $a>1$ and $D \mid \left(a^{m}-1\right)$, we have $(D, a)=1$, and thus $\left(D, a^{n v}\right)=1$. Therefore, from (2) and property (7), we derive $D \mid \left(a^{d}-1\right)$, i.e., $D \mid \left(a^{(m, n)}-1\right)$.
Combining the results proven above, we conclude that $D=a^{(m, n)}-1$.
|
D=a^{(m, n)}-1
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let $m>n \geqslant 0$, prove: $\left(2^{2^{n}}+1\right) \mid\left(2^{2^{m}}-1\right)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Prove that in the factorization (5) by taking $x=2^{2^{n+1}}, y=1$, and substituting $2^{m-n-1}$ for $n$ there, we get
$$2^{2^{m}}-1=\left(2^{2^{n+1}}-1\right)\left[\left(2^{2^{n+1}}\right)^{2^{m-n-1}-1}+\cdots+2^{2^{n+1}}+1\right]$$
Hence
$$\left(2^{2^{2+1}}-1\right) \mid\left(2^{2^{m}}-1\right)$$
Also, $2^{2^{n+1}}-1=\left(2^{2^{n}}-1\right)\left(2^{2^{n}}+1\right)$,
Thus
$$\left(2^{2^{n}}+1\right) \mid\left(2^{2^{2+1}}-1\right)$$
Therefore, by the divisibility property (1), we know $\left(2^{2^{z}}+1\right) \mid\left(2^{2^{m}}-1\right)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Let $m, n>0, m n \mid\left(m^{2}+n^{2}\right)$, then $m=n$.
|
Proof: Let $(m, n)=d$, then $m=m_{1} d, n=n_{1} d$, where $\left(m_{1}, n_{1}\right)=1$. Thus, the given condition becomes $m_{1} n_{1} \mid\left(m_{1}^{2}+n_{1}^{2}\right)$, hence we also have $m_{1} \mid\left(m_{1}^{2}+n_{1}^{2}\right)$, which implies $m_{1} \mid n_{1}^{2}$. But $\left(m_{1}, n_{1}\right)=1$, so $\left(m_{1}, n_{1}^{2}\right)=1$. Combining $m_{1} \mid n_{1}^{2}$, we must have $m_{1}=1$. Similarly, $n_{1}=1$, therefore $m=n$.
|
m=n
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Let positive integers $a, b, c$ have a greatest common divisor of 1, and
$$\frac{ab}{a-b}=c$$
Prove: $a-b$ is a perfect square.
|
Proof: Let $(a, b)=d$, then $a=d a_{1}, b=d b_{1}$, where $\left(a_{1}, b_{1}\right)=1$. Since $(a, b, c)=1$, it follows that $(d, c)=1$.
Now, the equation in the problem can be transformed into
$$d a_{1} b_{1}=c a_{1}-c b_{1}$$
From this, it is clear that $a_{1}$ divides $c b_{1}$. Since $\left(a_{1}, b_{1}\right)=1$, it follows that $a_{1} \mid c$. Similarly, $b_{1} \mid c$. By $\left(a_{1}, b_{1}\right)=1$, it follows that $a_{1} b_{1} \mid c$ (refer to properties (9) and (10)).
Let $c=a_{1} b_{1} k$, where $k$ is a positive integer. On one hand, it is clear that $k$ divides $c$; on the other hand, combining with equation (1) we get $d=k\left(a_{1}-b_{1}\right)$, so $k \mid d$. Thus, $k \mid (c, d)$ (see property (3)). But $(c, d)=1$, so $k=1$.
Therefore, $d=a_{1}-b_{1}$. Hence, $a-b=d\left(a_{1}-b_{1}\right)=d^{2}$. This proves that $a-b$ is a perfect square.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 7 Let $k$ be a positive odd number, prove: $1+2+\cdots+n$ divides $1^{k}+2^{k}+\cdots+n^{k}$.
|
To prove that since $1+2+\cdots+n=\frac{n(n+1)}{2}$, the problem is equivalent to proving: $n(n+1)$ divides $2\left(1^{k}+2^{k}+\cdots+n^{k}\right)$. Because $n$ and $n+1$ are coprime, this is equivalent to proving
$$n \mid 2\left(1^{k}+2^{k}+\cdots+n^{k}\right)$$
and
$$(n+1) \mid 2\left(1^{k}+2^{k}+\cdots+n^{k}\right)$$
In fact, since $k$ is odd, by the factorization formula (6) from the first unit, we know that
$$\begin{aligned}
& 2\left(1^{k}+2^{k}+\cdots+n^{k}\right) \\
= & {\left[1^{k}+(n-1)^{k}\right]+\left[2^{k}+(n-2)^{k}\right]+\cdots+\left[(n-1)^{k}+1^{k}\right]+2 n^{k} }
\end{aligned}$$
is a multiple of $n$. Similarly,
$$2\left(1^{k}+2^{k}+\cdots+n^{k}\right)=\left[1^{k}+n^{k}\right]+\left[2^{k}+(n-1)^{k}\right]+\cdots+\left[n^{k}+1^{k}\right]$$
is a multiple of $n+1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
I Let $n$ be an integer, prove: $(12 n+5,9 n+4)=1$.
|
1. We have $4(9 n+4)-3(12 n+5)=1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2 Let $m, n$ be positive integers, $m$ is odd, prove: $\left(2^{m}-1,2^{n}+1\right)=1$.
|
2. Let $d=\left(2^{m}-1,2^{n}+1\right)$. Then $2^{m}-1=d u, 2^{n}+1=d v$, where $u, v$ are integers. It is easy to see that $(d u+1)^{n}=(d v-1)^{m}$, expanding both sides (note that $m$ is odd), we get $d A+1=d B-1$ ($A, B$ are some two integers), from which it follows that $d \mid 2$, i.e., $d=1$ or 2. But clearly, $d$ can only be 1.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
3 Let $(a, b)=1$, prove: $\left(a^{2}+b^{2}, a b\right)=1$.
untranslated text remains the same as requested. However, if you need any further assistance or a different format, please let me know!
|
3. Since $(a, b)=1$, we have $\left(a^{2}, b\right)=1$, thus $\left(a^{2}+b^{2}, b\right)=1$. Similarly, $\left(a^{2}+b^{2}\right.$, $a)=1$. Therefore, $\left(a^{2}+b^{2}, a b\right)=1$ (using (6) of this unit).
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4 If the $k$-th power of a rational number is an integer $(k \geqslant 1)$, then this rational number must be an integer. More generally, prove: a rational root of an integer-coefficient polynomial with a leading coefficient of $\pm 1$ must be an integer.
|
4. Let the reduced rational number $\frac{p}{q}$ be a root of the integer-coefficient polynomial $f(x)=x^{n}+$ $a_{1} x^{n-1}+\cdots+a_{n-1} x+a_{n}$ (with the leading coefficient 1). From $f\left(\frac{p}{q}\right)=0$ we easily get
$$p^{n}+a_{1} p^{n-1} q+\cdots+a_{n-1} p q^{n-1}+a_{n} q^{n}=0$$
Since $a_{1} p^{n-1} q, \cdots, a_{n-1} p q^{n-1}, a_{n} q^{n}$ are all divisible by $q$, it follows that $q \mid p^{n}$. But $(p, q)=1$, hence $\left(q, p^{n}\right)=1$. Therefore, we must have $q= \pm 1$, which means the rational number $\frac{p}{q}$ is an integer.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Proof: The infinite sequence $10001,100010001, \cdots$ contains no prime numbers.
|
Let $a_{n}=\underbrace{10001 \cdots 10001}_{n \text { ones}}(n \geqslant 2)$, then
$$a_{n}=1+10^{4}+10^{8}+\cdots+10^{4(n-1)}=\frac{10^{4 n}-1}{10^{4}-1} .$$
To decompose the number on the right-hand side of the above equation into the product of two (greater than 1) integers, we distinguish two cases:
- $n$ is even. Let $n=2 k$, then
$$a_{2 k}=\frac{10^{8 k}-1}{10^{4}-1}=\frac{10^{8 k}-1}{10^{8}-1} \cdot \frac{10^{8}-1}{10^{4}-1}$$
It is easy to see that $\frac{10^{8}-1}{10^{4}-1}$ is an integer greater than 1, and for $k \geqslant 2, \frac{10^{8 k}-1}{10^{8}-1}$ is also an integer greater than 1. Therefore, $a_{2 k}(k=2,3, \cdots)$ are all composite numbers. Also, $a_{2}=10001=13 \times 137$ is a composite number.
- $n$ is odd. Let $n=2 k+1$, then
$$a_{2 k+1}=\frac{10^{4(2 k+1)}-1}{10^{4}-1}=\frac{10^{2(2 k+1)}-1}{10^{2}-1} \cdot \frac{10^{2(2 k+1)}+1}{10^{2}+1}$$
is the product of two integers greater than 1, hence $a_{2 k+1}$ are also all composite numbers. Therefore, all $a_{n}$ are composite.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 For a positive integer $n$, let $S(n)$ denote the sum of the digits in the decimal representation of $n$. Prove: $9 \mid n$ if and only if $9 \mid S(n)$.
|
Proof Let $n=a_{k} \times 10^{k}+\cdots+a_{1} \times 10+a_{0}$ (here $0 \leqslant a_{i} \leqslant 9$, and $a_{k} \neq 0$), then $S(n)=a_{0}+a_{1}+\cdots+a_{k}$. We have
$$n-S(n)=a_{k}\left(10^{k}-1\right)+\cdots+a_{1}(10-1)$$
For $1 \leqslant i \leqslant k$, by the factorization formula (5) we know $9 \mid\left(10^{i}-1\right)$, so each of the $k$ terms on the right side of (1) is a multiple of 9, and thus by the divisibility property (2), their sum is also divisible by 9, i.e., $9 \mid(n-$ $S(n))$. From this, it is easy to derive both aspects of the conclusion.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Proof: For any integer $n>1$, the number $n^{4}+4^{n}$ is not a prime.
untranslated text remains the same as requested. However, if you need the entire text to be translated, please let me know!
|
Prove that if $n$ is even, then $n^{4}+4^{n}$ is greater than 2 and divisible by 2, so it is not a prime. However, for odd $n$, it is easy to see that $n^{4}+4^{n}$ does not have a fixed divisor (greater than 1). We use a different approach:
Let the odd number $n=2 k+1, k \geqslant 1$, then
$$\begin{aligned}
n^{4}+4^{n} & =n^{4}+4 \cdot 4^{2 k}=n^{4}+4 \cdot\left(2^{k}\right)^{4} \\
& =n^{4}+4 n^{2} \cdot\left(2^{k}\right)^{2}+4 \cdot\left(2^{k}\right)^{4}-4 n^{2} \cdot\left(2^{k}\right)^{2} \\
& =\left(n^{2}+2 \cdot 2^{2 k}\right)^{2}-\left(2 \cdot n \cdot 2^{k}\right)^{2} \\
& =\left(n^{2}+2^{2+1} n+2^{2 k+1}\right)\left(n^{2}-2^{k+1} n+2^{2 k+1}\right) .
\end{aligned}$$
The first factor on the right side of the equation is clearly not 1, and the second factor is $\left(n-2^{k}\right)^{2}+2^{2 k}$, which is also not 1 (since $k \geqslant 1$), so $n^{4}+4^{n}$ is composite for $n>1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Proof: If integers $a, b$ satisfy $2a^2 + a = 3b^2 + b$, then $a - b$ and $2a + 2b + 1$ are both perfect squares.
|
Prove that the given relation is
$$(a-b)(2a+2b+1)=b^2$$
The first point of the argument is to prove that the integers $a-b$ and $2a+2b+1$ are coprime. Let $d=(a-b, 2a+2b+1)$. If $d>1$, then $d$ has a prime factor $p$, and thus from (1) we know $p \mid b^2$. Since $p$ is a prime, it follows that $p \mid b$. Combining this with $p \mid (a-b)$, we get $p \mid a$. Then, from $p \mid (2a+2b+1)$, we derive $p \mid 1$, which is impossible. Therefore, $d=1$. Since the right-hand side of (1) is $b^2$, a perfect square, it follows that $|a-b|$ and $|2a+2b+1|$ are both perfect squares (see (8) in Unit 2).
Now, we prove that $a-b \geqslant 0$. From (1), we know $2a+2b+1 \geqslant 0$, so $a-b$ and $2a+2b+1$ are both perfect squares.
Assume there are integers $a, b$ satisfying the equation in the problem, but $a-b < 0$. Let $a-b = -r^2$ where $r > 0$; combining this with (1) we get $r \mid b$, and from $b-a = r^2$ we get $r \mid a$. Let $b = b_1 r$ and $a = a_1 r$, substituting into the equation in the problem (note $r > 0$ and $b_1 = a_1 + r$), we get
$$a_1^2 + 6a_1 r + 3r^2 + 1 = 0$$
To prove that the above equation cannot hold, we can use the following method:
Consider (2) as a quadratic equation in $a_1$, and solve it using the quadratic formula:
$$a_1 = -3r \pm \sqrt{6r^2 - 1}$$
Since $a_1$ is an integer, it follows from the above equation that $6r^2 - 1$ must be a perfect square. However, it is easy to see that the remainder when a perfect square is divided by 3 can only be 0 or 1; but the remainder when $6r^2 - 1$ is divided by 3 is 2, leading to a contradiction.
Or more directly: Since $a_1^2$ leaves a remainder of 0 or 1 when divided by 3, the left-hand side of (2) leaves a remainder of 1 or 2 when divided by 3; but the right-hand side of (2) is 0, divisible by 3. This is a contradiction. Therefore, (2) does not hold for any integers $a_1$ and $r$, so we must have $a-b \geqslant 0$, which proves the conclusion of the problem.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Let $n, a, b$ be integers, $n>0$ and $a \neq b$. Prove: If $n \mid\left(a^{n}-b^{n}\right)$, then $n \left\lvert\, \frac{a^{n}-b^{n}}{a-b}\right.$.
|
Proof: Let $p$ be a prime, and $p^{a} \| n$. We will prove that $p^{a} \left\lvert\, \frac{a^{n}-b^{n}}{a-b}\right.$, which will lead to the conclusion of this problem (see the note below).
Let $t=a-b$. If $p \nmid t$, then $\left(p^{a}, t\right)=1$. Since $n \mid\left(a^{n}-b^{n}\right)$, it follows that $p^{a} \mid\left(a^{n}-b^{n}\right)$. Also, $a^{n}-b^{n}=t \cdot \frac{a^{n}-b^{n}}{t}$, thus $p^{a} \left\lvert\, \frac{a^{n}-b^{n}}{t}\right.$.
If $p \mid t$, using the binomial theorem, we get
$$\frac{a^{n}-b^{n}}{t}=\frac{(b+t)^{n}-b^{n}}{t}=\sum_{i=1}^{n} \mathrm{C}_{n}^{i} b^{n-i} t^{i-1}$$
Let $p^{\beta} \| i (i \geqslant 1)$, then $2 \beta \leqslant p^{\beta} \leqslant i$, from which it is easy to see that $\beta \leqslant i-1$. Therefore, in $\mathrm{C}_{n}^{i} t^{i-1}=\frac{n}{i} \mathrm{C}_{n-1}^{i-1} t^{i-1}$, the power of $p$ is at least $\alpha-\beta+(i-1) \geqslant \alpha$, so each term on the right side of (1) is divisible by $p^{\alpha}$. Hence, $p^{a} \left\lvert\, \frac{a^{n}-b^{n}}{t}\right.$, which means $p^{a} \left\lvert\, \frac{a^{n}-b^{n}}{a-b}\right.$ (see Note 1 in Example 3 of Unit 1).
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Let $m, n$ be non-negative integers, prove that $\frac{(2 m)!(2 n)!}{m!n!(m+n)!}$ is an integer.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
However, it seems the translation request was already fulfilled in the provided statement. If you need the proof or further explanation in English, please let me know!
|
To prove, we only need to show that for each prime $p$, the power of $p$ in the standard factorization of the denominator $m!n!(m+n)!$ does not exceed the power of $p$ in the numerator $(2m)!(2n)!$. By the formula in (7), this is equivalent to proving
$$\sum_{l=1}^{\infty}\left(\left[\frac{2 m}{p^{l}}\right]+\left[\frac{2 n}{p^{l}}\right]\right) \geqslant \sum_{l=1}^{\infty}\left(\left[\frac{m}{p^{l}}\right]+\left[\frac{n}{p^{l}}\right]+\left[\frac{m+n}{p^{l}}\right]\right)$$
In fact, we can prove a stronger result:
For any real numbers $x, y$, we have
$$[2 x]+[2 y] \geqslant[x]+[y]+[x+y] .$$
To prove (2), we note that for any integer $k$ and any real number $\alpha$, we have $[k+\alpha]=[\alpha]+k$. This easily implies that if $x$ or $y$ changes by an integer amount, then both sides of inequality (2) change by the same amount. Therefore, we only need to prove (2) for the case $0 \leqslant x<1,0 \leqslant y<1$, thus reducing the problem to proving the inequality
$$[2 x]+[2 y] \geqslant[x+y] .$$
Note that now $0 \leqslant[x+y] \leqslant 1$. If $[x+y]=0$, the conclusion is obviously true. If $[x+y]=1$, then $x+y \geqslant 1$, so at least one of $x, y$ is greater than or equal to $\frac{1}{2}$. Without loss of generality, assume $x \geqslant \frac{1}{2}$, thus $[2 x]+[2 y] \geqslant[2 x]=1$, which proves (2), and hence (1) is also true, thus proving the conclusion of this problem.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
I prove: For any given positive integer $n>1$, there exist $n$ consecutive composite numbers.
|
1. It is easy to verify that $(n+1)!+2, (n+1)!+3, \cdots, (n+1)!+(n+1)$ are $n$ consecutive composite numbers.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2 Prove: there are infinitely many primes of the form $4k-1$, and there are also infinitely many primes of the form $6k-1$ ($k$ being a positive integer).
|
2. The method used by Euclid to prove that there are infinitely many primes can be slightly modified to argue: Assume that there are only finitely many primes of the form $4k-1$, and let them all be $p_{1}, \cdots, p_{m}$. Consider the number $N = 4 p_{1} \cdots p_{m} - 1$. Clearly $m > 1$, so $N > 1$, and thus $N$ has a prime factor. Furthermore, since the product of two numbers of the form $4k+1$ is still of the form $4k+1$, and $N$ is of the form $4k-1$, $N$ must have a prime factor $p$ of the form $4k-1$. By the previous assumption, $p$ must be one of $p_{1}, \cdots, p_{m}$, and thus $N - 4 p_{1} \cdots p_{m}$ is divisible by $p$, i.e., $p \mid 1$, which is a contradiction. Similarly, it can be proven that there are infinitely many primes of the form $6k-1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
3 Prove: There are infinitely many odd numbers $m$, such that $8^{m}+9 m^{2}$ is a composite number.
|
3. Take $m=9 k^{3}(k=1,3, \cdots)$, then $8^{m}+9 m^{2}=\left(2^{m}\right)^{3}+\left(9 k^{2}\right)^{3}$. It is easy to see that it has a proper divisor $2^{m}+9 k^{2}$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4 Let integers $a, b, c, d$ satisfy $a>b>c>d>0$, and
$$a^{2}+a c-c^{2}=b^{2}+b d-d^{2},$$
Prove: $a b+c d$ is not a prime.
|
4. Proof by contradiction, suppose there exists a set of $a, b, c, d$ satisfying the conditions such that $ab + cd$ is a prime number, denoted as $p$. Substituting $a = \frac{p - cd}{b}$ into the given equation, we get
$$p(p - 2cd + bc) = (b^2 + c^2)(b^2 + bd - d^2)$$
Since $p$ is a prime number, $p$ must divide $b^2 + c^2$ or $p$ must divide $b^2 + bd - d^2$.
If $p \mid (b^2 + c^2)$, then from $0 < b^2 + c^2 < 2ab < 2(ab + cd) = 2p$, it follows that $b^2 + c^2 = p$, i.e., $ab + cd = b^2 + c^2$, thus $b \mid c(c - d)$. Clearly, $(b, c) = 1$ (since $ab + cd$ is a prime number), hence $b \mid (c - d)$, which contradicts $0 < c - d < c < b$.
If $p \mid (b^2 + bd - d^2)$, then from $0 < b^2 + bd - d^2 < 2(ab + cd) = 2p$ we know that $b^2 + bd - d^2 = p$, i.e., $ab + cd = b^2 + bd - d^2 = a^2 + ac - c^2$, so $a \mid (c + d)c$ and $b \mid (c + d)d$ both hold. But it is easy to see that $(ab, cd) = 1$, hence $c + d$ is divisible by both $a$ and $b$. Since $0 < c + d < 2a$ and $0 < c + d < 2b$, it must follow that $c + d = a$ and $c + d = b$, which is a contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Proof: The product of two consecutive positive integers cannot be a perfect square, nor can it be a perfect cube.
|
Proof by contradiction, we assume there exist positive integers $x, y$, such that
$$x(x+1)=y^{2} .$$
Multiplying both sides of the equation by 4, we transform it into $(2 x+1)^{2}=4 y^{2}+1$, which can be factored as
$$(2 x+1+2 y)(2 x+1-2 y)=1$$
Since the two factors on the left are positive integers, we have
$$\left\{\begin{array}{l}
2 x+1+2 y=1 \\
2 x+1-2 y=1
\end{array}\right.$$
Solving this, we get $x=y=0$, which is a contradiction. This proves the first assertion in the problem.
However, for the equation
$$x(x+1)=y^{3},$$
the above factorization method is not effective. We adopt another (property-based) factorization: Suppose the said equation has positive integer solutions $x, y$, then since $x$ and $x+1$ are coprime, and their product is a perfect cube, $x$ and $x+1$ must both be cubes of positive integers, i.e.,
$$x=u^{3}, x+1=v^{3}, y=u v,$$
where $u, v$ are positive integers, leading to $v^{3}-u^{3}=1$, hence
$$(v-u)\left(v^{2}+u v+u^{2}\right)=1,$$
which is clearly impossible. It is not difficult to see that a similar argument can prove that the product of two consecutive positive integers cannot be an integer's $k$-th power (where $k \geqslant 2$).
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Prove: The equation
$$y+y^{2}=x+x^{2}+x^{3}$$
has no integer solutions for $x \neq 0$.
|
Proof: Suppose the equation has an integer solution $x \neq 0$. We decompose it as
$$(y-x)(y+x+1)=x^{3}$$
We first prove that $(y-x, y+x+1)=1$. If this is not true, then there exists a prime $p$ that is a common divisor of $y-x$ and $y+x+1$. From (1), we know $p \mid x^{3}$, so the prime $p$ divides $x$. Combining $p \mid(y-x)$, we get $p \mid y$, but $p \mid(x+y+1)$, thus $p \mid 1$, which is impossible. Therefore, the two factors on the left side of (1) are coprime. Since the right side of (1) is a perfect cube, there exist integers $a, b$ such that
$$y-x=a^{3}, y+x+1=b^{3}, x=a b.$$
Eliminating $x, y$ yields
$$b^{3}-a^{3}=2 a b+1$$
Now we prove that equation (2) has no integer solutions, which leads to a contradiction. We factorize (2) as
$$(b-a)\left(b^{2}+a b+a^{2}\right)=2 a b+1$$
Note that $x=a b$ and $x \neq 0$, so $a b \neq 0$. If $a b>0$, then from (3) it is easy to see that $b-a>0$. Since $a, b$ are integers, we have $b-a \geqslant 1$, so the left side of (3) $\geqslant b^{2}+a b+a^{2}>3 a b>$ the right side; If $a b<0$, then the left side of (3) is $\geqslant 3|a b|$, while the absolute value of the right side of (3) $<2|a b|$, so (3) cannot hold. This proves that the equation in the problem has no integer solutions $x \neq 0$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Let $k$ be a given positive integer, $k \geqslant 2$, prove: the product of three consecutive positive integers cannot be an integer's $k$-th power.
保留了源文本的换行和格式。
|
Prove that if there are positive integers $x \geqslant 2$ and $y$, such that
$$(x-1) x(x+1)=y^{k}$$
Note that the three factors $x-1, x, x+1$ are not always pairwise coprime, so we cannot conclude from (1) that they are all $k$-th powers. One way to overcome this difficulty is to transform (1) into
$$\left(x^{2}-1\right) x=y^{k}$$
Since $x$ and $x^{2}-1$ are coprime, it follows from (2) that there exist positive integers $a, b$ such that
$$x=a^{k}, x^{2}-1=b^{k}, a b=y$$
Thus we have
$$\begin{aligned}
1 & =a^{2 k}-b^{k}=\left(a^{2}\right)^{k}-b^{k} \\
& =\left(a^{2}-b\right)\left(a^{2 k-2}+a^{2 k-4} b+\cdots+a^{2} b^{k-2}+b^{k-1}\right)
\end{aligned}$$
Since $x \geqslant 2$, it follows that $a \geqslant 2$, and since $k \geqslant 2$, the second factor in the above equation must be greater than 1, leading to a contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 7 Let positive integers $x, y, z$ satisfy $2 x^{x}=y^{y}+z^{z}$, then $x=y=z$.
|
Proof First, expand $(x+1)^{x+1}$ to know
$$(x+1)^{x+1}>x^{x+1}+(x+1) x^{x}>2 x^{x}$$
From this, we know that $y$ and $z$ must both be $\leqslant x$: for if $y$ or $z$ is greater than $x$, without loss of generality, assume $y>x$. Since $y$ and $x$ are integers, we have $y \geqslant x+1$, thus
$$y^{y}+z^{z}>y^{y} \geqslant(x+1)^{y} \geqslant(x+1)^{x+1}>2 x^{x} \text { (apply (1)), }$$
which leads to a contradiction.
Therefore, $y \leqslant x, z \leqslant x$, so
$$y^{y}+z^{z} \leqslant x^{x}+x^{x}=2 x^{x}$$
Combining with the original equation, we must have $y=x$, and $z=x$, hence $x=y=z$. Q.E.D.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
1 Prove: The product of four consecutive positive integers cannot be a perfect square.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the request is to translate the text and not to restate the instruction, here is the translation:
1 Prove: The product of four consecutive positive integers cannot be a perfect square.
|
1. Let $x(x+1)(x+2)(x+3)=y^{2}, x, y$ are positive integers. Then we have
$$\left(x^{2}+3 x+1\right)^{2}-y^{2}=1$$
It is evident that this is impossible.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2 Find all integers that can be expressed as the difference of squares of two integers.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
2. Let the integer $n$ be expressible as the difference of squares of two integers: $n=x^{2}-y^{2}$, i.e., $n=(x-y)(x+y)$. Since $x-y$ and $x+y$ have the same parity, either $n$ is odd, or $n$ is divisible by 4. Conversely, if $n$ is odd, we can take $x-y=1, x+y=n$, i.e., $x=\frac{n+1}{2}, y=\frac{n-1}{2}$; if $4 \mid n$, we can take $x-y=2, x+y=\frac{n}{2}$, i.e., $x=\frac{n}{4}+1, y=\frac{n}{4}-1$, then $x^{2}-y^{2}=n$.
| null |
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Let $k \geqslant 1$ be an odd number, prove: for any positive integer $n$, the number $1^{k}+2^{k}+\cdots+n^{k}$ cannot be divisible by $n+2$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Prove that the conclusion is obviously true when $n=1$. Let $n \geqslant 2$, and denote the said sum as $A$, then
$$2 A=2+\left(2^{k}+n^{k}\right)+\left(3^{k}+(n-1)^{k}\right)+\cdots+\left(n^{k}+2^{k}\right) .$$
Since $k$ is a positive odd number, by the factorization formula (6), for each $i \geqslant 2$, the number $i^{k}+(n+2-i)^{k}$ is divisible by $i+(n+2-i)=n+2$, hence the remainder when $2 A$ is divided by $n+2$ is 2, thus $A$ cannot be divisible by $n+2$ (note that $n+2>2$).
| null |
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Let $m \geqslant n \geqslant 1$, prove: $\frac{(m, n)}{m} \mathrm{C}_{m}^{n}$ is an integer.
|
Prove that $\frac{x}{m} \mathrm{C}_{m}^{n}$ is an integer when $x=m$, which is $\mathrm{C}_{m}^{n}$; when $x=n$, it is $\frac{n}{m}$. $\frac{m}{n} \mathrm{C}_{m-1}^{n-1}=\mathrm{C}_{m-1}^{n-1}$, which is also an integer. Furthermore, by Bézout's identity, there exist integers $u$ and $v$ such that
$$(m, n)=m u+n v,$$
thus $\frac{(m, n)}{m} \mathrm{C}_{m}^{n}=u \mathrm{C}_{m}^{n}+v \frac{n}{m} \mathrm{C}_{m}^{n}$ is an integer.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let $a, b$ be two distinct positive integers, and $ab(a+b)$ is a multiple of $a^{2}+ab+b^{2}$. Prove: $|a-b|>\sqrt[3]{ab}$.
|
To prove that since $a b(a+b)$ is divisible by $a^{2}+a b+b^{2}$, we first divide $a b(a+b)$ by $a^{2}+a b+b^{2}$, obtaining
$$a b(a+b)=\left(a^{2}+a b+b^{2}\right) a-a^{3}$$
Thus, $\left(a^{2}+a b+b^{2}\right) \mid a^{3}$. Similarly, $\left(a^{2}+a b+b^{2}\right) \mid b^{3}$, meaning $a^{2}+a b+b^{2}$ is a common divisor of $a^{3}$ and $b^{3}$, so $\left(a^{2}+a b+b^{2}\right) \mid\left(a^{3}, b^{3}\right)$. (See (3) in Unit 2.) Also, $\left(a^{3}, b^{3}\right)=(a, b)^{3}$ (see the note below), hence
$$\left(a^{2}+a b+b^{2}\right) \mid(a, b)^{3} .$$
Let $d=(a, b), a=a_{1} d, b=b_{1} d$, then (1) becomes $\left(a_{1}^{2}+a_{1} b_{1}+b_{1}^{2}\right) \mid d$. Therefore, $d \geqslant a_{1}^{2}+a_{1} b_{1}+b_{1}^{2}$, and more specifically, $d > a_{1} b_{1}$. Since $a \neq b$, the integers $a_{1} \neq b_{1}$, thus $\left|a_{1}-b_{1}\right| \geqslant 1$, leading to
$$|a-b|^{3}=d^{3}\left|a_{1}-b_{1}\right|^{3} \geqslant d^{3}>d^{2} a_{1} b_{1}=a b$$
Thus, $|a-b|>\sqrt[3]{a b}$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Take several different integers between two consecutive perfect squares $n^{2}$ and $(n+1)^{2}$, prove that their pairwise products are all different.
|
Proof: Let integers $a, b, c, d$ satisfy $n^{2}a$ and $c > a$, leading to $q > p$ and $v > u$. Since $p, q, u, v$ are all integers, we have $q \geqslant p+1, v \geqslant u+1$. Therefore, we get (note $a = p u > n^{2}$)
$$\begin{array}{l}
d = q v \geqslant (p+1)(u+1) = p u + (p+u) + 1 \\
> n^{2} + 2 \sqrt{p u} + 1 > n^{2} + 2 n + 1 = (n+1)^{2}
\end{array}$$
Contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 7 Proof: From $1,2, \cdots, 100$, if any 51 numbers are taken, there must be two numbers that are coprime.
|
The problem becomes extremely simple when broken down: We take consecutive pairs of numbers from $1,2, \cdots, 100$, forming the following 50 pairs
$$\{1,2\},\{3,4\}, \cdots,\{99,100\},$$
then any selection of 51 numbers must include one of the above pairs. Since these two numbers are consecutive, they are of course coprime.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 8 Proof: There exist 1000 consecutive positive integers, among which exactly 10 are prime.
|
The basis of this proof is Exercise 3, Question 1, which concludes that there exist 1000 consecutive positive integers
$$a, a+1, \cdots, a+999$$
where each number is not a prime.
Now, perform the following operation on (1): delete the rightmost number $a+999$ and add $a-1$ to the left. Clearly, the resulting sequence
$$a-1, a, \cdots, a+998$$
contains at most one prime number. Repeat this procedure until reaching $1,2, \cdots, 1000$ and then stop. Note that the number of primes in the 1000 consecutive positive integers obtained after one operation, compared to the number of primes in the 1000 consecutive positive integers before the operation, either remains the same or increases or decreases by 1. The final sequence $1,2, \cdots, 1000$ clearly contains more than 10 primes, so during the above process, there must be one instance where the 1000 consecutive integers contain exactly 10 primes.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Let $m, n$ be positive integers, $m>2$, prove: $\left(2^{m}-1\right) \nmid\left(2^{n}+1\right)$.
|
First, when $n \leqslant m$, it is easy to see that the conclusion holds. In fact, when $m=n$, the conclusion is trivial; when $n < m$, the conclusion is also trivial (and can be seen from the divisibility property (3)).
Finally, the case $n > m$ can be reduced to the above special case: by the division algorithm, $n = mq + r, 0 \leqslant r < m$. Since
$$2^{n} + 1 = (2^{mq} - 1)2^{r} + 2^{r} + 1$$
By the factorization (5), we know that $\left(2^{m} - 1\right) \mid \left(2^{mq} - 1\right)$; and $0 \leqslant r < m$. Therefore, $\left(2^{m} - 1\right) \nmid \left(2^{n} + 1\right)$ when $n > m$. This proves the conclusion of the problem.
We might as well mention that the condition $m > 2$ in Example 5 is necessary. Because when $m = 2$, $2^{m} - 1 = 3$, and by (6), for all odd $n \geqslant 1$, the number $2^{n} + 1$ is divisible by 3.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 9 If in the standard factorization of a positive integer, the exponent of each prime factor is greater than 1, then it is called a power number. Prove: There exist infinitely many distinct positive integers, such that neither they nor the sum of any different numbers among them are power numbers.
|
Proof Let $2=p_{1}<p_{2}<\cdots<p_{n}<\cdots$ be all the prime numbers, then
$$p_{1}, p_{1}^{2} p_{2}, p_{1}^{2} p_{2}^{2} p_{3}, \cdots, p_{1}^{2} p_{2}^{2} \cdots p_{n-1}^{2} p_{n}, \cdots$$
satisfies the requirement.
To verify this claim, we denote the $n$-th number in the sequence as $a_{n}$. First, each $a_{n}$ is not a power. For any $r, s, \cdots, n(1 \leqslant r<s<\cdots<n)$, by (1), $p_{r} \mid a_{r}$ but $p_{r}^{2} \nmid a_{r}$, and $p_{r}\left|\frac{a_{s}}{a_{r}}, \cdots, p_{r}\right| \frac{a_{n}}{a_{r}}$. Therefore, in
$$a_{r}+a_{s}+\cdots+a_{n}=a_{r}\left(\frac{a_{s}}{a_{r}}+\cdots+\frac{a_{n}}{a_{r}}+1\right)$$
the second factor is coprime with $p_{r}$, so the prime $p_{r}$ appears exactly once in the standard factorization of $a_{r}+a_{s}+\cdots+a_{n}$, hence $a_{r}+a_{s}+\cdots+a_{n}$ is not a power. Moreover, since there are infinitely many primes, there are also infinitely many numbers in (1).
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 11 Proof: There are infinitely many positive integers $n$, such that $n \mid\left(2^{n}+2\right)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Example 11 Proof: There are infinitely many positive integers $n$, such that $n \mid (2^{n} + 2)$.
|
At first glance, this problem seems very similar to Example 10, but it is actually much more challenging. We still use the method of inductive construction, with the key move being to strengthen the inductive hypothesis. Below is the proof: If $n$ satisfies
$$2|n, n|\left(2^{n}+2\right),(n-1) \mid\left(2^{n}+1\right),$$
then for $m=2^{n}+2$, we have
$$2|m, m|\left(2^{m}+2\right),(m-1) \mid\left(2^{m}+1\right)$$
In fact, since $2^{n}+2=2\left(2^{n-1}+1\right)$ is an odd number multiplied by 2 and $2 \mid n$, the integer $k$ in $2^{n}+2=n k$ is an odd number, so
$$2^{m}+1=2^{n}+1=\left(2^{n}\right)^{k}+1$$
is a multiple of $2^{n}+1=m-1$.
Similarly, from $2^{n}+1=(n-1) l$ we know that $l$ is an odd number, hence
$$2^{m}+2=2\left(2^{m-1}+1\right)=2\left(\left(2^{n-1}\right)^{l}+1\right)$$
is a multiple of $2\left(2^{n-1}+1\right)=2^{n}+2=m$. Also, $m=2^{n}+2$ is clearly even, so the above assertion is proven.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
1 Let $m$ be an integer greater than 4, and not a prime. Prove $m \mid (m-1)$!.
|
1. Since $m$ is not a prime number, $m$ can be expressed as $m=a b$, where $14$, so $a>2$, and thus $a^{2}>2 a$, which means $m>2 a$. Therefore, $a$ and $2 a$ are two distinct terms in the sequence $1,2, \cdots, m-1$, and hence $(m-1)!$ is divisible by $a \cdot 2 a=2 m$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2 Prove: A positive integer $n$ can be expressed as the sum of several consecutive positive integers (at least two) if and only if $n$ is not a power of 2.
|
2. Let $n=x+(x+1)+\cdots+(x+k-1), x$ be a positive integer, $k \geqslant 2$. That is,
$$(2 x+k-1) k=2 n$$
If $n$ is a power of 2, then $k$ and $2 x-1+k$ are both powers of 2, but $2 x-1$ is odd, so it must be $k=1$, which contradicts the problem's condition.
Conversely, if $n$ is not a power of 2, let $n=2^{m-1}(2 t+1), m \geqslant 1, t \geqslant 1$. When $t \geqslant 2^{m-1}$, we can take $k=2^{m}, x=t+1-2^{m-1}$; when $t<2^{m-1}$, we can take $k=2 t+1, x=2^{m-1}-t$. Then $k$ and $x$ are both positive integers and $k \geqslant 2$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
3 Prove: Any positive integer $n$ can be expressed as $a-b$, where $a$ and $b$ are positive integers, and the number of distinct prime factors of $a$ and $b$ is the same.
|
3. When $n$ is even, we can take $a=2n, b=n$. If $n$ is odd, let $p$ be the smallest odd prime that does not divide $n$, then $p-1$ either has no odd prime factors (i.e., is a power of 2), or its odd prime factors all divide $n$. Therefore, $a=pn, b=(p-1)n$ have the number of distinct prime factors equal to the number of distinct prime factors of $n$ plus 1.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
For any given integer $n \geqslant 3$, prove that there exists an $n$-term arithmetic sequence composed of positive integers (with a non-zero common difference), such that any two terms are coprime.
|
4. The sequence $\{k \cdot n!+1\}(k=1, \cdots, n)$ meets the requirement. Suppose there exist $s, t(1 \leqslant s<t \leqslant n)$ such that $s \cdot n!+1$ and $t \cdot n!+1$ are not coprime, then there is a prime $p$ that divides both numbers, and thus divides their difference, i.e., $p \mid (t-s) n!$. Since $p$ is a prime, we have $p \mid (t-s)$ or $p \mid n!$. But $1 \leqslant t-s<n$, so if $p \mid (t-s)$, then $p \mid n!$ as well. Therefore, we always have $p \mid n!$, and combining this with $p \mid s \cdot n!+1$ we get $p \mid 1$, which is a contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5 Proof: For each $n \geqslant 2$, there exist $n$ distinct positive integers $a_{1}, a_{2}, \cdots, a_{n}$, such that $\left(a_{i}-\right.$ $\left.a_{j}\right) \mid\left(a_{i}+a_{j}\right)(1 \leqslant i, j \leqslant n, i \neq j)$.
|
5. Using the inductive construction method. When $n=2$, we can take $a_{1}=1, a_{2}=2$. Assuming that for $n=k$, we already have $a_{1}, \cdots, a_{k}$ that meet the requirements, let $b_{0}$ be the least common multiple of $a_{1}, \cdots, a_{k}, a_{i}-a_{j}(1 \leqslant i, j \leqslant k, i \neq j)$, then the $k+1$ numbers
$$b_{0}, a_{1}+b_{0}, \cdots, a_{k}+b_{0}$$
meet the requirements.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Given any $n>1$, prove: there exists a positive integer $a$, such that all numbers in $a^{a}+1, a^{a^{a}}+1, \cdots$ are divisible by $n$.
---
The translation maintains the original text's line breaks and format.
|
We note that if $a$ is odd, then $a^{a}, a^{a^{a}}, \cdots$ are all odd, and thus from (6) we know that $a^{a}+1, a^{a^{a}}+1=a^{\left(a^{a}\right)}+1, \cdots$ all have the factor $a+1$. Therefore, taking $a=2n-1$ satisfies the requirement in the problem.
|
a=2n-1
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Let $a, b, c, d$ be positive integers, prove: $a^{4b+d}-a^{4c+d}$ is divisible by 240.
|
To prove that since $240=2^{4} \times 3 \times 5$, we will separately prove that $a^{4 b+d}-a^{4 c+d}$ is divisible by 3, 5, and 16, thereby proving the conclusion (see the note in Example 5 of Unit 3).
First, we prove $3 \mid\left(a^{4 b+d}-a^{4 c+d}\right)$. By the result in (12) that $a^{2} \equiv 0,1(\bmod 3)$, it follows that $a^{4 b} \equiv a^{4 c} \equiv 0,1(\bmod 3)$. Therefore,
$$a^{4 b+d}-a^{4 c+d}=a^{d}\left(a^{4 b}-a^{4 c}\right) \equiv 0(\bmod 3)$$
Similarly, by $a^{2} \equiv 0, \pm 1(\bmod 5)$, it follows that $a^{4} \equiv 0,1(\bmod 5)$, thus $a^{4 b} \equiv a^{4 c} \equiv 0,1(\bmod 5)$. Therefore, $a^{4 b+d}-a^{4 c+d} \equiv 0(\bmod 5)$.
Finally, by $a^{4} \equiv 0,1(\bmod 16)$, it follows that $a^{4 b} \equiv a^{4 c} \equiv 0,1(\bmod 16)$, hence $a^{4 b+d}-a^{4 c+d} \equiv 0(\bmod 16)$. This completes the proof of our conclusion.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let integers $a, b, c$ satisfy $a+b+c=0$, and let $d=a^{1999}+b^{1999}+c^{1999}$. Prove: $|d|$ is not a prime number.
|
To prove that $|d|$ has a non-trivial fixed divisor, we will use congruences here:
First, for any integer $u$, the number $u^{1999}$ has the same parity as $u$, i.e., $u^{1999} \equiv u(\bmod 2)$, hence $d \equiv a+b+c \equiv 0(\bmod 2)$, which means $2 \mid d$.
Furthermore, for any integer $u$, it is easy to verify (by distinguishing $3 \mid u$ and $3 \nmid u$)
$$u^{3} \equiv u(\bmod 3)$$
From this, we can deduce
$$\begin{aligned}
u^{1999}=u \cdot u^{1998} & \equiv u \cdot u^{666} \equiv u \cdot u^{222} \equiv u^{75} \\
& \equiv u^{25} \equiv u^{9} \equiv u^{3} \equiv u(\bmod 3)
\end{aligned}$$
Therefore, $d \equiv a+b+c \equiv 0(\bmod 3)$. Hence, $6 \mid d$, which means $d$ is not a prime number.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let integers $x, y, z$ satisfy
$$(x-y)(y-z)(z-x)=x+y+z,$$
Prove: $x+y+z$ is divisible by 27.
|
We will deduce from (1) that $x$, $y$, and $z$ must be pairwise congruent modulo 3, thus $27 \mid (x-y)(y-z)(z-x)$, and hence by (1) we know $27 \mid (x+y+z)$.
Proof by contradiction: First, assume that exactly two of $x$, $y$, and $z$ are congruent modulo 3. Without loss of generality, let $x \equiv y \pmod{3}$, but $x \not\equiv z \pmod{3}$. In this case, $3 \mid (x-y)$, but $3 \nmid (x+y+z)$. Therefore, the left-hand side of (1) $\equiv 0 \pmod{3}$, but the right-hand side $\not\equiv 0 \pmod{3}$, which is a contradiction. Hence, this scenario cannot occur.
Next, assume that the residues of $x$, $y$, and $z$ modulo 3 are all different. In this case, it is easy to see that $3 \mid (x+y+z)$, but $3 \nmid (x-y)(y-z)(z-x)$. Thus, the left-hand side and the right-hand side of (1) have different residues modulo 3, which is a contradiction. Therefore, this scenario also cannot occur.
Thus, our initial assertion is correct, proving the conclusion of the problem.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Let $n>1$, prove: $\underbrace{11 \cdots 1}_{n \uparrow 1}$ is not a perfect square.
|
Proof by contradiction, suppose there exists some $n>1$ and integer $x$, such that
$$\underbrace{11 \cdots 1}_{n \text { ones }}=x^{2}.$$
From (1), we know that $x$ is odd (this is obtained by taking (1) modulo 2, noting that $x^{2} \equiv x(\bmod 2)$). Furthermore, since $2 \nmid x$, it follows that $x^{2} \equiv 1(\bmod 4)$. However,
$$\underbrace{11 \cdots 1}_{n \text { ones }}-1=\underbrace{11 \cdots 10}_{n-1 \text { ones }}$$
is only divisible by 2, but not by 4, meaning the left side of (1) $\not \equiv 1(\bmod 4)$, leading to a contradiction!
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Use the digits $1, 2, 3, 4, 5, 6, 7$ to form seven-digit numbers, each digit being used exactly once. Prove: None of these seven-digit numbers is a multiple of another.
|
Prove that if there are two seven-digit numbers $a, b (a \neq b)$ such that
$$a = bc,$$
where $c$ is an integer greater than 1. Since the sum of the digits of $a$ and $b$ is $1+2+3+4+5+6+7 \equiv 1(\bmod 9)$, we have $a \equiv b \equiv 1(\bmod 9)$ (see Note 2 of Example 3 in Unit 1). Now, taking (1) modulo 9, we get $c \equiv 1(\bmod 9)$. Since $c > 1$, it follows that $c \geqslant 10$, and thus $a \geqslant 10b > 10^7$, which contradicts the fact that $a$ is a seven-digit number.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 The sequence $\left\{x_{n}\right\}$ is $1,3,5,11, \cdots$ and satisfies the recurrence relation
$$x_{n+1}=x_{n}+2 x_{n-1}, n \geqslant 2$$
The sequence $\left\{y_{n}\right\}$ is $7,17,55,161, \cdots$ and satisfies the recurrence relation
$$y_{n+1}=2 y_{n}+3 y_{n-1}, n \geqslant 2$$
Prove: These two sequences have no common terms.
|
Consider modulo 8. First, we prove that the sequence $\left\{x_{n}\right\}$ modulo 8 is a periodic sequence
$$1,3,5,3,5, \cdots$$
Since $x_{2} \equiv 3, x_{3} \equiv 5(\bmod 8)$. If we already have
$$x_{n-1} \equiv 3, x_{n} \equiv 5(\bmod 8)$$
then by the recurrence formula (1), we get
$$\begin{array}{l}
x_{n+1}=x_{n}+2 x_{n-1} \equiv 5+2 \times 3 \equiv 3(\bmod 8), \\
x_{n+2}=x_{n+1}+2 x_{n} \equiv 3+2 \times 5 \equiv 5(\bmod 8),
\end{array}$$
This completes the inductive proof of our claim.
Similarly, by (2) we can prove that the sequence $\left\{y_{n}\right\}$ modulo 8 becomes a periodic sequence
$$7,1,7,1,7,1, \cdots$$
From (3) and (4), it is clear that the two sequences $x_{2}, x_{3}, \cdots$ and $y_{1}, y_{2}, \cdots$ have no common terms modulo 8, hence these two sequences have no common terms. Since $\left\{y_{n}\right\}$ is increasing, $y_{1}, y_{2}, \cdots$ will never equal $x_{1}=1$, which proves that $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ have no common terms.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 8 Connect the vertices of a regular $n$-sided polygon to form a closed $n$-broken line. Prove: if $n$ is even, there are two parallel lines in the connections; if $n$ is odd, it is impossible to have exactly two parallel lines in the connections.
|
Prove that this is a geometric problem not suitable for a geometric solution, as it is related to the complete residue system modulo $n$. $\square$
Label the vertices in a counterclockwise order with the numbers $0,1, \cdots, n-1$. Let the closed broken line in the problem be $a_{0} \rightarrow a_{1} \rightarrow \cdots \rightarrow a_{n-1} \rightarrow a_{n}=a_{0}$, where $a_{0}, a_{1}, \cdots, a_{n-1}$ is a permutation of $0,1, \cdots, n-1$. $\square$
First, by the fact that the $a_{i}$ are vertices of a regular $n$-gon, we easily know that
$$\begin{array}{l}
a_{i} a_{i+1} / / a_{j} a_{j+1} \Leftrightarrow \overparen{a_{i+1} a_{j}}=\overparen{a_{j+1} a_{i}} \\
\Leftrightarrow a_{i}+a_{i+1} \equiv a_{j}+a_{j+1}(\bmod n) .
\end{array}$$
When $n$ is even, $2 \nmid(n-1)$, so the sum of any complete residue system modulo $n$ is $\equiv 0+1+\cdots+(n-1) = \frac{n(n-1)}{2} \not \equiv 0(\bmod n)$.
On the other hand, we always have
$$\begin{aligned}
\sum_{i=0}^{n-1}\left(a_{i}+a_{i+1}\right) & =\sum_{i=0}^{n-1} a_{i}+\sum_{i=0}^{n-1} a_{i+1}=2 \sum_{i=0}^{n-1} a_{i}=2 \times \frac{n(n-1)}{2} \\
& =n(n-1) \equiv 0(\bmod n)
\end{aligned}$$
Therefore, $a_{i}+a_{i+1}(i=0,1, \cdots, n-1)$ cannot form a complete residue system modulo $n$, i.e., there must be $i \neq j(0 \leqslant i, j \leqslant n-1)$ such that
$$a_{i}+a_{i+1} \equiv a_{j}+a_{j+1}(\bmod n)$$
Thus, there must be a pair of sides $a_{i} a_{i+1} / / a_{j} a_{j+1}$.
When $n$ is odd, if there is exactly one pair of sides $a_{i} a_{i+1} / / a_{j} a_{j+1}$, then among the $n$ numbers $a_{0}+a_{1}, a_{1}+a_{2}, \cdots, a_{n-1}+a_{0}$, exactly one residue class $r$ appears twice, and thus exactly one residue class $s$ is missing. Therefore (since $2 \mid(n-1)$ ),
$$\begin{aligned}
\sum_{i=0}^{n-1}\left(a_{i}+a_{i+1}\right) & \equiv 0+1+\cdots+(n-1)+r-s=\frac{n(n-1)}{2}+r-s \\
& \equiv r-s(\bmod n)
\end{aligned}$$
Combining with (1) we get $r \equiv s(\bmod n)$, a contradiction! This shows that when $n$ is odd, it is impossible to have exactly one pair of parallel sides. $\square$
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 9 Let $n>3$ be an odd number, prove: after removing any one element from the $n$-element set $S=\{0,1, \cdots, n-1\}$, the remaining elements can always be divided into two groups, each with $\frac{n-1}{2}$ numbers, such that the sums of the two groups are congruent modulo $n$.
|
To prove a key point of the argument, for any $x \in S, x \neq 0$, the set $S \backslash\{x\}$ can be obtained from $T=$ $\{1,2, \cdots, n-1\}$ by the transformation
$$T+x(\bmod n)=\{a+x(\bmod n), a \in T\}$$
This reduces the problem to proving a special case: $T=S \backslash\{0\}$ can be divided into two groups, each containing $\frac{n-1}{2}$ numbers, such that the sums of the two groups are congruent modulo $n$.
We consider two cases. When $n=4 k+1(k \geqslant 1)$, note that the $2 k$ pairs
$$\{1,4 k\},\{2,4 k-1\}, \cdots,\{2 k, 2 k+1\}$$
each have a sum that is congruent to 0 modulo $n$. Thus, by selecting any $k$ pairs to form one set and the remaining $k$ pairs to form another set, the requirement is satisfied.
If $n=4 k+3(k \geqslant 1)$, we first place $1, 2, 4 k$ in one set and $3, 4 k+1, 4 k+2$ in another set. Then, we take the remaining $2 k-2$ pairs
$$\{4,4 k-1\}, \cdots,\{2 k+1,2 k+2\}$$
and distribute $k-1$ pairs to each of the two sets.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 7 Given any $n \geqslant 2$, prove: there exist $n$ distinct positive integers, such that the sum of any two of them divides the product of these $n$ numbers.
|
We arbitrarily take $n$ distinct positive integers $a_{1}, \cdots, a_{n}$, and select a (positive integer) parameter $K$, hoping that the product $K^{n} a_{1} \cdots a_{n}$ of $K a_{1}, \cdots, K a_{n}$ is divisible by the sum of any two terms $K a_{i}+K a_{j}$ $(1 \leqslant i, j \leqslant n, i \neq j)$. Since $n \geqslant 2$, it is clear that taking
$$K=\prod_{1 \leq i<j \leqslant n}\left(a_{i}+a_{j}\right)$$
meets the requirement (note that $K a_{1}, \cdots, K a_{n}$ are distinct).
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 10 Proof: For any integer $n \geqslant 4$, there exists an $n$-degree polynomial
$$f(x)=x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$$
with the following properties:
(1) $a_{0}, a_{1}, \cdots, a_{n-1}$ are all positive integers;
(2) For any positive integer $m$, and any $k(k \geqslant 2)$ distinct positive integers $r_{1}, \cdots, r_{k}$, we have
$$f(m) \neq f\left(r_{1}\right) f\left(r_{2}\right) \cdots f\left(r_{k}\right)$$
|
To prove the basic idea of this problem is to require that two integers cannot be equal, and congruence is precisely useful for this (see the note below Example 3).
We wish to construct a polynomial of degree \( n \) with positive integer coefficients and leading coefficient 1, such that for any integer \( a \), we have \( f(a) \equiv 2 \pmod{4} \). This implies that for any \( k \) integers \( r_1, \cdots, r_k \) (where \( k \geq 2 \)), we have \( f(r_1) \cdots f(r_k) \equiv 0 \pmod{4} \), but \( f(m) \equiv 2 \pmod{4} \). Therefore, for any integer \( m \), the number \( f(m) \) and \( f(r_1) \cdots f(r_k) \) are not congruent modulo 4, and thus they cannot be equal.
We take
\[ f(x) = (x+1)(x+2) \cdots (x+n) + 2 \]
Expanding the right-hand side of (1) shows that \( f(x) \) is an \( n \)-degree polynomial with leading coefficient 1 and positive integer coefficients. On the other hand, for any integer \( a \), since \( n \geq 4 \), among the \( n \) consecutive integers \( a+1, \cdots, a+n \), there must be one that is a multiple of 4. Therefore, \( 4 \mid (a+1) \cdots (a+n) \), and by (1), we have \( f(a) \equiv 2 \pmod{4} \). This shows that the polynomial (1) meets the problem's requirements.
There are many ways to construct \( f(x) \). Here is a slightly different method:
We note that when \( n \geq 4 \) is even, then for any integer \( a \), we have \( 4 \mid a^n - a^2 \). This is because if \( a \) is even, then \( 4 \mid a^2 \), so 4 divides \( a^2(a^{n-2} - 1) = a^n - a^2 \); if \( a \) is odd, then since \( n-2 \) is even, \( a^{n-2} \) is the square of an odd number, so \( 4 \mid a^{n-2} - 1 \), and thus \( 4 \mid a^2(a^{n-2} - 1) \).
Similarly, it is not difficult to prove that when \( n \geq 5 \) is odd, \( a^n - a^3 = a^3(a^{n-3} - 1) \) is divisible by 4.
Therefore, for even \( n \geq 4 \), we take
\[ \begin{aligned}
f(x) & = x^n + 4(x^{n-1} + \cdots + x^3) + 3x^2 + 4x + 2 \\
& = x^n - x^2 + 4(x^{n-1} + \cdots + x) + 2
\end{aligned} \]
For odd \( n \geq 5 \), we take
\[ \begin{aligned}
f(x) & = x^n + 4(x^{n-1} + \cdots + x^4) + 3x^3 + 4x^2 + 4x + 2 \\
& = x^n - x^3 + 4(x^{n-1} + \cdots + x) + 2
\end{aligned} \]
From (2) and (4), it is clear that \( f(x) \) is an \( n \)-degree polynomial with leading coefficient 1 and positive integer coefficients; and by (3) and (5) and the results mentioned earlier, for any integer \( a \), we have \( f(a) \equiv 2 \pmod{4} \). Therefore, the polynomials (2) or (4) meet the requirements. Proof complete.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
1 Label the vertices of a cube with the numbers +1 or -1, and label a face with a number that equals the product of the numbers at the four vertices of that face. Prove: the sum of the 14 numbers thus labeled cannot be 0.
|
1. Let $S$ be the sum in question. If we change any -1 at a vertex to +1, then four numbers in $S$, denoted as $a, b, c, d$, will have their signs changed. Let $S^{\prime}$ represent the sum of the 14 numbers after the change. Since $a+b+c+d \equiv 0(\bmod 2)$, we have
$$S-S^{\prime}=2(a+b+c+d) \equiv 0(\bmod 4)$$
Repeating this process of changing numbers until all numbers at the vertices are +1, we find that $S \equiv 1+1+\cdots+1=14 \equiv 2(\bmod 4)$, so $S \neq 0$.
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
number_theory
| false |
3 Let $p$ be a prime, $a \geqslant 2, m \geqslant 1, a^{m} \equiv 1(\bmod p), a^{p-1} \equiv 1\left(\bmod p^{2}\right)$. Prove: $a^{m} \equiv 1\left(\bmod p^{2}\right)$.
|
3. From $a^{m} \equiv 1(\bmod p)$ we get $a^{m}=1+p x$. Therefore,
$$a^{p m}=(1+p x)^{p}=1+p^{2} x+C_{p}^{2} p^{2} x^{2}+\cdots \equiv 1\left(\bmod p^{2}\right) .$$
Also, $a^{p-1} \equiv 1\left(\bmod p^{2}\right)$, so $a^{(p-1) m} \equiv 1\left(\bmod p^{2}\right)$, thus $a^{m m} \equiv a^{m}\left(\bmod p^{2}\right)$. Combining with (1), we know $a^{m} \equiv 1\left(\bmod p^{2}\right)$.
|
a^{m} \equiv 1\left(\bmod p^{2}\right)
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4. Let $m$ be a given positive integer. Prove: the sequence $\left\{x_{n}\right\}$ defined by
$$x_{1}=x_{2}=1, x_{k+2}=x_{k+1}+x_{k}(k=1,2, \cdots)$$
contains at least one term among its first $m^{2}$ terms that is divisible by $m$.
|
4. Without loss of generality, let $m>1$. We use $\bar{x}_{k}$ to denote the remainder when $x_{k}$ is divided by $m$. Consider the ordered pairs
$$\left\langle\bar{x}_{1}, \bar{x}_{2}\right\rangle,\left\langle\bar{x}_{2}, \bar{x}_{3}\right\rangle, \cdots,\left\langle\bar{x}_{n}, \bar{x}_{n+1}\right\rangle, \cdots$$
Since the remainders when divided by $m$ form $m^{2}$ distinct ordered pairs, among the first $m^{2}+1$ pairs in the sequence (1), there must be two that are the same. Let $\left\langle\bar{x}_{i}, \bar{x}_{i+1}\right\rangle$ be the pair with the smallest index that is equal to some $\left\langle\bar{x}_{j}, \bar{x}_{j+1}\right\rangle \left(j \leqslant m^{2}+1\right)$. We prove that $i$ must be 1; otherwise, from
$$x_{i-1}=x_{i+1}-x_{i}, x_{j-1}=x_{j+1}-x_{j}$$
we deduce $x_{i-1} \equiv x_{j-1}(\bmod m)$, hence $\left\langle\bar{x}_{i-1}, \bar{x}_{i}\right\rangle=\left\langle\bar{x}_{j-1}, \bar{x}_{j}\right\rangle$, which contradicts the minimality of $i$, so $i=1$. Now, by $\left\langle\bar{x}_{j}, \bar{x}_{j+1}\right\rangle=\left\langle\bar{x}_{1}, \bar{x}_{2}\right\rangle=\langle 1,1\rangle$. We know $x_{j-1} \equiv x_{j+1}-x_{j} \equiv 1-1 \equiv 0(\bmod m)$, i.e., $m \mid x_{j-1}\left(1<j-1 \leqslant m^{2}\right)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Fermat's Little Theorem: Let $p$ be a prime, and $a$ be any integer coprime to $p$, then
$$a^{p-1} \equiv 1(\bmod p)$$
Fermat's Little Theorem has a variant form, which is sometimes more applicable:
For any integer $a$, $a^{p} \equiv a(\bmod p)$.
(When $p \nmid a$, the two propositions are equivalent; when $p \mid a$, the latter is obviously true.)
|
It is not difficult to give a proof of Fermat's Little Theorem by induction: It is easy to see that we only need to prove the proposition for $a=0$, $1, \cdots, p-1$. When $a=0$, the conclusion is obviously true. If we already have $a^{p}=a(\bmod p)$, then since $p \mid C_{p}^{i}(i=1,2, \cdots, p-1)$, we have
$$(a+1)^{p}=a^{p}+\mathrm{C}_{p}^{1} a^{p-1}+\cdots+\mathrm{C}_{p}^{p-1} a+1 \equiv a^{p}+1 \equiv a+1(\bmod p),$$
This shows that the proposition also holds when $a$ is replaced by $a+1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
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