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Example 1 If $a|n, b| n$, and there exist integers $x, y$, such that $a x+b y=1$, prove: $a b \mid n$.
|
Proof: From the conditions, we can set \( n = a u, n = b v, \) where \( u \) and \( v \) are integers. Thus,
\[
\begin{aligned}
n & = n(a x + b y) \\
& = n a x + n b y \\
& = a b v x + a b u y \\
& = a b(v x + u y),
\end{aligned}
\]
Therefore,
\[
a b \mid n.
\]
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 A roundtable conference has 2012 participants. After a break, they sit around the table again in a different order. Prove: There are at least two people such that the number of people between them is the same before and after the break.
|
Let $n=1006$, we label each seat, and the seat numbers are marked in a clockwise direction as
$$1,2,3, \cdots, 2 n$$
Thus, each person can be represented by a pair $(i, j)$, where $i, j$ are the seat numbers before and after the break, respectively. Clearly, all "x-coordinates" $i$ and "y-coordinates" $j$ take on all values (1), forming a complete residue system modulo $2 n$.
If for any two people $\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right)$, the number of people sitting between them before and after the break is different, then we should have
$$j_{2}-j_{1} \neq i_{2}-i_{1}$$
Note that in the above equation, when $j_{2}<j_{1}$ (or $i_{2}<i_{1}$), $j_{2}$ should be replaced by $2 n+j_{2}$ (or $i_{2}$ should be replaced by $2 n+i_{2}$). A better way to write it is
$$j_{2}-j_{1} \not \equiv i_{2}-i_{1}(\bmod 2 n)$$
That is,
$$j_{2}-i_{2} \not \equiv j_{1}-i_{1}(\bmod 2 n)$$
The meaning of the above equation is that the differences between the x-coordinates and y-coordinates of any two people are not congruent modulo $2 n$, thus
$$j_{1}-i_{1}, j_{2}-i_{2}, \cdots, j_{2 n}-i_{2 n}$$
is also a complete residue system modulo $2 n$.
Considering that the sum of the numbers in any complete residue system modulo $2 n$ should be congruent to
$$1+2+3+\cdots+2 n=\frac{2 n(2 n+1)}{2}=n(2 n+1)$$
modulo $2 n$, but $n(2 n+1) \not=0(\bmod 2 n)$, hence the sum of the numbers in (2) cannot be divisible by $2 n$, and thus cannot equal 0. This contradicts
$$\sum_{k=1}^{2 n}\left(j_{k}-i_{k}\right)=\sum_{k=1}^{2 n} j_{k}-\sum_{k=1}^{2 n} i_{k}=\sum_{k=1}^{2 n} j-\sum_{k=1}^{2 n} i=0$$
This indicates that there must be at least two people for whom the number of people sitting between them is the same before and after the break.
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Fermat's Little Theorem: Let $p$ be a prime, $a$ an integer, then $a^{p} \equiv a(\bmod p)$. In particular, if $p \nmid a$, then $a^{p-1} \equiv 1(\bmod p)$.
|
Prove that when $p \mid a$, the conclusion is obviously true.
When $p \nmid a$, let $x_{1}, x_{2}, \cdots, x_{p-1}$ be a permutation of $1,2, \cdots, p-1$. We first prove that $a x_{1}, a x_{2}, \cdots, a x_{p-1}$ are pairwise incongruent modulo $p$.
In fact, if there exist $1 \leqslant i<j \leqslant p-1$ such that $a x_{i} \equiv a x_{j}(\bmod p)$, then $p \mid a(x_{i}-x_{j})$. Since $p \nmid a$, it follows that $p \mid x_{i}-x_{j}$ (note that $p$ is prime), but $x_{i}$ and $x_{j}$ are incongruent modulo $p$, a contradiction.
Moreover, none of $a x_{1}, a x_{2}, \cdots, a x_{p-1}$ is a multiple of $p$, so the remainders of $a x_{1}$, $a x_{2}, \cdots, a x_{p-1}$ when divided by $p$ form a permutation of $1,2, \cdots, p-1$. Using the properties of congruences, we have
$$\left(a x_{1}\right)\left(a x_{2}\right) \cdots\left(a x_{p-1}\right) \equiv x_{1} x_{2} \cdots x_{p-1}(\bmod p)$$
Since $x_{1} x_{2} \cdots x_{p-1} = (p-1)!$, which is not a multiple of $p$ (note that $p$ is prime again), it follows that $a^{p-1} \equiv 1(\bmod p)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Let $n$ be a positive integer. Prove: $7 \mid 3^{n}+n^{3}$ if and only if $7 \mid 3^{n} n^{3}+1$.
|
\begin{tabular}{lr}
& If $7 \mid 3^{n}+n^{3}$, then $7 \nmid n$, \\
thus, by Fermat's Little Theorem, we know & $n^{6} \equiv 1(\bmod 7)$, \\
therefore, by & $7 \mid 3^{n}+n^{3}$, \\
we know & $7 \mid\left(3^{n}+n^{3}\right) n^{3}$, \\
hence & $7 \mid 3^{n} n^{3}+1$. \\
\multicolumn{1}{c}{ Conversely, if } & $7 \mid 3^{n} n^{3}+1$, \\
then & $7 \nmid n$, \\
and & $7 \mid\left(3^{n} n^{3}+1\right) \cdot n^{3}$, \\
i.e., & $7 \mid 3^{n} \cdot n^{6}+n^{3}$, \\
using Fermat's Little Theorem, we know & $n^{6} \equiv 1(\bmod 7)$, \\
thus & $7 \mid 3^{n}+n^{3}$. \\
\end{tabular}
The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let $x$ be an integer, $p$ is an odd prime factor of $x^{2}+1$, prove: $p \equiv 1(\bmod 4)$.
|
Proof that since $p$ is an odd prime, if $p \neq 1(\bmod 4)$, then $p \equiv 3(\bmod 4)$, we can set $p=$ $4k+3$. In this case, from $x^{2} \equiv -1(\bmod p)$, we get
$$x^{p-1}=x^{4k+2}=\left(x^{2}\right)^{2k+1} \equiv (-1)^{2k+1} \equiv -1(\bmod p)$$
By Fermat's Little Theorem, we should have
$$x^{p-1} \equiv 1(\bmod p)$$
Combining the above, it leads to $p \mid 2$. Contradiction.
Therefore, $p \equiv 1(\bmod 4)$.
|
p \equiv 1(\bmod 4)
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let $x$ be an integer, and $p$ a prime factor of the number $x^{6}+x^{5}+\cdots+1$. Prove: $p=7$ or $p \equiv$ $1(\bmod 7)$.
|
Prove that when $x=1$, $p=7$; when $x \neq 1$, $p$ is a prime factor of $\frac{x^{7}-1}{x-1}$, hence, $x^{7} \equiv 1(\bmod p)$, which implies $p \nmid x$. Therefore, by Fermat's Little Theorem, we have $x^{p-1} \equiv 1(\bmod p)$, and thus $x^{(7, p-1)} \equiv 1(\bmod p)$.
If $7 \nmid p-1$, i.e., $p \neq 1(\bmod 7)$, then $(7, p-1)=1$, leading to $x \equiv 1(\bmod p)$. Thus,
$$0 \equiv x^{6}+x^{5}+\cdots+1 \equiv 1^{6}+1^{5}+\cdots+1=7(\bmod p)$$
This implies $p=7$.
Therefore, the proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Let $p$ be a prime. Prove: There exist infinitely many positive integers $n$, such that $p \mid 2^{n}-n$.
untranslated text remains the same as requested.
|
Prove that if $p=2$, then taking $n$ as an even number, we have $p \mid 2^{n}-n$, and the proposition holds. Suppose $p>2$, then by Fermat's Little Theorem, we know
$$2^{p-1} \equiv 1(\bmod p)$$
Therefore, for any positive integer $k$, we have
$$2^{k(p-1)} \equiv 1(\bmod p)$$
So, it suffices to prove that there are infinitely many positive integers $k$ such that
$$\left.k(p-1) \equiv 1(\bmod p) \text { (so, let } n=k(p-1) \text {, then } p \mid 2^{n}-n\right) \text {. }$$
And this only requires $k \equiv-1(\bmod p)$, and there are certainly infinitely many such $k$.
Therefore, the proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Let $n$ be a positive integer greater than 1. Prove: The number $n^{5}+n^{4}+1$ is not a prime.
|
Notice that
$$\begin{aligned}
& n^{5}+n^{4}+1 \\
= & n^{5}+n^{4}+n^{3}-\left(n^{3}-1\right) \\
= & n^{3}\left(n^{2}+n+1\right)-(n-1)\left(n^{2}+n+1\right) \\
= & \left(n^{3}-n+1\right)\left(n^{2}+n+1\right)
\end{aligned}$$
Therefore, if $n^{5}+n^{4}+1$ is a prime number, then $n^{3}-n+1=1$, which requires $n=0$ or $\pm 1$. Hence, when $n>1$, $n^{5}+n^{4}+1$ is not a prime number.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 In decimal notation, add a 17-digit number to its reverse. Prove: the sum must contain an even digit.
Also, if 17 is replaced by a general positive integer $n$, does the proposition hold? For which $n$ does it hold?
|
Prove that if there exists a 17-digit number $\overline{a_{1} a_{2} \cdots a_{17}}$, such that $M=\overline{a_{1} \cdots a_{17}}+\overline{a_{17} a_{16} \cdots a_{1}}$ has all digits odd, then by examining the units digit, we can see that $a_{1}+a_{17}$ is odd. Now, examining the sum of the leading digit, if $a_{2}+a_{16}$ produces a carry, then by $a_{1}+a_{17}$ being odd, we know that $M$ has a digit that is even, which is a contradiction. Therefore, $a_{2}+a_{16}$ does not produce a carry. Similarly, we can conclude that $\overline{a_{3} a_{4} \cdots a_{15}}+\overline{a_{15} a_{14} \cdots a_{3}}$ has all digits odd. The same reasoning shows that $\overline{a_{5} \cdots a_{13}}+\overline{a_{13} \cdots a_{5}}$ has all digits odd, $\cdots$, and finally $a_{9}+a_{9}$ is odd, which is a contradiction. Therefore, $M$ must have a digit that is even.
For a general positive integer $n$, the same discussion shows that the proposition still holds when $n \equiv 1(\bmod 4)$. When $n$ is even, let $n=2 m$, then the number $\underbrace{}_{m \uparrow} \cdots \underbrace{5 \cdots 5}_{m \uparrow}$ and its reverse number have all digits odd in their sum; when $n \equiv 3(\bmod 4)$, let $n=4 k+3$, then the number $\underbrace{6464 \cdots 645}_{k+1 \uparrow 64} \underbrace{45 \cdots 45}_{k \uparrow 45}$ and its reverse number have all digits odd in their sum.
Therefore, the proposition holds if and only if $n \equiv 1(\bmod 4)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 (1) Given that there are $n$ integers, their sum is zero, and their product is $n$. Prove: $4 \mid n$;
(2) Let the positive integer $n$ be a multiple of 4. Prove: there exist $n$ integers, whose sum is zero, and whose product is $n$.
|
(1) Let integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy:
$$\left\{\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{n}=0 \\
a_{1} a_{2} \cdots a_{n}=n
\end{array}\right.$$
If $n$ is odd, then by (2) we know that $a_{1}, a_{2}, \cdots, a_{n}$ are odd, so $a_{1}+a_{2}+\cdots+a_{n}$ is the sum of an odd number ($n$) of odd numbers, which contradicts (1). Therefore, $n$ is even.
Now, if $n$ is not a multiple of 4, then by (2) we know that among $a_{1}, a_{2}, \cdots, a_{n}$, exactly one number is even. In this case, $a_{1}+a_{2}+\cdots+a_{n}$ is the sum of one even number and an odd number of $(n-1)$ odd numbers, which results in an odd number, again contradicting (1). Therefore, $4 \mid n$.
(2) It is sufficient to provide an example, handling $n \equiv 0(\bmod 8)$ and $n \equiv 4(\bmod 8)$ separately.
When $n \equiv 0(\bmod 8)$, let $n=8 k$, then there exist $n$ integers
$$4 k, 2, \underbrace{1, \cdots, 1}_{2 k-2 \uparrow}, \underbrace{-1,-1, \cdots,-1}_{6 k \uparrow}$$
satisfying the sum is zero and the product is $n$.
When $n \equiv 4(\bmod 8)$, let $n=8 k+4$, then there exist $n$ integers
$$4 k+2,-2, \underbrace{1, \cdots, 1}_{2 k+1 \uparrow}, \underbrace{-1, \cdots,-1}_{6 k+1 \uparrow}$$
satisfying the sum is zero and the product is $n$.
Therefore, the proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Let the prime numbers be arranged in ascending order as $p_{1}, p_{2}, \cdots$. Prove: For any positive integer $n$ greater than 1, the numbers $p_{1} p_{2} \cdots p_{n}-1$ and $p_{1} p_{2} \cdots p_{n}+1$ are not perfect squares.
|
Notice that, for $n \geqslant 2$, $3 \mid p_{1} p_{2} \cdots p_{n}$, hence
$$p_{1} p_{2} \cdots p_{n}-1 \equiv 2(\bmod 3)$$
Therefore, $p_{1} p_{2} \cdots p_{n}-1$ is not a perfect square.
Also, for $n \geqslant 2$, $p_{2} \cdots p_{n}$ is odd. Let $p_{2} \cdots p_{n}=2 k+1$, then we have
$$p_{1} p_{2} \cdots p_{n}+1=2(2 k+1)+1=4 k+3 \equiv 3(\bmod 4) .$$
Therefore, $p_{1} p_{2} \cdots p_{n}+1$ is not a perfect square either.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Given positive integers $a, b$ satisfy the equation
$$2 a^{2}+a=3 b^{2}+b$$
Prove: $a-b$ and $2 a+2 b+1$ are both perfect squares.
|
Proof: From the condition, we know
$$b^{2}=2 a^{2}+a-\left(2 b^{2}+b\right)=(a-b)(2 a+2 b+1) .$$
The left side of the above equation is greater than zero, and on the right side, \(2 a+2 b+1\) is greater than zero, so \(a-b\) is greater than zero.
From (1), to prove that \(a-b\) and \(2 a+2 b+1\) are both perfect squares, it is only necessary to prove
$$(a-b, 2 a+2 b+1)=1$$
Let \((a-b, 2 a+2 b+1)=d\), then from (1) we know \(d^{2} \mid b^{2}\), so \(d \mid b\). Furthermore, combining \(d \mid a-b\), we know \(d \mid a\), hence \(d \mid 2(a+b)\). Since \(d \mid 2 a+2 b+1\), it follows that \(d \mid 1\), thus \(d=1\).
The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let positive integers $x, y, z$ satisfy $(x, y, z)=1$, and $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$. Prove: $x+y, x-z, y-z$ are all perfect squares.
|
Proof: Let $(x, y)=m$, and set $x=m n, y=m l$, where $m, l, n$ are all positive integers, and $(l, n)=1$. Thus, from the condition we have
$$(l+n) z=m l n .$$
Using $(x, y, z)=1$, we know $(m, z)=1$, hence, from (1) we know $z \mid ln$. Since $(l, n)=1$, it follows that $(l, l+n)=1, (n, l+n)=1$, therefore, from (1) we know $l|z, n| z$, and by $(l, n)=1$, we have $ln \mid z$. So, $z=ln$, and thus $m=l+n$. This way, we have
$$\begin{array}{c}
x+y=m(l+n)=(l+n)^{2} \\
x-z=m n-l n=n(m-l)=n^{2} \\
y-z=m l-l n=l(m-n)=l^{2}
\end{array}$$
The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Given that $n$ is a positive integer, and $2n+1$ and $3n+1$ are both perfect squares. Prove: $40 \mid n$.
---
The translation maintains the original text's line breaks and format.
|
Proof: Let $2n+1=x^2, 3n+1=y^2$, where $x, y$ are positive integers.
By Property 1, we know $x^2 \equiv 1 \pmod{8}$ (since $x^2$ is odd, hence $x$ is odd), thus
$$n \equiv 0 \pmod{4}$$
Furthermore, $3n+1$ is odd, hence
i.e. $\square$
$$\begin{array}{c}
y^2 \equiv 1 \pmod{8} \\
3n+1 \equiv 1 \pmod{8}
\end{array}$$
Thus,
$$n \equiv 0 \pmod{8}$$
On the other hand, for any integer $a$, we have
$$a \equiv 0, \pm 1, \pm 2 \pmod{5}$$
Hence
$$a^2 \equiv 0, 1 \text{ or } 4 \pmod{5} \text{.}$$
By the given conditions,
$$x^2 + y^2 = 5n + 2 \equiv 2 \pmod{5}$$
Combining the conclusions derived earlier, we know
Thus
$$\begin{array}{c}
x^2 \equiv y^2 \equiv 1 \pmod{5} \\
2n + 1 \equiv 1 \pmod{5} \\
n \equiv 0 \pmod{5}
\end{array}$$
Using $(5,8)=1$, we know $40 \mid n$.
|
40 \mid n
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 If $a, b$ are positive integers such that $ab+1$ is a perfect square, then we denote $a \sim b$. Prove: If $a \sim b$, then there exists a positive integer $c$, such that $a \sim c, b \sim c$.
untranslated text remains the same as requested.
|
Proof: Given $a \sim b$, we can set $ab + 1 = x^2$, where $x$ is a positive integer. The next perfect square related to $a$, $b$, and $x$ is $(a + x)^2$ or $(b + x)^2$. Therefore, we take $c = 2x + a + b$, then
$$\begin{aligned}
ac + 1 & = a(2x + a + b) + 1 \\
& = 2ax + a^2 + ab + 1 \\
& = 2ax + a^2 + x^2 = (x + a)^2, \\
bc + 1 & = (x + b)^2.
\end{aligned}$$
The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
For each positive integer $n$, let $S(n)$ denote the sum of the digits of $n$ in its decimal representation. Prove: for any positive integer $m$, there exists a positive integer $n$ such that $S(n) = m S(3 n)$.
|
3. Notice that, for any positive integer $k, S(\underbrace{1}_{k \uparrow} \underbrace{0 \cdots 0})=9$, thus, let $1 \underbrace{0 \cdots 08}_{k \uparrow}=3 n$, then
$$n=\underbrace{3 \cdots 36}_{k \uparrow},$$
so $S(n)=3 k+6$. Therefore, for any positive integer $m$, take $k=3 m-2$, then
$$S(n)=m S(3 n) \text {. }$$
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5I Let $n$ be a positive integer. Prove: there exists a positive integer $m$ whose decimal representation contains only the digits 0 and 1, such that $n \mid m$.
|
5. Consider the sequence
$$1,11,111, \cdots, \underbrace{1 \cdots 1}_{n+1 \uparrow},$$
where there must be two numbers that are congruent modulo $n$ (because the remainder of any integer divided by $n$ can only be 0, 1, 2, ..., $n-1$, a total of $n$ cases), their difference (the larger minus the smaller) is the required $m$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
6. Let $n$ be a positive odd number. Prove: There exists a positive integer $m$ whose decimal representation consists entirely of odd digits, such that $n \mid m$.
|
6. If $(5, n)=1$, then by the conclusion of the previous problem, there exists $m=\underbrace{1 \cdots 1}_{i \uparrow} \underbrace{0 \cdots 0}_{i \uparrow}$, such that $n \mid m$, and $n$ is odd. Combining this with $5 \nmid n$, we know $(n, 10)=1$, hence $n \mid \underbrace{1 \cdots 1}_{i \uparrow}$. The proposition is proved.
If $5 \mid n$, let $5^{\alpha} \| n$, then we can write $n=5^{\alpha} \cdot n_{1}$, where $5 \nmid n_{1}$. Using the conclusion of Example 5 in Section 2.2, we know there exists an $\alpha$-digit positive integer $m_{1}$, such that $5^{a} \mid m_{1}$, and each digit of $m_{1}$ is odd. At this point, consider the number
$$m_{1}, \overline{m_{1} m_{1}}, \cdots, \underbrace{\overline{m_{1} \cdots m_{1}}}_{n_{1}+1 \uparrow},$$
where $\underbrace{\overline{m_{1} \cdots m_{1}}}_{i \uparrow}$ represents the decimal number formed by writing $i$ copies of $m_{1}$ (so the numbers listed above are all multiples of $5^{\alpha}$). Then there exist $1 \leqslant i<j \leqslant n_{1}+1$, such that
$$\underbrace{\overline{m_{1} \cdots m_{1}}}_{j \uparrow} \equiv \underbrace{\overline{m_{1} \cdots m_{1}}}_{i \uparrow}\left(\bmod n_{1}\right)$$
Combining this with $\left(n_{1}, 10\right)=1$, we know
$$n_{1} \mid \underbrace{\overline{m_{1} \cdots m_{1}}}_{j-i \uparrow}$$
Thus, let $m=\underbrace{\overline{m_{1} \cdots m_{1}}}_{j-i \uparrow}$, then each digit of $m$ is odd, and
$$\begin{array}{l}
5^{a}\left|m, n_{1}\right| m \\
\left(5^{a}, n_{1}\right)=1
\end{array}$$
Hence $5^{a} \cdot n_{1} \mid m$, i.e., $n \mid m$.
The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
7 Prove: For each positive integer $n$, the number $19 \times 8^{n}+17$ is composite.
|
7. If $n$ is even, then
$$19 \times 8^{n}+17 \equiv 1 \times(-1)^{n}+2 \equiv 0(\bmod 3)$$
If $n \equiv 1(\bmod 4)$, write $n=4 k+1$, then
$$\begin{aligned}
19 \times 8^{n}+17 & =19 \times 64^{2 k} \times 8+17 \\
& \equiv 6 \times(-1)^{2 k} \times 8+4 \\
& \equiv 0(\bmod 13)
\end{aligned}$$
If $n \equiv 3(\bmod 4)$, then
$$\begin{aligned}
19 \times 8^{n}+17 & =19 \times 64^{2 k+1} \times 8+17 \\
& \equiv(-1) \times(-1)^{2 k+1} \times 3+2 \\
& \equiv 0(\bmod 5)
\end{aligned}$$
Therefore, for any positive integer $n$, the number $19 \times 8^{n}+17$ is composite.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
8 Fibonacci sequence $\left\{F_{n}\right\}$ is defined as follows: $F_{1}=F_{2}=1, F_{n+2}=F_{n+1}+F_{n}, n=1$, $2, \cdots$.
(1) Prove: The sum of any 10 consecutive terms of this sequence is a multiple of 11;
(2) Find the smallest positive integer $k$, such that the sum of any $k$ consecutive terms of this sequence is a multiple of 12.
|
8. Consider the sequence $\left\{F_{n}\right\}$ where each term is taken modulo 11 (or 12).
(1) $\left\{F_{n}(\bmod 11)\right\}: 1,1,2,3,5,-3,2,-1,1,0,1,1, \cdots$, so $\left\{F_{n}(\bmod 11)\right\}$ is a purely periodic sequence with a period of 10. Therefore,
the sum of any 10 consecutive terms in $\left\{F_{n}\right\}$
$$\begin{array}{l}
\equiv 1+1+2+3+5+(-3)+2+(-1)+1+0 \\
=11 \equiv 0(\bmod 11)
\end{array}$$
The proposition is proved. $\square$
(2) $\left\{F_{n}(\bmod 12)\right\}: 1,1,2,3,5,-4,1,-3,-2,-5,5,0,1, 1, \cdots$ is a purely periodic sequence with a period of 12. Direct verification shows that the smallest positive integer $k$ satisfying the condition is $k=36$.
Note: If $k$ is the smallest positive integer satisfying (2), and $n$ is any positive integer satisfying (2), then $k \mid n$ (this conclusion is left for the reader to prove). Therefore, after finding that $n=36$ (where the sum of the numbers in each period of $\left\{F_{n}(\bmod 12)\right\}$ is $\equiv 4(\bmod 12)$), it is only necessary to verify that the positive divisors of 36 do not satisfy the condition to conclude that 36 is the smallest positive integer satisfying the condition.
|
36
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
9 Let integers $a, b$ satisfy: $21 \mid a^{2}+b^{2}$. Prove: $441 \mid a^{2}+b^{2}$.
|
9. First prove separately:
(1) If $a^{2}+b^{2} \equiv 0(\bmod 3)$, then $a \equiv b \equiv 0(\bmod 3)$;
(2) If $a^{2}+b^{2} \equiv 0(\bmod 7)$, then $a \equiv b \equiv 0(\bmod 7)$.
This only requires noting that, for any integer $x$, we have
$$x^{2} \equiv 0 \text { or } 1(\bmod 3),$$
and
$$x^{2} \equiv 0,1,2 \text { or } 4(\bmod 7);$$
this can prove the statements. $\square$
Now from $21 \mid a^{2}+b^{2}$ we can deduce $21|a, 21| b$, hence $21^{2} \mid a^{2}+b^{2}$, so the proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
10 positive integers $a$, $b$, $c$ satisfy: $c^{2}=a^{2}+b^{2}+ab$. Prove: $c$ has a prime factor greater than 5.
|
10. We prove respectively:
(1) If $2 \mid c$, then $2|a, 2| b$;
(2) If $3 \mid c$, then $3|a, 3| b$;
(3) If $5 \mid c$, then $5|a, 5| b$.
The proof of (1) is trivial.
For (2), it suffices to note that
$$c^{2}=a^{2}+a b+b^{2}=(a-b)^{2}+3 a b,$$
which makes the proof straightforward.
For (3), from the condition, we know
$$4 c^{2}=4 a^{2}+4 a b+4 b^{2}=3 a^{2}+(a+2 b)^{2},$$
and for any integer $x$, we know
Thus, from
we know
$$\begin{array}{c}
x^{2} \equiv 0,1,4(\bmod 5) \\
3 x^{2}+y^{2} \equiv 0(\bmod 5) \\
x^{2} \equiv y^{2} \equiv 0(\bmod 5) \\
x \equiv y \equiv 0(\bmod 5)
\end{array}$$
Therefore, from $5 \mid c$, we know
$$3 a^{2}+(a+2 b)^{2} \equiv 0(\bmod 5)$$
Thus,
$$a \equiv a+2 b \equiv 0(\bmod 5)$$
We can get
$$a \equiv b \equiv 0(\bmod 5)$$
So (3) holds.
Returning to the original problem, when $c$ is a multiple of $2$, $3$, or $5$, both sides of $c^{2}=a^{2}+a b+b^{2}$ can be divided by $2^{2}$, $3^{2}$, or $5^{2}$ respectively, and the form of the equation remains unchanged. Therefore, $c$ has a prime factor greater than 5.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 For any positive integer $n$, prove: there exist $n$ consecutive positive integers, all of which are composite.
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 instruction. Here is the translation again for clarity:
Example 4 For any positive integer $n$, prove: there exist $n$ consecutive positive integers, all of which are composite.
|
Proof: Let $n$ be a positive integer, then
$$(n+1)!+2,(n+1)!+3, \cdots,(n+1)!+(n+1)$$
are $n$ consecutive positive integers, and the $k$-th number is a multiple of $k+1$ (and greater than $k+1$), so they are $n$ consecutive composite numbers.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
16 Let $a_{1}, a_{2}, \cdots, a_{100}$ be a permutation of $1,2, \cdots, 100$, and let $b_{i}=a_{1}+a_{2}+\cdots+a_{i}$, $i=1,2, \cdots, 100$, and let $r_{i}$ be the remainder when $b_{i}$ is divided by 100. Prove that $r_{1}, r_{2}, \cdots, r_{100}$ contains at least 11 different numbers.
|
16. If $r_{1}, r_{2}, \cdots, r_{100}$ contain only 10 different numbers, then for $i=1,2, \cdots, 99$, $r_{i+1}-r_{i}$ has only $10^{2}-9=91$ (here subtracting 9 is because when $r_{i+1}=r_{i}$, the resulting values are all zero) different values. However, under modulo 100, $r_{i+1}-r_{i}$ are sequentially $a_{2}, a_{3}, \cdots, a_{100}$, which have 99 different values, a contradiction.
Therefore, $r_{1}, r_{2}, \cdots, r_{100}$ must contain at least 11 different values.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
18 Let $m, n$ be positive integers, $m>1$, prove: $m\left(2^{m}-1\right) \mid n$ if and only if $\left(2^{m}-1\right)^{2} \mid$ $2^{n}-1$.
|
18. If \( m\left(2^{m}-1\right) \mid n \), let \( n=m\left(2^{m}-1\right) k \), then
\[
\begin{aligned}
2^{n}-1 & =2^{m\left(2^{m}-1\right) k}-1 \\
& =\left(2^{m k}\right)^{\left(2^{m}-1\right)}-1 \\
& =\left(2^{m k}-1\right) A,
\end{aligned}
\]
where
\[
A=\left(2^{m k}\right)^{2^{m}-2}+\left(2^{m k}\right)^{2^{m}-3}+\cdots+\left(2^{m k}\right)^{1}+1
\]
Notice that
\[
\begin{array}{c}
2^{m k}-1=\left(2^{m}\right)^{k}-1 \equiv 1^{k}-1 \equiv 0\left(\bmod 2^{m}-1\right) \\
A \equiv 1^{2^{m}-2}+1^{2^{m}-3}+\cdots+1^{1}+1=2^{m}-1 \equiv 0\left(\bmod 2^{m}-1\right)
\end{array}
\]
Therefore,
\[
\left(2^{m}-1\right)^{2} \mid 2^{n}-1
\]
Conversely, if \(\left(2^{m}-1\right)^{2} \mid 2^{n}-1\), we first prove \( m \mid n \). If not, let \( n=m q+r, 0< r < m \), then by
we know
\[
\begin{array}{c}
2^{n} \equiv 1\left(\bmod 2^{m}-1\right) \\
\left(2^{m}\right)^{q} \cdot 2^{r} \equiv 1\left(\bmod 2^{m}-1\right) \\
2^{r} \equiv 1\left(\bmod 2^{m}-1\right) \\
1 \leqslant 2^{r}-1<2^{m}-1
\end{array}
\]
So \( 2^{m}-1 \nmid 2^{r}-1 \), which is a contradiction. Therefore, \( m \mid n \).
Now let \( n=m q \), then
\[
2^{n}-1=\left(2^{m}-1\right) \times B
\]
where
\[
B=\left(2^{m}\right)^{q-1}+\left(2^{m}\right)^{q-2}+\cdots+2^{m}+1
\]
By
we know
\[
\begin{array}{c}
\left(2^{m}-1\right)^{2} \mid 2^{n}-1 \\
2^{m}-1 \mid B
\end{array}
\]
Also,
\[
B \equiv 1^{q-1}+1^{q-2}+\cdots+1=q\left(\bmod 2^{m}-1\right)
\]
So
\[
2^{m}-1 \mid q
\]
Thus,
\[
m\left(2^{m}-1\right) \mid n
\]
The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
19 Let $a, b$ be coprime positive integers, and $p$ be an odd prime. Prove: $\left(a+b, \frac{a^{p}+b^{p}}{a+b}\right)=1$ or $p$.
|
19. Let $A=\frac{a^{p}+b^{p}}{a+b}=a^{p-1}-a^{p-2} b+\cdots-a b^{p-2}+b^{p-1}$, combining the fact that $p$ is odd and $b \equiv-a(\bmod a+b)$, we know
$$A \equiv \underbrace{a^{p-1}+a^{p-1}+\cdots+a^{p-1}}_{p \uparrow}=p a^{p-1}(\bmod a+b)$$
and
$$\begin{array}{c}
(a, b)=1 \\
(a, a+b)=1
\end{array}$$
Therefore,
$$\begin{aligned}
\left(a+b, \frac{a^{p}+b^{p}}{a+b}\right) & =(a+b, A) \\
& =\left(a+b, p a^{p-1}\right) \\
& =(a+b, p)=1 \text { or } p .
\end{aligned}$$
|
1 \text{ or } p
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
24 Let $m, n$ be positive integers, and $m n \mid m^{2}+n^{2}+m$. Prove: $m$ is a perfect square.
|
24. From the condition, we can set \( m^{2}+n^{2}+m=k m n \), where \( k \) is a positive integer. Thus, the quadratic equation in \( n \)
\[ n^{2}-k m n+m^{2}+m=0 \]
has positive integer solutions, so
\[ \Delta=(k m)^{2}-4\left(m^{2}+m\right)=m\left(k^{2} m-4 m-4\right) \]
is a perfect square.
If \( m \) is odd, then
\[ \left(m, k^{2} m-4 m-4\right)=(m,-4)=1 \]
Hence, from \(\Delta\) being a perfect square, we know that \( m \) is a perfect square.
If \( m \) is even, then from (1) we know \( n \) is even (otherwise the left side of (1) would be odd, a contradiction), so \( 4 \mid n^{2} \), \( 4 \mid k m n \), \( 4 \mid m^{2} \), and thus from (1) we know \( 4 \mid m \). Let \( m=4 m_{1} \), then
\[ \Delta=16 m_{1}\left(k^{2} m_{1}-m_{1}-1\right) \]
So, \( m_{1}\left(k^{2} m_{1}-m_{1}-1\right) \) is a perfect square, at this time
\[ \left(m_{1}, k^{2} m_{1}-m_{1}-1\right)=\left(m_{1},-1\right)=1 \]
Hence, \( m_{1} \) is a perfect square. Therefore, \( m=4 m_{1} \) is also a perfect square. The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
25 Prove: If a positive integer $n$ can be expressed as the sum of squares of three positive integers, then $n^{2}$ can also be expressed as the sum of squares of three positive integers.
|
25. Let $n=x^{2}+y^{2}+z^{2}, x \geqslant y \geqslant z$ be positive integers, then
$$\begin{aligned}
n^{2} & =\left(x^{2}+y^{2}+z^{2}\right)^{2} \\
& =\left(x^{2}+y^{2}\right)^{2}+2\left(x^{2}+y^{2}\right) z^{2}+z^{4} \\
& =\left(x^{2}+y^{2}-z^{2}\right)^{2}+4\left(x^{2}+y^{2}\right) z^{2} \\
& =\left(x^{2}+y^{2}-z^{2}\right)^{2}+(2 x z)^{2}+(2 y z)^{2} .
\end{aligned}$$
Notice that, $x^{2}+y^{2}-z^{2}>0$, so $n^{2}$ can be expressed as the sum of squares of 3 positive integers.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
27 Prove: There exist infinitely many integers $n$, such that the numbers $n$, $n+1$, and $n+2$ can all be expressed as the sum of two integers (not necessarily distinct) squared. For example: $0=0^{2}+0^{2}$, $1=0^{2}+1^{2}$, $2=1^{2}+1^{2}$, so $n=0$ is an integer that satisfies the condition.
|
27. Find a positive integer $l$ such that $l^{2}-1=x^{2}+y^{2}$ has positive integer solutions. Let $x=2 m^{2}$, $y=2 m$, and $l=2 m^{2}+1$, then $l^{2}-1=x^{2}+y^{2}$. Therefore, for any positive integer $m$, take
$$n=\left(2 m^{2}+1\right)^{2}-1=4 m^{4}+4 m^{2},$$
then $\square$
$$\begin{array}{c}
n=\left(2 m^{2}\right)^{2}+(2 m)^{2}, \\
n+1=\left(2 m^{2}+1\right)^{2}+0^{2}, \\
n+2=\left(2 m^{2}+1\right)^{2}+1^{2}
\end{array}$$
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
29 Let $n>1$ be a positive integer, prove: the number $2^{n}-1$ is neither a perfect square nor a perfect cube.
|
29. Since $n \geqslant 2$, we have $2^{n}-1 \equiv -1 \pmod{4}$, while a perfect square $\equiv 0$ or $1 \pmod{4}$. Therefore, $2^{n}-1$ is not a perfect square.
On the other hand, if there exist $n>1$ and a positive integer $x$ such that
$$2^{n}-1=x^{3}$$
then
$$2^{n}=(x+1)\left(x^{2}-x+1\right)$$
Since
$$x^{2}-x+1=x(x-1)+1$$
where $x(x-1)$ is even (one of two consecutive integers is even), $x^{2}-x+1$ is odd, which requires
$$x^{2}-x+1=1$$
Thus, $x=1$, leading to $n=1$, a contradiction. Therefore, $2^{n}-1$ is not a perfect cube.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
30 Let $a$, $b$, $c$ be positive integers, and $\sqrt{a}+\sqrt{b}+\sqrt{c}$ be an integer. Prove: $a$, $b$, $c$ are all perfect squares.
|
30. First, we prove: For any positive integer $a$, if $\sqrt{a}$ is a rational number, then $a$ is a perfect square.
Indeed, if $\sqrt{a}=\frac{q}{p}$, where $p, q$ are positive integers and $(p, q)=1$, then $a=\frac{q^{2}}{p^{2}}$. Since $a$ is a positive integer, it follows that $p^{2} \mid q^{2}$. But $(p, q)=1$, so $p=1$, which means $a=q^{2}$.
Next, we prove the original problem: Suppose $\sqrt{a}+\sqrt{b}+\sqrt{c}=m$, where $m$ is an integer, then
$$\begin{aligned}
(\sqrt{a}+\sqrt{b})^{2} & =(m-\sqrt{c})^{2} \\
a+b+2 \sqrt{a b} & =m^{2}-2 \sqrt{c}+c
\end{aligned}$$
Thus, $\sqrt{a b}+\sqrt{c}$ is a rational number. We can set $\sqrt{a b}+\sqrt{c}=n$, where $n$ is a positive rational number, then
$$a b=(n-\sqrt{c})^{2}=n^{2}-2 n \sqrt{c}+c$$
Therefore, $\sqrt{c}$ is a rational number. Using the previous conclusion, we know that $c$ is a perfect square, so $\sqrt{a}+\sqrt{b}=m-\sqrt{c}$ is a positive integer. Similarly, we can conclude that $a$ and $b$ are also perfect squares.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Given that the positive integer $c$ is an odd composite number. Prove: there exists a positive integer $a$, such that $a \leqslant \frac{c}{3}-1$, and $(2 a-1)^{2}+8 c$ is a perfect square.
|
31. By completing the perfect square. From the condition, we can set $c=p q, 3 \leqslant p \leqslant q, p, q$ are both odd numbers. Now we need to find $a$ such that $(2 a-1)^{2}+8 p q$ is a perfect square. A natural choice is: let $2 a-1=2 q-p$, then
$$\begin{aligned}
(2 a-1)^{2}+8 p q & =(2 q-p)^{2}+8 p q \\
& =(2 q+p)^{2}
\end{aligned}$$
At this time
$$\begin{aligned}
a & =\frac{1}{2}(2 q-p+1) \\
& =q-\frac{p-1}{2} \\
& \leqslant q-1=\frac{c}{p}-1 \\
& \leqslant \frac{c}{3}-1
\end{aligned}$$
This meets the requirement in the problem.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
32 Let integers $a, b$ satisfy: for any positive integer $n$, the number $2^{n} \cdot a + b$ is a perfect square. Prove $a=0$.
|
32. If $a \neq 0$, note that in $a0$. Now let $2^{n} a+b=x_{n}^{2}$, where $x_{n}$ is a positive integer, then for any positive integer $n$, we have $x_{n}<x_{n+1}$.
Since $4 x_{n}^{2}-x_{n+2}^{2}=4\left(2^{n} a+b\right)-\left(2^{n+2} a+b\right)=3 b$, thus
$$3|b|=\left|2 x_{n}-x_{n+2}\right| \cdot\left|2 x_{n}+x_{n+2}\right|$$
And $2 x_{n}+x_{n+2}$ increases as $n$ increases, so it can only be
$$\left|2 x_{n}-x_{n+2}\right|=0$$
That is
$$|b|=0$$
But in this case, $2^{n} a$ and $2^{n+1} a$ must both be perfect squares, which is impossible, leading to a contradiction. Therefore, $a=0$.
|
a=0
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Let $a, b, c, d, e, f$ all be positive integers, $S=a+b+c+d+e+f$ is a factor of $abc+def$ and $ab+bc+ca-de-ef-ed$. Prove: $S$ is a composite number.
The text is translated while preserving the original line breaks and format.
|
Consider the polynomial
$$f(x)=(x+a)(x+b)(x+c)-(x-d)(x-e)(x-f) .$$
After expanding, we can see that
$$f(x)=S x^{2}+(a b+b c+c a-d e-e f-f d) x+(a b c+d e f)$$
From the condition, for any $x \in \mathbf{Z}$, we have $S \mid f(x)$. In particular, taking $x=d$, we have $S \mid f(d)$, i.e., $S \mid(d+a)(d+b)(d+c)$. Since $a, b, c, d, e, f$ are all positive integers, $d+a, d+b,$ and $d+c$ are all less than $S$, so $S$ is a composite number.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
36 Let $p$ be a prime, $a$ and $n$ be positive integers, and $2^{p}+3^{p}=a^{n}$. Prove: $n=1$.
The text has been translated while preserving the original line breaks and format.
|
36. When $p=2$, $a^{n}=13$, we know $a=13, n=1$. When $p>2$, since $p$ is a prime, $p$ is an odd number, at this time
$$2^{p}+3^{p}=(2+3)\left(2^{p-1}-2^{p-2} \times 3+\cdots-2 \times 3^{p-2}+3^{p-1}\right),$$
thus $5 \mid a^{n}$, which means $5 \mid a$. If $n>1$, then $5^{2} \mid a^{n}$, in this case, we should have
$$2^{p-1}-2^{p-2} \times 3+\cdots-2 \times 3^{p-2}+3^{p-1} \equiv 0(\bmod 5)$$
Using $3 \equiv-2(\bmod 5)$, $p$ is an odd number, and the above equation, we know
$$\begin{aligned}
& 2^{p-1}-2^{p-2} \times 3+\cdots-2 \times 3^{p-2}+3^{p-1} \\
\equiv & \underbrace{2^{p-1}+2^{p-1}+\cdots+2^{p-1}}_{p \uparrow 2^{p-1}} \\
= & p \cdot 2^{p-1} \equiv 0(\bmod 5)
\end{aligned}$$
Therefore, $5 \mid p$, and since $p$ is a prime, $p=5$, which leads to $a^{n}=2^{5}+3^{5}=275=5^{2} \times 11, n$ can only be 1, a contradiction.
Thus, $n=1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
30 Let $m$ be a positive integer, and $m \equiv 2(\bmod 4)$. Prove: there exists at most one pair of positive integers $(a, b)$, such that $m=a b$, and $0<a-b<\sqrt{5+4 \sqrt{4 m+1}}$.
|
39. If the value of $a+b$ can be determined (considering $m$ as a constant), then by the inverse of Vieta's theorem, there is at most one pair of positive integers $(a, b)$ that satisfies the condition.
From the condition, we know that $(a+b)^{2}=(a-b)^{2}+4 a b$ satisfies
$$\begin{aligned}
1+4 m & \leqslant(a+b)^{2} \\
& <5+4 \sqrt{4 m+1}+4 m \\
& =(\sqrt{4 m+1}+2)^{2}
\end{aligned}$$
Thus,
$$\sqrt{4 m+1} \leqslant a+b<\sqrt{4 m+1}+2$$
Therefore,
$$a+b=\left\{\begin{array}{l}
\sqrt{4 m+1} \text { or } \sqrt{4 m+1}+1, \text { if } \sqrt{4 m+1} \text { is an integer; } \\
{[\sqrt{4 m+1}]+1 \text { or }[\sqrt{4 m+1}]+2, \text { if } \sqrt{4 m+1} \text { is not an integer. }}
\end{array}\right.$$
In summary, $a+b$ can only take values from two consecutive positive integers. Since $a b=m \equiv 2(\bmod 4)$, it follows that $a$ and $b$ are one odd and one even, meaning $a+b$ is odd. This way, we know that the value of $a+b$ is uniquely determined, and the proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
40 Let $n$ be a positive integer greater than 10, and each digit of $n$ is $1$, $3$, $7$, or $9$. Prove: $n$ has a prime factor greater than 10.
|
40. By contradiction, if every prime factor of $n$ is no greater than 10, using the given conditions, we know that $n$ is odd and $n$ is not a multiple of 5. Therefore, there exist non-negative integers $i, j$ such that
$$n=3^{i} \cdot 7^{j},$$
Consider the remainders of $3^{i}$ and $7^{j}$ when divided by 20. For $i=0,1,2, \cdots, j=0,1,2, \cdots$, we have respectively
$$\begin{array}{l}
\left\{3^{i}(\bmod 20)\right\}: 1,3,9,7,1,3, \cdots \\
\left\{7^{j}(\bmod 20)\right\}: 1,7,9,3,1,7, \cdots
\end{array}$$
Both of these are periodic sequences with a period of 4. Therefore,
$$3^{i} \cdot 7^{j} \equiv a b(\bmod 20)$$
Here, $a, b$ are all $1, 3, 7$ or 9. By calculating each case, we find that
$$3^{i} \cdot 7^{j} \equiv 1,3,7 \text{ or } 9(\bmod 20),$$
This indicates that the tens digit of all numbers of the form $3^{i} \cdot 7^{j}$ is even. However, every digit of $n$ is 1, 3, 7, or 9, which is a contradiction.
Therefore, $n$ has a prime factor greater than 10.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
42 Let $f(n)=1+n+n^{2}+\cdots+n^{2010}$. Prove: for any integer $m$, if $2 \leqslant m \leqslant 2010$, then there does not exist a positive integer $n$, such that $m \mid f(n)$.
|
42. If there exists $2 \leqslant m \leqslant 2010$, such that for some positive integer $n$, we have $m \mid f(n)$. Since $f(1)=2011$ is a prime number (here 2011 is a prime number, which needs to be verified by dividing 2011 by all primes not greater than $\sqrt{2011}$), hence $n \neq 1$, and at this point we can write $f(n)=\frac{n^{2011}-1}{n-1}$.
For a prime factor $p$ of $m$, from $m \mid f(n)$ we know $n^{2011} \equiv 1(\bmod p)$, and by Fermat's Little Theorem we know $n^{p-1} \equiv 1(\bmod p)$, so we have
$$n^{(2011, p-1)} \equiv 1(\bmod p)$$
Combining $p-1<2011$, and 2011 being a prime number, we get $(2011, p-1)=1$, thus, $n \equiv 1(\bmod p)$, hence
$$0 \equiv f(n) \equiv 1+1^{2}+\cdots+1^{2010}=2011(\bmod p)$$
This requires $p=2011$, which contradicts $m \leqslant 2010$.
Therefore, the proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Property 1 Let $d=(a, b)$, then there exist integers $x, y$, such that
$$a x+b y=d \text {. }$$
|
We use the division algorithm to handle this, and the proof process of this conclusion is also the process of finding the greatest common divisor (GCD) of $a$ and $b$, which is called "Euclidean algorithm".
Assume $a$ and $b$ are both non-zero (when one of $a$ and $b$ is zero, the conclusion is obvious), and $|a| \leqslant |b|$.
Let $b = a q_{1} + r_{1}$, where $0 \leqslant r_{1} < |a|$. Then, we have $a = r_{1} q_{2} + r_{2}$, where $0 \leqslant r_{2} < r_{1}$. Continuing this process, we get $r_{1} = r_{2} q_{3} + r_{3}$, where $0 \leqslant r_{3} < r_{2}$, and so on. Since $r_{1} > r_{2} > r_{3} > \cdots$, eventually, after a certain number of steps, we will get $r_{k+1} = 0$. Thus, we obtain the following series of equations:
$$\begin{array}{l}
b = a q_{1} + r_{1}, 0 < r_{1} < |a| \\
a = r_{1} q_{2} + r_{2}, 0 < r_{2} < r_{1} \\
r_{1} = r_{2} q_{3} + r_{3}, 0 < r_{3} < r_{2} \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
r_{k-2} = r_{k-1} q_{k} + r_{k}, 0 < r_{k} < r_{k-1} \\
r_{k-1} = r_{k} q_{k+1}
\end{array}$$
Notice that from the first equation to the $k$-th equation, we have
$$d \mid r_{1}, d \mid r_{2}, \cdots, d \mid r_{k}$$
From the $(k+1)$-th equation, working backwards, we also have
$$r_{k} \mid r_{k-1}, r_{k} \mid r_{k-2}, \cdots, r_{k} \mid r_{1}, r_{k} \mid a, r_{k} \mid b$$
Therefore, $r_{k}$ is a common divisor of $a$ and $b$. Combining this with the fact that $d$ is the greatest common divisor of $a$ and $b$, we have $r_{k} \leqslant d$. Since $d \mid r_{k}$, it follows that $d \leqslant r_{k}$. Thus, $d = r_{k}$. In other words, we have found the greatest common divisor of $a$ and $b$.
Now, using $d = r_{k}$ and the $k$-th equation, we know
$$d = r_{k-2} - r_{k-1} q_{k}$$
From $r_{k-1} = r_{k-3} - r_{k-2} q_{k-1}$ (the $(k-1)$-th equation rearranged),
substituting into the above equation, we see that $d$ can be expressed as a "linear combination" of $r_{k-2}$ and $r_{k-3}$ (see Property 2 in Section 1.1). Continuing this process backwards, we see that $d$ can be expressed as a "linear combination" of $a$ and $b$, i.e., there exist integers $x$ and $y$ such that
$$d = a x + b y$$
|
d = a x + b y
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Theorem 1 The necessary and sufficient condition for the indeterminate equation (2) to have integer solutions is $(a, b) \mid c$.
|
The necessity is obvious, while the sufficiency can be obtained by Bézout's theorem.
In fact, by Bézout's theorem, we know there exist integers $x_{0}, y_{0}$ such that $a x_{0} + b y_{0} = (a, b)$. Let $c = (a, b) c_{1}$, then $\left(c_{1} x_{0}, c_{1} y_{0}\right)$ is a solution to (2), and the sufficiency is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Theorem 2 If the indeterminate equation (2) has an integer solution $\left(x_{0}, y_{0}\right)$, then all integer solutions of (2) are
$$\left\{\begin{array}{l}
x=x_{0}+\frac{b}{(a, b)} t, \\
y=y_{0}-\frac{a}{(a, b)} t .
\end{array}\right.$$
|
Proof: Let $(x, y)$ be an integer solution to (2), and combine with $\left(x_{0}, y_{0}\right)$ being a solution to (2), we have
$$\left\{\begin{array}{l}
a x+b y=c, \\
a x_{0}+b y_{0}=c,
\end{array}\right.$$
Thus,
$$a\left(x-x_{0}\right)+b\left(y-y_{0}\right)=0$$
That is,
$$a\left(x-x_{0}\right)=b\left(y_{0}-y\right)$$
Hence,
$$b \mid a\left(x-x_{0}\right),$$
Therefore,
$$\left.\frac{b}{(a, b)} \right\rvert\, x-x_{0}$$
We can set
$$x-x_{0}=\frac{b}{(a, b)} t$$
Then
$$y-y_{0}=-\frac{a}{(a, b)} t$$
where $t$ is an integer, which shows that the solutions to equation (2) have the form (3).
Conversely, the integer pairs $(x, y)$ determined by (3) are solutions to equation (2) can be directly verified by substituting into (2).
Therefore, Theorem 2 holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let $a, b$ be coprime positive integers. Prove: the indeterminate equation
$$a x+b y=a b-a-b$$
has no non-negative integer solutions.
|
If there exists a pair of non-negative integers $\left(x_{0}, y_{0}\right)$ satisfying (1), then
$$a\left(x_{0}+1\right)+b\left(y_{0}+1\right)=a b,$$
then, we should have
and
thus
$$\begin{array}{l}
a \mid b\left(y_{0}+1\right) \\
(a, b)=1 \\
a \mid y_{0}+1
\end{array}$$
Since $a$ and $y_{0}+1$ are both positive integers, we have $a \leqslant y_{0}+1$.
Similarly, we can prove: $b \mid x_{0}+1$, hence $b \leqslant x_{0}+1$. But in this case, the left side of (2) $\geqslant a b+b a=$ $2 a b>$ the right side of (2), which is a contradiction.
Therefore, (1) has no non-negative integer solutions.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let positive integers $a, b$ be coprime, and positive integer $c$ be greater than $a b-a-b$. Prove: the indeterminate equation
$$a x+b y=c$$
has non-negative integer solutions.
|
Proof: Let the general solution of (1) be
$$\left\{\begin{array}{l}
x=x_{0}+b t, \\
y=y_{0}-a t,
\end{array}(\text { where } t \text { is an integer })\right.$$
where $\left(x_{0}, y_{0}\right)$ is a particular solution of (1).
Thus, by adjusting the value of $t$ (adding or subtracting $b$ from $x_{0}$), we can find a solution $\left(x_{1}, y_{1}\right)$ of (1) such that $0 \leqslant x_{1} \leqslant b-1$, then
$$b y_{1}=c-a x_{1}>a b-a-b-a x_{1} \geqslant a b-a-b-a(b-1)=-b \text {, }$$
hence
$$\text { i.e. } \square$$
$$\begin{array}{l}
y_{1}>-1 \\
y_{1} \geqslant 0 .
\end{array}$$
This shows that $\left(x_{1}, y_{1}\right)$ is a non-negative integer solution of (1), and the proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 10 Prove: the indeterminate equation
$$x^{2}+y^{2}-8 z^{3}=6$$
has no integer solutions.
|
Prove that if $(x, y, z)$ is an integer solution to equation (1), taking both sides of (1) modulo 2, we can see that $x$ and $y$ have the same parity; then taking both sides of (1) modulo 4, we can see that $x$ and $y$ are both odd, thus $x^{2} \equiv y^{2} \equiv 1(\bmod 8)$, which requires
$$6=x^{2}+y^{2}-8 z^{3} \equiv 2(\bmod 8)$$
This is a contradiction. Therefore, equation (1) has no integer solutions.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 13 Proof: The equation $x^{2}+y^{5}=z^{3}$ has infinitely many integer solutions satisfying $x y z \neq 0$.
|
Prove that taking $x=2^{156+10}, y=2^{6 k+4}, z=2^{10 k+7}, k$ as a non-negative integer, then such $x$, $y$, $z$ satisfy $x^{2}+y^{5}=z^{3}$, so the equation has infinitely many integer solutions satisfying $x y z \neq 0$.
Another proof First find a particular solution to the equation, it is easy to see that $x=10, y=3, z=7$ is a solution to the equation $x^{2}+y^{5}=z^{3}$. Therefore, $x=10 a^{15 k}, y=3 a^{6 k}, z=7 a^{10 k}$ ($a, k$ as non-negative integers) are solutions to the equation.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 14 Proof: For any integer $n$, the equation
$$x^{2}+y^{2}-z^{2}=n$$
has infinitely many integer solutions $(x, y, z)$.
|
Prove that the proposition "when $m$ is an odd number or a multiple of 4, the equation $a^{2}-b^{2}=m$ has integer solutions $(a, b)$" is useful for solving this problem. This proposition is based on the following two identities:
$$\begin{array}{c}
(k+1)^{2}-k^{2}=2 k+1 \\
(k+1)^{2}-(k-1)^{2}=4 k
\end{array}$$
For equation (1), it is only necessary to choose $x$ such that $x$ has the opposite parity to $n$ (there are infinitely many such $x$), thus using the above proposition, the equation
$$y^{2}-z^{2}=n-x^{2}$$
has integer solutions, which means equation (1) has infinitely many integer solutions.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 16 Proof: There exist infinitely many sets of positive integers $(x, y, z)$ such that $x, y, z$ are all distinct, and
$$x^{x}=y^{3}+z^{3}$$
|
To prove an idea: take $x$ as a number of the form $3k+1$, at this time
$$\begin{aligned}
x^{x} & =(3 k+1)^{3 k+1} \\
& =(3 k+1)(3 k+1)^{3 k} \\
& =3 k(3 k+1)^{3 k}+(3 k+1)^{3 k}
\end{aligned}$$
Therefore, if we make $3k$ a perfect cube, then the positive integers $x, y, z$ that meet the requirements can be found.
For this, let $k=3^{3m+2}$, where $m$ is a positive integer, then let
$$x=3 k+1, y=3^{m+1}(3 k+1)^{k}, z=(3 k+1)^{k},$$
then $x, y, z$ are all different, and satisfy $x^{x}=y^{3}+z^{3}$. The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Let $(x, y, z)$ be an integer solution of the Pythagorean equation (1). Prove: $x$, $y$, $z$ must include one number that is a multiple of 3, one number that is a multiple of 4, and one number that is a multiple of 5.
|
Prove that using the fact that perfect squares $\equiv 0,1(\bmod 3)$, if $x$ and $y$ are not multiples of 3, then
this leads to
$$\begin{array}{c}
x^{2}+y^{2} \equiv 2(\bmod 3) \\
z^{2} \equiv 2(\bmod 3)
\end{array}$$
which is a contradiction. Hence, one of $x$ or $y$ must be a multiple of 3.
If $x$, $y$, and $z$ are not multiples of 5, then
$$x^{2}+y^{2} \equiv 0,2 \text { or } 3(\bmod 5),$$
while
$$z^{2} \equiv 1 \text { or } 4(\bmod 5),$$
which is a contradiction. Hence, one of $x$, $y$, or $z$ must be a multiple of 5.
If $x$, $y$, and $z$ are all even, then divide both sides of (1) by 4 until one of $x$ or $y$ is odd. Suppose $x$ is odd, then $y$ must be even (otherwise $z^{2}=x^{2}+y^{2} \equiv 1+1 \equiv 2(\bmod 4)$, which is a contradiction), in this case $z$ is odd. Taking both sides of (1) modulo 8, we find that $y^{2} \equiv 0(\bmod 8)$, hence $4 \mid y$.
In summary, the proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let $(x, y, z)$ be a Pythagorean triple $\left(x^{2}+y^{2}=z^{2}\right)$. Prove: $z^{2}+x y$ and $z^{2}-$ $x y$ can both be expressed as the sum of squares of two positive integers.
|
Notice that $z^{2} \pm x y=\frac{2 z^{2} \pm 2 x y}{2}$.
$$=\frac{z^{2}+(x \pm y)^{2}}{2}$$
And
$$2 a^{2}+2 b^{2}=(a+b)^{2}+(a-b)^{2}$$
Therefore
$$\begin{aligned}
z^{2} \pm x y & =\frac{(z+x \pm y)^{2}+(-z+x \pm y)^{2}}{4} \\
& =\left(\frac{x \pm y+z}{2}\right)^{2}+\left(\frac{x \pm y-z}{2}\right)^{2}
\end{aligned}$$
From $x^{2}+y^{2}=z^{2}$, we know that $x \pm y$ and $z$ have the same parity, so $\frac{x \pm y \pm z}{2}$ are all integers. Furthermore, by $x<z, y<z$, and $z^{2}=x^{2}+y^{2}<(x+y)^{2}$, we get $z<x+y$, which means $\frac{x \pm y \pm z}{2}$ are all non-zero integers.
Therefore, $z^{2} \pm x y$ can both be expressed as the sum of squares of two positive integers.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let $n$ be a positive integer greater than 2. Prove: there exists a right-angled triangle with integer side lengths, one of whose legs has length exactly $n$.
The above text is translated into English, preserving the original text's line breaks and format.
|
To prove that the indeterminate equation $x^{2}+n^{2}=z^{2}$ has positive integer solutions.
Using $(z-x)(z+x)=n^{2}$, and noting that $z-x$ and $z+x$ have the same parity, when $n$ is odd, from $(z-x, z+x)=\left(1, n^{2}\right)$, we can obtain a set of positive integer solutions
$$(x, z)=\left(\frac{n^{2}-1}{2}, \frac{n^{2}+1}{2}\right) ;$$
When $n$ is even, by the condition, we know $n \geqslant 4$. Using
$$(z-x, z+x)=\left(2, \frac{n^{2}}{2}\right)$$
we can obtain a set of positive integer solutions
$$(x, z)=\left(\frac{n^{2}-4}{4}, \frac{n^{2}+4}{4}\right)$$
In summary, the proposition is established.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Let $n$ be a positive integer. Prove: there exist $n$ pairwise non-congruent Pythagorean triangles (right triangles with integer side lengths) whose perimeters are all equal.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
To prove that if we can find $n$ mutually dissimilar right-angled triangles, then by multiplying each triangle by an appropriate positive integer, we can obtain right-angled triangles with the same perimeter but not congruent to each other. This is a starting point for solving this problem.
First, we prove that any two different primitive Pythagorean triples determine non-similar right-angled triangles.
In fact, let $(a, b, c)$ and $(x, y, z)$ be two different primitive Pythagorean triples, where $a<b<c$, $x<y<z$. If the right-angled triangles determined by them are similar, then
$$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$$
Let this ratio be $k$, then $k$ is a rational number. Suppose $k=\frac{q}{p}, p, q$ are positive integers, and $(p, q)=1$, then
$$x=\frac{a q}{p}, y=\frac{b q}{p}$$
Since $x, y$ are positive integers, we have
and
$$\begin{array}{l}
p|a, p| b \\
(a, b)=1
\end{array}$$
Thus, $p=1$, at this time $(x, y)=(a q, b q)=(a, b) q=q$.
Hence, $q=1$, which leads to $x=a, y=b$, and consequently $z=c$. This is a contradiction.
Using the above conclusion, we take $n$ primitive Pythagorean triples
$$\left(x_{k}, y_{k}, z_{k}\right), k=1,2, \cdots, n$$
Here $x_{k}<y_{k}<z_{k}$, and $\left(x_{k}, y_{k}\right)=1$, then these $n$ triples determine $n$ triangles that are mutually dissimilar. Let
$$S_{k}=x_{k}+y_{k}+z_{k}$$
and assume the least common multiple of $S_{1}, S_{2}, \cdots, S_{n}$ is $S$. Now let
$$a_{k}=\frac{S}{S_{k}} \cdot x_{k}, b_{k}=\frac{S}{S_{k}} \cdot y_{k}, c_{k}=\frac{S}{S_{k}} \cdot z_{k},$$
then $\left(a_{k}, b_{k}, c_{k}\right)(k=1,2, \cdots, n)$ determine $n$ right-angled triangles that are not congruent to each other, and their perimeters are all equal to $S$.
Therefore, the proposition is true.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Property 5 Let $a \mid b c$ , and $(a, b)=1$ , then $a \mid c$.
|
By property 3, we know there exist integers $x, y$ such that
$$a x+b y=1$$
Thus, $a c x+b c y=c$. By $a \mid b c$ and $a \mid a c x$, we can conclude that $a \mid c$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
16. Let $x, y$ be positive integers, and $y>3, x^{2}+y^{4}=2\left((x-6)^{2}+(y+1)^{2}\right)$. Prove: $x^{2}+y^{4}=1994$.
|
16. Transpose and expand, we get
that is
$$\begin{array}{c}
y^{4}-2 y^{2}-4 y-2=x^{2}-24 x+72 \\
(x-12)^{2}=y^{4}-2 y^{2}-4 y+70
\end{array}$$
Notice that
$$\begin{aligned}
\left(y^{2}-2\right)^{2} & =y^{4}-4 y^{2}+4 \\
& 3 \text{ holds, and } y^{4}-2 y^{2}-4 y+70 \text{ is a perfect square, so it can only be or }
$$\begin{array}{c}
y^{4}-2 y^{2}-4 y+70=\left(y^{2}-1\right)^{2} \\
y^{4}-2 y^{2}-4 y+70=\left(y^{2}\right)^{2}
\end{array}$$
The former has no positive integer solutions, while the latter has a unique positive integer solution \( y=5 \). In this case, we require \((x-12)^{2}=5^{4}\), and the positive integer \( x=37 \). Therefore, \( x^{2}+y^{4}=37^{2}+5^{4}=1994 \).
|
1994
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
17 Prove: For any positive integer $n \geqslant 3$, there exists a perfect cube that can be expressed as the sum of the cubes of $n$ distinct positive integers.
|
17. Using a recursive construction method, it is based on the following two equations:
$$\begin{array}{c}
6^{3}=3^{3}+4^{3}+5^{3} \\
13^{3}=5^{3}+7^{3}+9^{3}+10^{3}
\end{array}$$
Generally, suppose there exist positive integers \(x_{1}<x_{2}<\cdots<x_{n}(n \geqslant 3)\) and a positive integer \(y\), such that
then \(\square\)
$$\begin{aligned}
& y^{3}=x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3} \\
(6 y)^{3}= & \left(6 x_{1}\right)^{3}+\left(6 x_{2}\right)^{3}+\cdots+\left(6 x_{n}\right)^{3} \\
= & \left(3 x_{1}\right)^{3}+\left(4 x_{1}\right)^{3}+\left(5 x_{1}\right)^{3}+\left(6 x_{2}\right)^{3}+\cdots+\left(6 x_{n}\right)^{3}
\end{aligned}$$
This shows that the proposition holds for \(n+2\).
Combining (1) and (2) (i.e., the proposition holds for \(n=3, 4\)), we can conclude that the proposition holds for all \(n \geqslant 3\).
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Property 6 Let $p$ be a prime, $p \mid a b$, then $p \mid a$ or $p \mid b$.
|
Proof: Since $p$ has only two positive divisors, hence $(p, a)=1$ or $(p, a)=p$. If $(p, a)=1$, then by property 5 we know $p \mid b$; if $(p, a)=p$, then $p \mid a$.
Below we introduce some concepts and properties of common multiples.
Let $a, b$ be integers not equal to zero. If the integer $c$ satisfies $a \mid c$ and $b \mid c$, then $c$ is called a common multiple of $a, b$. Among all positive common multiples of $a, b$, the smallest one is called the least common multiple of $a, b$, denoted as $[a, b]$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
27 Prove: There exists a positive integer $n$, which appears in exactly 2012 Pythagorean triples.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
27. Consider only numbers $n$ in the form of powers of prime numbers. For this, take a prime $p$ such that $p \equiv 3(\bmod 4)$ (for example, $7, 11, 19 \cdots$), and consider the number $p^{2012}$.
Using the method from Example 2 in Section 3.3, we know that the number $p^{2012}$ cannot be the hypotenuse of a Pythagorean triangle. We now prove that the indeterminate equation $x^{2} + p^{4024} = z^{2}$ has exactly 2012 solutions. In fact, because
$$(z-x)(z+x) = p^{4024}$$
we have $(z-x, z+x) = \left(1, p^{4024}\right), \left(p, p^{4023}\right), \cdots, \left(p^{2011}, p^{2013}\right)$.
Since $p$ is odd, each pair $(x, z)$ obtained is a positive integer, so $p^{2012}$ appears exactly in 2012 Pythagorean triples.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
28 Find all right-angled triangles with integer side lengths where the perimeter equals twice the area (numerically).
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
28. Equivalent to finding the Pythagorean triple $(x, y, z)$ that satisfies the condition
$$x+y+z=xy$$
Noting that $z=\sqrt{x^{2}+y^{2}}$, (1) becomes
$$x^{2}+y^{2}=(xy-x-y)^{2}$$
i.e.,
$$x^{2}+y^{2}=x^{2} y^{2}-2 xy(x+y)+x^{2}+2 xy+y^{2},$$
Thus,
$$x^{2} y^{2}-2 xy(x+y)+2 xy=0$$
Therefore,
$$xy-2(x+y)+2=0$$
i.e.,
$$(x-2)(y-2)=2$$
We get
$$(x, y)=(3,4),(4,3)$$
In summary, the sides of the right triangle are $3,4,5$.
| null |
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Property 7 Let $a, b$ be non-zero integers, $d, c$ be a common divisor and a common multiple of $a, b$ respectively, then $d|(a, b), [a, b] | c$.
|
Prove that this property essentially reflects the attributes of the greatest common divisor (GCD) and the least common multiple (LCM). The former is the conclusion of property 2, listed here again for comparison.
For the latter, we use proof by contradiction.
If $[a, b] \nmid c$, let $c=[a, b] \cdot q+r, 0<r<[a, b]$, then by $a \mid c$ and $a \mid [a, b]$, we know $a \mid r$, similarly $b \mid r$, i.e., $r$ is a common multiple of $a$ and $b$, but $r<[a, b]$, which contradicts the fact that $[a, b]$ is the least common multiple of $a$ and $b$. Therefore, $[a, b] \mid c$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
31 Let integers $a, b$ satisfy $5a \geqslant 7b \geqslant 0$. Prove: The system of equations for $x, y, z, w$
$$\left\{\begin{array}{l}
x+2 y+3 z+7 w=a \\
y+2 z+5 w=b
\end{array}\right.$$
has non-negative integer solutions.
|
31. Let integers $a, b$ satisfy $5 a \geqslant 7 b \geqslant 0$. Let
$$w=\left[\frac{b}{5}\right] \text { (representing the greatest integer not exceeding } \frac{b}{5} \text {), } v=b-5 w \text {, }$$
then when $v=0,1,2,3,4$ (there are only 5 possible values for $v$), the original system of equations, viewed as a system of indeterminate equations in $x, y, z$, respectively has non-negative integer solutions
$$\begin{aligned}
& (x, y, z) \\
= & (a-7 w, 0,0),(a-7 w-2,1,0),(a-7 w-3,0,1) \\
& (a-7 w-5,1,1),(a-7 w-6,0,2)
\end{aligned}$$
Therefore, the proposition holds.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
34 Prove: There exist infinitely many triples of positive integers $(a, b, c)$, such that $a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+$ $a^{2}$ are all perfect squares.
|
34. We use Pythagorean triples to construct. Take any Pythagorean triple \((x, y, z)\) (not necessarily primitive), and let
$$a=x\left|4 y^{2}-z^{2}\right|, b=y\left|4 x^{2}-z^{2}\right|, c=4 x y z,$$
then we have
$$\begin{aligned}
a^{2}+b^{2} & =x^{2}\left(3 y^{2}-x^{2}\right)^{2}+y^{2}\left(3 x^{2}-y^{2}\right)^{2} \\
& =x^{6}+3 x^{2} y^{4}+3 x^{4} y^{2}+y^{6} \\
& =\left(x^{2}+y^{2}\right)^{3}=\left(z^{3}\right)^{2}, \\
& a^{2}+c^{2}=x^{2}\left(4 y^{2}+z^{2}\right)^{2}, \\
& b^{2}+c^{2}=y^{2}\left(4 x^{2}+z^{2}\right)^{2} .
\end{aligned}$$
Since there are infinitely many Pythagorean triples, there are infinitely many sets of positive integer triples that satisfy the conditions.
For example: when \(x=3, y=4, z=5\), we get
$$a=117, b=44, c=240,$$
$$117^{2}+44^{2}=125^{2}, 117^{2}+240^{2}=267^{2}, 44^{2}+240^{2}=244^{2} .$$
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
35 Prove: The indeterminate equation $x^{2}=y^{5}-4$ has no integer solutions.
|
35. Taking both sides modulo 11, by Fermat's Little Theorem, when $11 \nmid y$, we have
$$y^{10} \equiv 1(\bmod 11)$$
Thus, in this case,
$$y^{5} \equiv \pm 1(\bmod 11)$$
Therefore, we always have
$$y^{5} \equiv -1,0,1(\bmod 11)$$
That is,
$$y^{5}-4 \equiv 6,7,8(\bmod 11)$$
On the other hand, for $x^{2}$, since
$$x \equiv 0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5(\bmod 11)$$
Thus,
$$x^{2} \equiv 0,1,4,9,5,3(\bmod 11)$$
Therefore, $x^{2}=y^{5}-4$ has no integer solutions.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
36 Prove: There do not exist positive integers $m, n$, such that $19^{19}=m^{3}+n^{4}$.
|
36. Handle both sides modulo 13. By Fermat's Little Theorem, when $13 \nmid m$, we have $m^{12} \equiv 1(\bmod 13)$, which implies $m^{6} \equiv 1$ or $25(\bmod 13)$, so $m^{3} \equiv \pm 1$ or $\pm 5(\bmod 13)$, meaning $m^{3} \equiv 0,1, 5,8,12(\bmod 13)$.
Similarly, we know that $n^{2} \equiv 0,1,4,9,3,12(\bmod 13)$, thus $n^{4} \equiv 0,1,3$ or $9(\bmod 13)$.
Using the above conclusions, we can see that for positive integers $m$ and $n$, we always have
$$m^{3}+n^{4} \equiv 0,1,2,3,4,5,6,8,9,10,11,12(\bmod 13)$$
That is, $m^{3}+n^{4} \neq 7(\bmod 13)$.
However, $19^{19} \equiv 6^{19} \equiv 6^{7} \equiv-6 \equiv 7(\bmod 13)$ (here we use Fermat's Little Theorem and $6^{6} \equiv-1(\bmod 13)$).
Therefore, there do not exist positive integers $m$ and $n$ such that $19^{19}=m^{3}+n^{4}$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Property 8 Let $a$ and $b$ be positive integers, then $[a, b]=\frac{a b}{(a, b)}$.
|
Let $c=\frac{a b}{(a, b)}$, then from $(a, b) \mid a$ and $(a, b) \mid b$ we know $b \mid c, a \mid c$. That is, $c$ is a common multiple of $a$ and $b$, hence $[a, b] \mid c$.
Conversely, by Bézout's theorem, there exist integers $x, y$ such that
$$a x + b y = (a, b)$$
That is,
Thus
$$\begin{array}{c}
\frac{a}{(a, b)} x + \frac{b}{(a, b)} y = 1 \\
\frac{a[a, b]}{(a, b)} x + \frac{b[a, b]}{(a, b)} y = [a, b]
\end{array}$$
By $b \mid [a, b]$ and $a \mid [a, b]$, we know
$$c \left| \frac{a[a, b]}{(a, b)}, c \right| \frac{b[a, b]}{(a, b)}$$
Therefore
$$c \mid [a, b] .$$
In summary, we have
$$[a, b] = \frac{a b}{(a, b)}$$
In general, for $n$ integers (non-zero) $a_{1}, a_{2}, \cdots, a_{n}$, we can similarly introduce the concepts of greatest common divisor and least common multiple, denoted as $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ and $\left[a_{1}, a_{2}, \cdots, a_{n}\right]$, respectively. It is easy to obtain the following conclusions:
|
[a, b] = \frac{a b}{(a, b)}
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Let $a, b$ be positive integers, and $\frac{ab}{a+b}$ is also a positive integer. Prove: $(a, b)>1$.
|
Prove that if $(a, b)=1$, then $(a, a+b)=1$ (this can be deduced from property 3), hence, from $a+b \mid a b$ and $(a, a+b)=1$, we get $a+b \mid b$, but $a+b>b$, so $a+b \mid b$ cannot hold. Therefore, $(a, b)>1$.
This is the translation of the provided text into English, maintaining the original formatting and line breaks.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let positive integers $a, b, c$ satisfy $b^{2}=a c$. Prove: $(a, b)^{2}=a(a, c)$.
The text is translated while preserving the original line breaks and format.
|
To prove: If we can prove $(a, b)^{2}=\left(a^{2}, b^{2}\right)$, then by property 11, we know
$$(a, b)^{2}=\left(a^{2}, b^{2}\right)=\left(a^{2}, a c\right)=a(a, c)$$
The proposition is proved. $\square$
For this, let $d=(a, b)$, and assume $a=d u, b=d v$, then by property 11, $u$ and $v$ are two coprime positive integers. To prove $\left(a^{2}, b^{2}\right)=d^{2}$, it suffices to prove: $\left(u^{2}, v^{2}\right)=1$.
Using Bézout's theorem, we know there exist integers $x, y$ such that $u x+v y=1$, hence $u^{2} x^{2}=(1-v y)^{2}=1+v\left(v y^{2}-2 y\right)$, combining property 3 we know $\left(u^{2}, v\right)=1$, swapping the positions of $u^{2}$ and $v$, doing the same again, we have $\left(v^{2}, u^{2}\right)=1$.
Therefore, the proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 The Fibonacci sequence is defined as follows: $F_{1}=F_{2}=1, F_{n+2}=F_{n+1}+F_{n}, n=1$, $2, \cdots$ Prove: For any positive integers $m$ and $n$, we have $\left(F_{m}, F_{n}\right)=F_{(m, n)}$.
|
Prove that when $m=n$, the proposition is obviously true. Now, without loss of generality, assume $m<n$, and note that
$$\begin{aligned}
F_{n} & =F_{2} F_{n-1}+F_{1} F_{n-2} \\
& =F_{2}\left(F_{n-2}+F_{n-3}\right)+F_{1} F_{n-2} \\
& =\left(F_{2}+F_{1}\right) F_{n-2}+F_{2} F_{n-3} \\
& =F_{3} F_{n-2}+F_{2} F_{n-3} \\
& =F_{3}\left(F_{n-3}+F_{n-4}\right)+F_{2} F_{n-3} \\
& =F_{4} F_{n-3}+F_{3} F_{n-4} \\
& =\cdots \\
& =F_{m} F_{n-m+1}+F_{m-1} F_{n-m},
\end{aligned}$$
Therefore, if $d \mid F_{m}$ and $d \mid F_{n}$, then by the above equation, $d \mid F_{m-1} F_{n-m}$. For any positive integer $m$, we have $\left(F_{m}, F_{m-1}\right)=\left(F_{m-1}+F_{m-2}, F_{m-1}\right)=\left(F_{m-1}, F_{m-2}\right)=\cdots=\left(F_{2}, F_{1}\right)=1$, so $\left(d, F_{m-1}\right)=1$, hence $d \mid F_{n-m}$; conversely, if $d^{\prime} \mid F_{n-m}$ and $d^{\prime} \mid F_{m}$, then by the above equation, $d^{\prime} \mid F_{n}$. Thus, $\left(F_{n}, F_{m}\right)=\left(F_{n-m}, F_{m}\right)$.
Using the above conclusion, performing the Euclidean algorithm on the indices, we can prove that $\left(F_{n}, F_{m}\right)=F_{(m, n)}$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Let $n$ be a positive integer greater than 1. Prove: there exist $n$ positive integers, which when arranged in ascending order form an arithmetic sequence (i.e., from the second term onwards, the difference between each term and the one before it is a constant), such that any two of them are coprime.
|
Consider the following $n$ numbers:
$$n!+1,2 \times(n!)+1, \cdots, n \times(n!)+1$$
These $n$ positive integers form an arithmetic sequence with a common difference of $n!$.
We prove that any two of them are coprime.
In fact, if there exist $1 \leqslant i < j \leqslant n$ such that $\gcd(i \times(n!)+1, j \times(n!)+1) = d > 1$. Consider a prime factor $p$ of $d$, we have
$$p \mid(j \times(n!)+1)-(i \times(n!)+1),$$
which means $p \mid(j-i) \times n!$. By property 6, $p \mid j-i$ or $p \mid n!$, combined with $1 \leqslant j-i<n$, we know $(j-i) \mid n!$, so, in any case, $p \mid n!$. However, $p|d, d| i \times(n!)+1$, hence $p \mid i \times(n!)+1$, combined with $p \mid n!$, leads to $p \mid 1$, a contradiction.
Therefore, the proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Fundamental Theorem of Arithmetic Let $n$ be a positive integer greater than 1, then $n$ can be decomposed into the product of several prime numbers, and this decomposition is unique when the order of these prime factors is not considered. That is, for any positive integer $n$ greater than 1, there exists a unique prime factorization form:
$$n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{q_{k}},$$
where $p_{1}<p_{2}<\cdots<p_{k}$ are prime numbers, and $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{k}$ are positive integers.
|
To prove the existence, we can use the previous analysis. Below, we prove the uniqueness.
If \( n \) has two prime factorization forms:
\[ n = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}} = q_{1}^{\beta_{1}} q_{2}^{\beta_{2}} \cdots q_{l}^{\beta_{l}} \]
where \( p_{1} < p_{2} < \cdots < p_{k} \) and \( q_{1} < q_{2} < \cdots < q_{l} \), and all are primes, \( \alpha_{i} \) and \( \beta_{j} \) are positive integers, \( 1 \leq i \leq k \), \( 1 \leq j \leq l \).
We need to prove that \( k = l \) and \( p_{i} = q_{i} \), \( \alpha_{i} = \beta_{i} \).
In fact, from (1) we know \( p_{i} \mid q_{1}^{\beta_{1}} q_{2}^{\beta_{2}} \cdots q_{l}^{\beta_{l}} \). Using the property 6 from the previous section, we know there exists some \( j \) such that \( p_{i} \mid q_{j}^{\beta_{j}} \). Using property 6 again, we know \( p_{i} \mid q_{j} \), which requires \( p_{i} = q_{j} \). That is, for \( 1 \leq i \leq k \) and each \( p_{i} \), there is some \( q_{j} \) in \( q_{1}, q_{2}, \cdots, q_{l} \) such that \( p_{i} = q_{j} \). Conversely, for each \( q_{j} \), there is some \( p_{i} \) in \( p_{1}, p_{2}, \cdots, p_{k} \) such that \( q_{j} = p_{i} \). This shows that \( k = l \), and \( q_{1}, q_{2}, \cdots, q_{l} \) is a permutation of \( p_{1}, p_{2}, \cdots, p_{k} \). Combining \( p_{1} < p_{2} < \cdots < p_{k} \) and \( q_{1} < q_{2} < \cdots < q_{l} \), we know \( p_{i} = q_{i} \), \( 1 \leq i \leq k \). Further proving \( \alpha_{i} = \beta_{i} \) is straightforward.
Using the prime factorization of a positive integer \( n \), we can easily derive the following conclusions.
1. Let the number of all positive divisors of \( n \) (including 1 and \( n \)) be \( d(n) \), then
\[ d(n) = (\alpha_{1} + 1)(\alpha_{2} + 1) \cdots (\alpha_{k} + 1) \]
From this formula, it is easy to see that \( n \) is a perfect square if and only if \( d(n) \) is odd.
2. Let the sum of all positive divisors of \( n \) be \( \sigma(n) \), then
\[ \sigma(n) = (1 + p_{1} + \cdots + p_{1}^{\alpha_{1}})(1 + p_{2} + \cdots + p_{2}^{\alpha_{2}}) \cdots (1 + p_{k} + \cdots + p_{k}^{\alpha_{k}}) \]
From this, we know that \( \sigma(n) \) is odd if and only if \( n \) is a perfect square or twice a perfect square.
3. Let the prime factorizations of \( n \) and \( m \) be
\[ n = p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}, \quad m = p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdots p_{k}^{\beta_{k}} \]
where \( p_{1} < p_{2} < \cdots < p_{k} \), all are primes, \( \alpha_{i} \) and \( \beta_{i} \) are non-negative integers, and for each \( 1 \leq i \leq k \), \( \alpha_{i} \) and \( \beta_{i} \) are not both zero, then we have \( (m, n) = p_{1}^{\gamma_{1}} p_{2}^{\gamma_{2}} \cdots p_{k}^{\gamma_{k}} \); \( [m, n] = p_{1}^{\delta_{1}} p_{2}^{\delta_{2}} \cdots p_{k}^{\delta_{k}} \), where \( \gamma_{i} = \min \{\alpha_{i}, \beta_{i}\} \), \( \delta_{i} = \max \{\alpha_{i}, \beta_{i}\} \), \( 1 \leq i \leq k \).
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let $n$ be a positive integer. Prove: The number $2^{2^{n}}+2^{2^{n-1}}+1$ has at least $n$ distinct prime factors.
|
We make the following decomposition:
$$\begin{aligned}
& 2^{2^{n}}+2^{2^{n-1}}+1 \\
= & \left(2^{2^{n}}+1\right)^{2}-2^{2^{n-1}} \\
= & \left(2^{2^{n-1}}+2^{2^{n-2}}+1\right)\left(2^{2^{n-1}}-2^{2^{n-2}}+1\right) \\
= & \left(2^{2^{n-2}}+2^{2^{n-3}}+1\right)\left(2^{2^{n-2}}-2^{2^{n-3}}+1\right)\left(2^{2^{n-1}}-2^{2^{n-2}}+1\right) \\
= & \cdots \\
= & \left(2^{2^{1}}+2^{2^{0}}+1\right)\left(2^{2^{1}}-2^{2^{0}}+1\right)\left(2^{2^{2}}-2^{2^{1}}+1\right) \cdots\left(2^{2^{n-1}}-2^{2^{n-2}}+1\right) .
\end{aligned}$$
Thus, we express \(2^{2^{n}}+2^{2^{n-1}}+1\) as the product of \(n\) positive integers greater than 1. To prove that it has \(n\) distinct prime factors, we only need to prove that these \(n\) positive integers greater than 1 are pairwise coprime.
Notice that, when \(m > l\), \(2^{2^{l}}+2^{2^{l-1}}+1\) and \(2^{2^{l}}-2^{2^{l-1}}+1\) are both factors of \(2^{2^{m}}+2^{2^{m-1}}+1\), so
$$\begin{aligned}
& \left(2^{2^{m}}-2^{2^{m-1}}+1,2^{2^{l}} \pm 2^{2^{l-1}}+1\right) \\
\leqslant & \left(2^{2^{m}}-2^{2^{m-1}}+1,2^{2^{m}}+2^{2^{m-1}}+1\right) \\
= & \left(2^{2^{m}}-2^{2^{m-1}}+1,2 \times 2^{2^{m-1}}\right)
\end{aligned}$$
Since \(2 \times 2^{2^{m-1}}\) has only one prime factor 2, and \(2^{2^{m}}-2^{2^{m-1}}+1\) is odd, we have
$$\left(2^{2^{m}}-2^{2^{m-1}}+1,2 \times 2^{2^{m-1}}\right)=1$$
Therefore,
$$\left(2^{2^{m}}-2^{2^{m-1}}+1,2^{2^{l}} \pm 2^{2^{l-1}}+1\right)=1$$
Thus, \(2^{2^{1}}+2^{2^{0}}+1, 2^{2^{1}}-2^{2^{0}}+1, 2^{2^{2}}-2^{2^{1}}+1, \cdots, 2^{2^{n-1}}-2^{2^{n-2}}+1\) are pairwise coprime, and hence \(2^{2^{n}}+2^{2^{n-1}}+1\) has at least \(n\) distinct prime factors.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Proof: Regardless of how many 3s are added between the two 0s in the number 12008, the resulting number is always a multiple of 19.
---
The translation maintains the original text's line breaks and format.
|
Let $a_{0}=12008, a_{n}=120 \underbrace{3 \cdots 3}_{n \text { 3's}} 08, n=1,2, \cdots$.
First, since
$$a_{0}=19 \times 632,$$
it follows that
$$19 \mid a_{0} .$$
Second, suppose $19 \mid a_{n}$, then from
$$\begin{array}{c}
a_{n+1}-10 a_{n}=228=19 \times 12 \\
19 \mid a_{n+1}
\end{array}$$
Therefore, for all integers $n$, the number $a_{n}$ is a multiple of 19.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Given a positive integer $n>1$, let $d_{1}, d_{2}, \cdots, d_{n}$ be positive integers satisfying: $\left(d_{1}, d_{2}, \cdots, d_{n}\right)=1$, and for $j=1,2, \cdots, n$ we have $d_{j} \mid \sum_{i=1}^{n} d_{i}\left(\right.$ here $\sum_{i=1}^{n} d_{i}=d_{1}+d_{2}+\cdots+ d_{n}$ ).
(1) Prove: $d_{1} d_{2} \cdots d_{n} \mid\left(\sum_{i=1}^{n} d_{i}\right)^{n-2}$;
(2) Provide an example to show that when $n>2$, the exponent on the right-hand side cannot be reduced.
|
(1) Let $p$ be a prime factor of $d_{1} d_{2} \cdots d_{n}$, and let $k$ be the maximum power of $p$ in the prime factorization of each $d_{i}$. Then, from $d_{j} \mid \sum_{i=1}^{n} d_{i}$, it follows that $p^{k} \mid \sum_{i=1}^{n} d_{i}$, hence $p^{k(n-2)} \mid\left(\sum_{i=1}^{n} d_{i}\right)^{n-2}$.
Since $\left(d_{1}, d_{2}, \cdots, d_{n}\right)=1$, there exists a $d_{i}$ such that $p \nmid d_{i}$. Combining this with $p \mid \sum_{i=1}^{n} d_{i}$, we know that at least two of the numbers $d_{1}$, $d_{2}, \cdots, d_{n}$ are not multiples of $p$. Therefore, the power of $p$ in $d_{1} d_{2} \cdots d_{n}$ does not exceed $k(n-2)$, which confirms the conclusion.
(2) Let $d_{1}=1, d_{2}=n-1, d_{i}=n, 3 \leqslant i \leqslant n$, then $\sum_{i=1}^{n} d_{i}=n(n-1)$ is a multiple of each $d_{i}$, and $\left(d_{i}, d_{2}, \cdots, d_{n}\right)=1$.
In this case, $d_{1} d_{2} \cdots d_{n}=n^{n-2}(n-1)$, and combining $(n, n-1)=1$, we know that the smallest positive integer $m$ satisfying $n^{n-2}(n-1) \mid (n(n-1))^{m}$ is $m=n-2$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
1 Let $n$ be a positive integer greater than 1. Prove: $n^{4}+4^{n}$ is a composite number.
|
1. When $n$ is even, $n^{4}+4^{n}$ is an even number greater than 2, hence it is a composite number. When $n$ is odd, let $n=2 k+1$, then
$$n^{4}+4^{n}=n^{4}+4 \times\left(2^{k}\right)^{4}$$
Using
$$\begin{aligned}
x^{4}+4 y^{4} & =\left(x^{2}+2 y^{2}\right)^{2}-4 x^{2} y^{2} \\
& =\left(x^{2}-2 x y+2 y^{2}\right)\left(x^{2}+2 x y+2 y^{2}\right)
\end{aligned}$$
we can conclude that $n^{4}+4 \times\left(2^{k}\right)^{4}$ is a composite number.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
3 Let $m$ be a positive integer greater than 1, and $m \mid (m-1)! + 1$. Prove: $m$ is a prime number.
|
3. If $m$ is a composite number, then there exists a positive integer $p$, such that $2 \leqslant p<m$, and $p \mid m$. In this case, $p \mid (m-1)!$, but $m \mid (m-1)!+1$, so $p \mid (m-1)!+1$, which leads to $p \mid 1$, a contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5 Let $p$ be a positive integer, and $2^{p}-1$ be a prime number. Prove: $p$ is a prime number.
|
5. If $p$ is a composite number, let $p=q r, 2 \leqslant q \leqslant r$, then
$$2^{p}-1=\left(2^{q}\right)^{r}-1=\left(2^{q}-1\right)\left(\left(2^{q}\right)^{r-1}+\left(2^{q}\right)^{r-2}+\cdots+1\right),$$
this implies $2^{q}-1 \mid 2^{p}-1$, which contradicts that $2^{p}-1$ is a prime number. Therefore, $p$ is a prime.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
6 Let $n$ be a positive integer, and $2^{n}+1$ be a prime number. Prove: there exists a non-negative integer $k$, such that $n=2^{k}$.
The text above is translated into English, preserving the original text's line breaks and format.
|
6. By the Fundamental Theorem of Arithmetic, we can write \( n = 2^k \cdot q, k \geqslant 0, q \) as an odd number. If \( q > 1 \), then
\[
\begin{aligned}
2^n + 1 & = \left(2^{2^k}\right)^q + 1 \\
& = (x + 1)\left(x^{q-1} - x^{q-2} + \cdots - x + 1\right)
\end{aligned}
\]
is the product of two positive integers greater than 1, and thus is not a prime number, where \( x = 2^{2^k} \). Therefore, if \( 2^n + 1 \) is a prime number, it follows that \( q = 1 \), which proves the proposition.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
8 Let $a, b, c, d$ be integers, and $a \neq c, a-c \mid ab+cd$. Prove: $a-c \mid ad+bc$.
The translation is provided while preserving the original text's line breaks and format.
|
8. Using
$$\begin{aligned}
& (a d+b c)-(a b+c d) \\
= & d(a-c)-b(a-c) \\
= & (d-b)(a-c),
\end{aligned}$$
and $a-c \mid a b+c d$, we can get $a-c \mid a d+b c$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
9 Let $a, b, c, d$ be integers, and $ac, bc + ad, bd$ are all multiples of some integer $u$. Prove: the numbers $bc$ and $ad$ are also multiples of $u$.
|
9. From the identity
$$(b c+a d)^{2}+(b c-a d)^{2}=4 a b c d=4(a c)(b d)$$
combined with the condition, we know that $u^{2} \mid (b c-a d)^{2}$, hence $u \mid b c-a d$.
Now, let $b c+a d=u x, b c-a d=u y$, then from (1) we have,
$$x^{2}+y^{2}=4\left(\frac{a c}{u}\right)\left(\frac{b d}{u}\right)$$
Thus, $x^{2}+y^{2}$ is even, and consequently, both $x+y$ and $x-y$ are even. Therefore, from
$$b c=\frac{x+y}{2} \cdot u, a d=\frac{x-y}{2} \cdot u$$
we can conclude that $b c$ and $a d$ are both multiples of $u$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Given that any 10 consecutive digits of a 1000-digit positive integer form a 10-digit number that is a multiple of $2^{10}$. Prove: the positive integer is a multiple of $2^{1000}$.
|
Proof Let the positive integer $x=a_{1} a_{2} \cdots a_{1000}$, where $a_{i}$ are decimal digits. From the condition, we know
$$2^{10}\left|\overline{a_{991} \cdots a_{1000}}, 2^{10}\right| \overline{a_{990} \cdots a_{999}},$$
Therefore,
$$2^{10} \mid \overline{a_{990} \cdots a_{999}} \times 10$$
Let $y=\overline{a_{991} \cdots a_{999}}$, then we have
$$\begin{array}{c}
2^{10} \mid a_{990} \times 10^{10}+10 y, \\
2^{10} \mid 10 y .
\end{array}$$
Combining $2^{10} \mid \overline{a_{991} \cdots a_{1000}}$, we know
$$2^{10} \mid 10 y+a_{1000},$$
Thus, it requires
$$\begin{array}{l}
2^{10} \mid a_{1000} \\
a_{1000}=0
\end{array}$$
Similarly, by working backwards, we can get
that is $\square$
$$\begin{array}{l}
a_{11}=\cdots=a_{1000}=0, \\
x=a_{1} \cdots a_{10} \times 10^{998} .
\end{array}$$
Combining the condition $2^{10} \mid \overline{a_{1} \cdots a_{10}}$, we can obtain
$$2^{1000} \mid x$$
|
2^{1000} \mid x
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
10 Let $a$, $b$, $n$ be given positive integers, and for any positive integer $k(\neq b)$, we have $b-k \mid a-k^{n}$. Prove: $a=b^{n}$.
|
10. Notice that for any positive integer $k(\neq b)$, we have $b-k \mid b^{n}-k^{n}$, combined with $b-k \mid a-k^{n}$, it follows that $b-k \mid a-b^{n}$. This indicates that $a-b^{n}$ is a multiple of every positive integer, hence $a-b^{n}=0$, yielding $a=b^{n}$.
|
a=b^{n}
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
14 Let the prime numbers in ascending order be $p_{1}, p_{2}, p_{3}, \cdots$. Prove: When $n \geqslant 2$, the number $p_{n}+p_{n+1}$ can be expressed as the product of 3 integers greater than 1 (which can be the same).
保留源文本的换行和格式，翻译结果如下：
14 Let the prime numbers in ascending order be $p_{1}, p_{2}, p_{3}, \cdots$. Prove: When $n \geqslant 2$, the number $p_{n}+p_{n+1}$ can be expressed as the product of 3 integers greater than 1 (which can be the same).
|
14. When $n \geqslant 2$, $p_{n}$ and $p_{n+1}$ are both odd numbers, so $q=\frac{p_{n}+p_{n+1}}{2}$ is a positive integer. Also, $p_{n}<q<p_{n+1}$, and $p_{n}$ and $p_{n+1}$ are two consecutive primes, hence $q$ must be a composite number. Therefore, $q$ can be written as the product of two positive integers greater than 1, which proves the proposition.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
15 Let $n$ be a positive integer greater than 1. Prove: $n$ is a composite number if and only if there exist positive integers $a, b, x, y$, such that $n=a+b, \frac{x}{a}+\frac{y}{b}=1$.
|
15. If there exist $a, b, x, y$, such that
$$n=a+b \text{, and } \frac{x}{a}+\frac{y}{b}=1 \text{.}$$
We denote $d=(a, b)$, if $d=1$, then
we know
so
$$\begin{array}{l}
\frac{x}{a}+\frac{y}{b}=1 \\
b x+a y=a b \\
a|b x, b| a y
\end{array}$$
Combining $(a, b)=1$, it follows that $a|x, b| y$, thus
$$a b=b x+a y \geqslant a b+b a=2 a b,$$
which is a contradiction. Therefore, $d>1$, in this case $n=a+b=d\left(\frac{a}{d}+\frac{b}{d}\right)$ is a composite number.
Conversely, suppose $n$ is a composite number, let $n=p q, 2 \leqslant p \leqslant q$, then let $(a, b, x, y)=$ $(p, p(q-1), 1,(p-1)(q-1))$, we have
$$n=a+b \text{, and } \frac{x}{a}+\frac{y}{b}=1 \text{.}$$
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
16 Prove: In the sequence $10001,100010001,1000100010001, \cdots$, every number is composite.
|
16. Notice that $10001=73 \times 137$ is a composite number, and from the second term onward, we have
$$\begin{aligned}
a_{n} & =1 \underbrace{100010001 \cdots 001}_{n \uparrow 0001} \\
& =10^{4 n}+10^{4(n-1)}+\cdots+10^{4}+1 \\
& =\frac{10^{4(n+1)}-1}{10^{4}-1} \\
& =\frac{\left(10^{2(n+1)}-1\right)\left(10^{2(n+1)}+1\right)}{10^{4}-1}
\end{aligned}$$
Since when $n \geqslant 2$, $10^{4}-1<10^{2(n+1)}-1<10^{2(n+1)}+1$, therefore, $a_{n}$ is a composite number.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Let $m$ be a positive integer greater than 2, prove: for any positive integer $n$, we have $2^{m}-1 \nmid$
$$2^{n}+1$$
|
Prove that if there exists a positive integer $n$, such that $2^{m}-1 \mid 2^{n}+1$, then take the smallest such $n$.
Since $m>2$, we know $n>1$, and further, we should have $2^{n}+1 \geqslant 2^{m}-1$, so $n \geqslant m$. If $n=m$, it would lead to $2^{m}-1 \mid 2$ (since $2=\left(2^{n}+1\right)-\left(2^{n}-1\right)$, and each term on the right is a multiple of $2^{n}-1$), which is a contradiction. Hence, $n>m$.
Now, let $2^{n}+1=\left(2^{m}-1\right) q$, where $q$ is a positive integer. Then,
$$2^{n}+2^{m}=\left(2^{n}+1\right)+\left(2^{m}-1\right)=\left(2^{m}-1\right)(q+1),$$
i.e., $\square$
$$2^{m}\left(2^{n-m}+1\right)=\left(2^{m}-1\right)(q+1)$$
Thus,
$$\left(2^{n-m}+1\right)+\left(2^{m}-1\right)\left(2^{n-m}+1\right)=\left(2^{m}-1\right)(q+1)$$
We get $2^{n-m}+1=\left(2^{m}-1\right)\left(q-2^{n-m}\right)$, so $2^{m}-1 \mid 2^{n-m}+1$, which contradicts the minimality of $n$. Therefore, the proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
20 Let $p$ be an odd prime, and positive integers $m, n$ satisfy $\frac{m}{n}=1+\frac{1}{2}+\cdots+\frac{1}{p-1}$. Prove: $p \mid m$.
|
20. From the condition, we know
$$\begin{aligned}
\frac{2 m}{n} & =\left(1+\frac{1}{2}+\cdots+\frac{1}{p-1}\right)+\left(\frac{1}{p-1}+\frac{1}{p-2}+\cdots+1\right) \\
& =\left(1+\frac{1}{p-1}\right)+\left(\frac{1}{2}+\frac{1}{p-2}\right)+\cdots+\left(\frac{1}{p-1}+1\right) \\
& =\frac{p}{1 \times(p-1)}+\frac{p}{2 \times(p-2)}+\cdots+\frac{p}{(p-1) \times 1}
\end{aligned}$$
After combining the terms on the right side, we know there exists a positive integer $M$ such that $\frac{2 m}{n}=\frac{p M}{(p-1)!}$, i.e., $p n M = 2 m(p-1)!$. Since $p$ is an odd prime, we know $p \nmid 2, p \nmid (p-1)!$, so, $p \mid m$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
21 Let $a$, $m$, $n$ be positive integers, $a>1$, and $a^{m}+1 \mid a^{n}+1$. Prove: $m \mid n$.
|
21. If $m \nmid n$, from $a^{m}+1 \mid a^{n}+1$ and $a>1$, we know that $m1$, $a^{m}+1>a^{r}+1$, which is a contradiction.
Therefore, $m \mid n$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
22 Prove: For any positive integer $n$ and positive odd integer $m$, we have $\left(2^{m}-1,2^{n}+1\right)=1$.
|
22. Let $d=\left(2^{m}-1,2^{n}+1\right)$, then
hence
that is
$$\begin{array}{c}
d \mid 2^{m}-1 \\
d \mid\left(2^{m}\right)^{n}-1^{n} \\
d \mid 2^{n}-1
\end{array}$$
Additionally, $d \mid 2^{n}+1$, and since $m$ is odd, we have
$$2^{n}+1 \mid\left(2^{n}\right)^{m}+1^{m},$$
thus
$$d \mid 2^{m}+1$$
Comparing the two derived equations, we know $d \mid 2$, and since $2^{m}-1$ is odd, it follows that $d=1$.
|
1
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
24 Given positive integers $a, b, c, d$ whose least common multiple is $a+b+c+d$. Prove: $abcd$ is a multiple of 3 or 5.
|
24. From the conditions, we know that $a, b, c, d$ are not all equal. Without loss of generality, assume $d$ is the largest number, then
$$d3$ cases.
When $z \geqslant 6$, we have
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leqslant \frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$$
Therefore, it can only be that $x=y=z=6$, in which case $a b c d$ is a multiple of 3.
Thus, we only need to consider the cases where $z=4$ or 5. When $z=5$, we have $5 \mid 2 d$, which means $a b c d$ is a multiple of 5. Therefore, we only need to consider the case where $z=4$, in which case
we have
$$\begin{array}{c}
\frac{1}{x}+\frac{1}{y}=\frac{1}{4} \\
x y-4 x-4 y=0 \\
(x-4)(y-4)=16
\end{array}$$
Given $x>y$, we have
$$(x-4, y-4)=(16,1),(8,2),(4,4)$$
which correspond to
$$\begin{array}{l}
2 d=20 a=5 b=4 c \\
2 d=12 a=6 b=4 c \\
2 d=8 a=8 b=4 c
\end{array}$$
In the first case, $5 \mid d$ is required; in the second case, $3 \mid d$ is required; in the third case,
$$a=b, c=2 a, d=4 a$$
In this case, the least common multiple of $a, b, c, d$ is $d$, not $a+b+c+d$, which is a contradiction.
In summary, $a b c d$ is a multiple of 3 or 5.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
26 Let $a, m, n$ be positive integers, $a>1$. Prove: $\left(a^{m}-1, a^{n}-1\right)=a^{(m, n)}-1$.
|
26. Suppose $m>n$, then
and
$$\begin{array}{l}
\left(a^{m}-1, a^{n}-1\right)=\left(a^{m}-a^{n}, a^{n}-1\right) \\
=\left(a^{n}\left(a^{m-n}-1\right), a^{n}-1\right), \\
\left(a^{n}, a^{n}-1\right)=1, \\
\left(a^{m}-1, a^{n}-1\right)=\left(a^{m-n}-1, a^{n}-1\right),
\end{array}$$
By repeatedly applying this, performing the "Euclidean algorithm" on the exponents, we can see that the conclusion holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
27 Let $a, n$ be positive integers, $a>1$, and $a^{n}+1$ is a prime. Prove: $d\left(a^{n}-1\right) \geqslant n$.
|
27. Given that $a^{n}+1$ is a prime number, we know that $a$ is even. Similar to problem 6, there exists a non-negative integer $k$ such that $n=2^{k}$. Therefore,
$$\begin{aligned}
& a^{n}-1 \\
= & a^{2^{k}}-1 \\
= & \left(a^{2^{k-1}}-1\right)\left(a^{2^{k-1}}+1\right) \\
= & \cdots \\
= & (a-1)(a+1)\left(a^{2}+1\right) \cdots\left(a^{2^{k-1}}+1\right)
\end{aligned}$$
Furthermore, $\left(a^{2^{k-1}}-1, a^{2^{k-1}}+1\right)=\left(a^{2^{k-1}}-1,2\right)=1$ (the last step uses the fact that $a$ is even). Tracing back, we know that $a+1, a^{2}+1, a^{2^{2}}+1, \cdots, a^{2^{k-1}}+1$ are pairwise coprime. Thus, any product of these numbers forms $2^{k}$ distinct numbers, and these $2^{k}$ numbers are all divisors of $a^{n}-1$. Therefore, $d\left(a^{n}-1\right) \geqslant 2^{k}=n$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
29 Let positive integers $a, b, m, n$ satisfy: $(a, b)=1, a>1$, and $a^{m}+b^{m} \mid a^{n}+b^{n}$. Prove: $m \mid n$.
|
29. Utilize
$$a^{n}+b^{n}=\left(a^{n-m}+b^{n-m}\right)\left(a^{m}+b^{m}\right)-\left(a^{m} b^{n-m}+a^{n-m} b^{m}\right),$$
we know that if $n \geqslant 2 m$, then
$$a^{n}+b^{n}=\left(a^{n-m}+b^{n-m}\right)\left(a^{m}+b^{m}\right)-a^{m} b^{m}\left(a^{n-2 m}+b^{n-2 m}\right),$$
thus
$$a^{m}+b^{m} \mid a^{m} b^{m}\left(a^{n-2 m}+b^{n-2 m}\right)$$
Given
$$(a, b)=1$$
we have
$$\left(a^{m}, b^{m}\right)=1$$
Furthermore,
$$\left(a^{m}+b^{m}, a^{m}\right)=\left(a^{m}+b^{m}, b^{m}\right)=1$$
Hence,
$$\left(a^{m}+b^{m}, a^{m} b^{m}\right)=1$$
Therefore,
$$a^{m}+b^{m} \mid a^{n-2 m}+b^{n-2 m}$$
Replace $n$ with $n-2 m$ and repeat the above discussion, eventually reducing $n$ to a positive integer less than $2 m$. At this point, by $a^{m}+b^{m} \mid a^{n}+b^{n}$ and $a>1$, we know $n \geqslant m$. If $n=m$, the proposition is already established; if $m1$ is impossible.
In summary, $m \mid n$ (Note: In fact, it has been shown that $n$ is an odd multiple of $m$).
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
30 Proof: There exist 2012 different positive integers, such that for any two different numbers $a$ and $b$, the condition $(a-b)^{2} \mid a b$ is satisfied.
|
30. Generalizing the proposition, it can be proven: For any $n(\geqslant 2)$, there exist $n$ distinct positive integers such that for any two different numbers $a, b$, we have $(a-b)^{2} \mid a b$. The proof is as follows:
When $n=2$, take $a_{1}=1, a_{2}=2$, then they satisfy the condition.
Now assume $a_{1}<a_{2}<\cdots<a_{n}$ are $n(\geqslant 2)$ positive integers that meet the requirement, i.e., for $1 \leqslant i< j \leqslant n$, we have $\left(a_{i}-a_{j}\right)^{2} \mid a_{i} a_{j}$.
Consider the following $n+1$ numbers
$$a_{n}!, a_{n}!+a_{1}, a_{n}!+a_{2}, \cdots, a_{n}!+a_{n}$$
It is easy to prove that these $n+1$ positive integers satisfy the requirement.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
31 Let $a, b$ be positive integers, and $(a, b)=1$. Prove: For any positive integer $m$, the sequence
$$a, a+b, a+2b, \cdots, a+nb, \cdots$$
contains infinitely many numbers that are coprime to $m$.
|
31. For any positive integer $m$, given that $(a, b)=1$, we can write $m=m_{1} m_{2}$, such that the prime factors of $m_{1}$ are all prime factors of $a$, and
$$\left(a, m_{2}\right)=1,\left(m_{1}, b\right)=1,\left(m_{1}, m_{2}\right)=1$$
(This can be achieved by appropriately distributing the prime factorizations of $m$, $a$, and $b$).
Take a positive integer $k$, such that $\left(k, m_{1}\right)=1$, there are infinitely many such $k$, let $n=m_{2} k$, we will prove: $\left(a+n b, m_{1}\right)=1$.
In fact, let $d=\left(a+n b, m_{1}\right)$, if $d>1$, take a prime factor $p$ of $d$, then $p \mid m_{1}$, and thus $p \mid a$, so, $p \mid n b$.
But since $\quad\left(m_{1}, k\right)=\left(m_{1}, m_{2}\right)=\left(m_{1}, b\right)=1$,
we know $p \nmid m_{2} k b$, i.e., $p \nmid n b$. This is a contradiction. Therefore,
$$\left(a+n b, m_{1}\right)=1$$
Also, $\quad\left(a+n b, m_{2}\right)=\left(a+m_{2} k b, m_{2}\right)=\left(a, m_{2}\right)=1$,
thus
$$\left(a+n b, m_{1} m_{2}\right)=1$$
i.e.,
$$(a+n b, m)=1$$
The proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
34 Prove: Every positive integer can be expressed as the difference of two positive integers, and these two positive integers have the same number of prime factors.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
34. Let $n$ be a positive integer. If $n$ is even, then the expression $n=(2 n)-n$ satisfies the requirement. If $n$ is odd, let $p$ be the smallest odd prime that does not divide $n$. Then the expression $n=p n-(p-1) n$ shows that the number of prime factors of $p n$ is one more than the number of prime factors of $n$; and $p-1$ is even, and by the definition of $p$, every odd prime factor of $p-1$ is a prime factor of $n$, so the number of prime factors of $(p-1) n$ is also one more than the number of prime factors of $n$. The proposition is thus proved.
| null |
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
36 Let $p(k)$ denote the greatest odd divisor of the positive integer $k$. Prove: for any positive integer $n$, we have $\frac{2}{3} n<$
$$\sum_{k=1}^{n} \frac{p(k)}{k}<\frac{2}{3}(n+1) \text {. }$$
|
36. Let $S_{n}=\sum_{k=1}^{n} \frac{p(k)}{k}$, then by the definition of $p(k)$ we have
$$\begin{aligned}
S_{2 n} & =\sum_{k=1}^{2 n} \frac{p(k)}{k}=\sum_{k=1}^{n} \frac{p(2 k-1)}{2 k-1}+\sum_{k=1}^{n} \frac{p(2 k)}{2 k} \\
& =n+\frac{1}{2} \sum_{k=1}^{n} \frac{p(2 k)}{k}=n+\frac{1}{2} S_{n}
\end{aligned}$$
Similarly, we have
$$S_{2 n+1}=n+1+\frac{1}{2} S_{n}$$
Returning to the original problem, when $n=1$, the proposition is clearly true. Now assume the proposition holds for $1 \leqslant n \leqslant m$, and consider the case $n=m+1$.
If $m+1$ is even, then by (1) and the induction hypothesis, we have
$$\begin{array}{l}
\quad \frac{m+1}{2}+\frac{1}{2} \cdot \frac{2\left(\frac{m+1}{2}\right)}{3}<\frac{m+1}{2}+\frac{1}{2} S_{\frac{m+1}{2}}=S_{m+1}<\frac{m+1}{2}+\frac{1}{2} \\
\frac{2\left(\frac{m+1}{2}+1\right)}{3}
\end{array}$$
Thus, $\frac{2}{3}(m+1)<S_{m+1}<\frac{2}{3}(m+2)$, so the proposition holds for $m+1$.
If $m+1$ is odd, similarly using (2) we can also show that the proposition holds for $m+1$.
Therefore, the conclusion is true.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Property 2 If for any prime $p$ between 1 and $\sqrt{n}$, $p \nmid n$, then $n$ is a prime. Here $n(>1)$ is a positive integer.
|
Proof: In fact, if $n$ is a composite number, then it can be written as $n=p q, 2 \leqslant p \leqslant q$. Therefore, $p^{2} \leqslant n$, which means $p \leqslant \sqrt{n}$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
40. Let $n$ be a positive integer. Prove: The number $1+5^{n}+5^{2 n}+5^{3 n}+5^{4 n}$ is a composite number.
|
When $n$ is even, let $n=2 m, x=5^{m}$, then
$$\begin{aligned}
A & =1+5^{n}+5^{2 n}+5^{3 n}+5^{4 n}=1+x^{2}+x^{4}+x^{6}+x^{8} \\
& =\frac{x^{10}-1}{x^{2}-1}=\frac{\left(x^{5}-1\right)\left(x^{5}+1\right)}{(x-1)(x+1)} \\
& =\left(x^{4}+x^{3}+x^{2}+x+1\right)\left(x^{4}-x^{3}+x^{2}-x+1\right)
\end{aligned}$$
Since $x=5^{m}>1$, it is known that both numbers in the right-hand side of the above equation are greater than 1, thus, $A$ is a composite number.
When $n$ is odd, let $n=2 m+1, y=5^{m}, z=5 y^{2}$, then
$$\begin{aligned}
A & =1+z+z^{2}+z^{3}+z^{4} \\
& =\left(1+3 z+z^{2}\right)^{2}-5 z^{3}-10 z^{2}-5 z \\
& =\left(1+3 z+z^{2}\right)^{2}-5 z(z+1)^{2} \\
& =\left(1+5 y^{2}+25 y^{4}\right)^{2}-25 y^{2}\left(1+5 y^{2}\right)^{2} \\
& =\left(1+5 y^{2}+25 y^{4}-5 y\left(1+5 y^{2}\right)\right)\left(1+5 y^{2}+25 y^{4}+5 y\left(1+5 y^{2}\right)\right)
\end{aligned}$$
When $m>0$, i.e., $y \geqslant 5$, both expressions on the right-hand side of the above equation are greater than 1, at this time, $A$ is a composite number. When $m=0$, $A=1+5+5^{2}+5^{3}+5^{4}=781=11 \times 71$ is also a composite number.
Therefore, for any positive integer $n, A$ is a composite number, the proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Property 2 If $a \equiv b(\bmod m), c \equiv d(\bmod m)$, then $a+c \equiv b+d(\bmod m)$, $a-c \equiv b-d(\bmod m), a c \equiv b d(\bmod m)$
|
To prove: these conclusions and equations are very similar, and they are all easy to prove. We only provide the proof for the 3rd equation.
We only need to prove: $m \mid a c-b d$.
Since
$$\begin{aligned}
a c-b d & =a c-b c+b c-b d \\
& =(a-b) c+b(c-d)
\end{aligned}$$
By the condition $m|a-b, m| c-d$, we know $m \mid a c-b d$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let $n$ be a positive integer greater than 1, and suppose that any two numbers from $1!, 2!, \cdots, n!$ yield different remainders when divided by $n$. Prove: $n$ is a prime number.
|
Notice that, $n! \equiv 0 \pmod{n}$, and when $n=4$, we have $2! \equiv 3! \pmod{4}$. Therefore, if we can prove that for any composite number $n$ greater than 4, $(n-1)! \equiv 0 \pmod{n}$, we can derive a contradiction from the conditions given in the problem. Thus, we can prove that $n$ is a prime number.
In fact, if $n$ is a composite number greater than 4, we can decompose $n$ into the following two cases.
Case one: We can write $n = pq, 2 \leqslant p < q$, and since $n > 4$, we know $p \geqslant 3$, so $1 < p < 2p < (n-1)$, thus $p \cdot (2p) \mid (n-1)!$, hence, $(n-1)! \equiv 0 \pmod{n}$.
In summary, $n$ can only be a prime number.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let integers $x, y, z$ satisfy
$$(x-y)(y-z)(z-x)=x+y+z .$$
Prove: $x+y+z$ is a multiple of 27.
|
Consider the remainders of $x$, $y$, and $z$ when divided by 3. If any two of $x$, $y$, and $z$ are not congruent modulo 3, then
$$x+y+z \equiv 0+1+2 \equiv 0(\bmod 3)$$
However, $3 \nmid (x-y)(y-z)(z-x)$, which contradicts (1).
Now, at least two of $x$, $y$, and $z$ must be congruent modulo 3. By symmetry, assume $x \equiv y(\bmod 3)$. In this case, by (1) we have
$$3 \mid x+y+z$$
Thus,
$$z \equiv -(x+y) \equiv -2x \equiv x(\bmod 3)$$
This shows that
$$x \equiv y \equiv z(\bmod 3)$$
Therefore, by (1) we have
$$27 \mid x+y+z$$
|
27 \mid x+y+z
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Property 3 There are infinitely many prime numbers.
|
Proof: If there are only a finite number of primes, let them be $p_{1}<p_{2}<\cdots<p_{n}$. Consider the number
$$x=p_{1} p_{2} \cdots p_{n}+1$$
Its smallest divisor greater than 1, $p$, is a prime, so $p$ should be one of $p_{1}, p_{2}, \cdots, p_{n}$. Let $p=p_{i}, 1 \leqslant i \leqslant n$, and $x=p_{i} y$, then $p_{1} p_{2} \cdots p_{n}+1=p_{i} y$, i.e., $p_{i}(y-$ $\left.p_{1} p_{2} \cdots p_{i-1} p_{i+1} \cdots p_{n}\right)=1$. This leads to $p_{i} \mid 1$. Contradiction.
Therefore, there are infinitely many primes.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
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