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(2) Euler's Theorem Let $m>1$ be an integer, $a$ any integer coprime to $m$, and $\varphi(m)$ the Euler's function (see Unit 6), then
$$a^{\phi(m)} \equiv 1(\bmod m)$$
|
Euler's theorem can be proved as follows: Take $r_{1}, r_{2}, \cdots, r_{\phi(m)}$ as a reduced residue system modulo $m$. Since $(a, m)=1$, it follows that $a r_{1}, a r_{2}, \cdots, a r_{\phi(m)}$ is also a reduced residue system modulo $m$ (see Unit 6). Because two complete (reduced) residue systems modulo $m$ are permutations of each other modulo $m$, we have in particular
$$r_{1} \cdots r_{\phi(m)} \equiv a r_{1} \cdot a r_{2} \cdots \cdot a r_{\phi(m)}=a^{\phi(m)} r_{1} r_{2} \cdots r_{\phi(m)}(\bmod m)$$
Since $\left(r_{i}, m\right)=1$, it follows that $\left(r_{1} r_{2} \cdots r_{\phi(m)}, m\right)=1$, so the term $r_{1} \cdots r_{\phi(m)}$ can be canceled from both sides of the above equation, yielding $a^{\phi(m)} \equiv 1(\bmod m)$.
|
a^{\phi(m)} \equiv 1(\bmod m)
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Let $p$ be a given prime. Prove: The sequence $\left\{2^{n}-n\right\}(n \geqslant 1)$ has infinitely many terms divisible by $p$.
|
Prove that the conclusion is obviously true when $p=2$. Let $p>2$, then by Fermat's Little Theorem, we have $2^{p-1} \equiv$ $1(\bmod p)$, hence for any positive integer $m$,
$$2^{m(p-1)} \equiv 1(\bmod p)$$
We take $m \equiv-1(\bmod p)$, then by (1), we get
$$2^{m(p-1)}-m(p-1) \equiv 1+m \equiv 0(\bmod p)$$
Therefore, if $n=(k p-1)(p-1)$, then $2^{n}-n$ is divisible by $p$ (where $k$ is any positive integer), so there are infinitely many terms in the sequence that are divisible by $p$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
1 Let $n$ and $k$ be positive integers, then there are exactly $\left[\frac{n}{k}\right]$ numbers in $1,2, \cdots, n$ that are divisible by $k$.
|
1. Among $1,2, \cdots, n$, the numbers divisible by $k$ are $k, 2 k, \cdots, d k$, where the positive integer $d$ satisfies $d k \leqslant n$ but $(d+1) k>n$, thus $\frac{n}{k}-1<d \leqslant \frac{n}{k}$, i.e., $d=\left[\frac{n}{k}\right]$, hence there are $\left[\frac{n}{k}\right]$ numbers divisible by $k$.
|
\left[\frac{n}{k}\right]
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Proof: The sequence $1,31,331,3331, \cdots$ contains infinitely many composite numbers.
|
Prove that since 31 is a prime, by Fermat's Little Theorem, we know that $10^{30} \equiv 1(\bmod 31)$, hence for any positive integer $k$, we have $10^{30 k} \equiv 1(\bmod 31)$, thus
$$\frac{.1}{3}\left(10^{30 k}-1\right) \equiv 0(\bmod 31)$$
This indicates that the number composed of $30 k$ threes is divisible by 31. Multiplying this number by 100 and then adding 31, the result is also divisible by 31, meaning the $(30 k + 2)$-th term in the sequence is divisible by 31, hence it is not a prime. Therefore, the sequence contains infinitely many composite numbers.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Proof: For any given positive integer $n$, there exist $n$ consecutive positive integers, each of which has a square factor greater than 1.
|
Proof: Since there are infinitely many primes, we can take $n$ distinct primes $p_{1}$, $p_{2}$, ..., $p_{n}$, and consider the system of congruences
$$x \equiv -i \left(\bmod p_{i}^{2}\right), i=1,2, \cdots, n$$
Since $p_{1}^{2}, p_{2}^{2}, \cdots, p_{n}^{2}$ are clearly pairwise coprime, by the Chinese Remainder Theorem, the above system of congruences has a positive integer solution. Thus, the $n$ consecutive numbers $x+1, x+2, \cdots, x+n$ are divisible by the squares $p_{1}^{2}$, $p_{2}^{2}$, ..., $p_{n}^{2}$, respectively.
If we do not use primes directly, we can also adopt the following variant method. Since Fermat numbers $F_{k}=2^{2^{k}}+1(k \geqslant 0)$ are pairwise coprime (Example 3 in Unit 2), replacing $p_{i}^{2}$ in (1) with $F_{i}^{2}(i=1,2, \cdots, n)$, the corresponding system of congruences also has a solution, leading to the same proof.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 (1) Prove: For any positive integer $n$, there exist $n$ consecutive positive integers, none of which is a power number; $\square$
(2) Prove that there exist infinitely many distinct positive integers, such that neither they nor the sum of any distinct subset of them is a power number.
(Power number is defined as in Example 9 of Unit 5.)
|
Proof (1) We prove that there exist $n$ consecutive positive integers, each of which has at least one prime factor that appears exactly once in its prime factorization, making the number not a power.
Since there are infinitely many primes, we can take $n$ distinct primes $p_{1}, \cdots, p_{n}$. Consider the system of congruences
$$x \equiv -i + p_{i} \left(\bmod p_{i}^{2}\right), \quad i=1,2, \cdots, n$$
Since $p_{1}^{2}, p_{2}^{2}, \cdots, p_{n}^{2}$ are pairwise coprime, by the Chinese Remainder Theorem, the above system of congruences has a positive integer solution $x$. For $1 \leqslant i \leqslant n$, since $x + i \equiv p_{i} \left(\bmod p_{i}^{2}\right)$, it follows that $p_{i} \mid (x + i)$; but by (1), $p_{i}^{2} \nmid (x + i)$, meaning $p_{i}$ appears exactly once in the prime factorization of $x + i$, hence $x + 1, x + 2, \cdots, x + n$ are not powers.
(2) We construct inductively an infinite sequence of positive integers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$, all of which are not powers, such that for each $n$, the sum of any subset of $a_{1}, \cdots, a_{n}$ is not a power, thereby proving the conclusion in (2).
First, $a_{1}$ can be any non-power number, for example, take $a_{1} = 2$. Suppose $a_{1}, \cdots, a_{n}$ have been determined, we prove that we can choose $a_{n+1}$ to be a non-power number, $a_{n+1} > a_{n}$, and $a_{n+1}$ plus any subset of $a_{1}, \cdots, a_{n}$ is not a power.
Let $s_{1}, \cdots, s_{m}$ be all the distinct sums of any subset of $a_{1}, \cdots, a_{n}$, where $m = 2^{n} - 1$. Since there are infinitely many primes, we can take $m + 1$ distinct primes $p, p_{1}, \cdots, p_{m}$, and consider the system of congruences
$$x \equiv p \left(\bmod p^{2}\right), \quad x \equiv -s_{i} + p_{i} \left(\bmod p_{i}^{2}\right), \quad i=1, \cdots, m$$
Since $p^{2}, p_{1}^{2}, \cdots, p_{m}^{2}$ are pairwise coprime, the system of congruences (2) has infinitely many positive integer solutions $x$. Take any solution greater than $a_{n}$, and denote it as $a_{n+1}$. Then $a_{n+1} \equiv p \left(\bmod p^{2}\right)$ implies that $a_{n+1}$ is divisible by $p$ but not by $p^{2}$, hence $a_{n+1}$ is not a power. Also, $a_{n+1} \equiv -s_{i} + p_{i} \left(\bmod p_{i}^{2}\right)$ implies that $a_{n+1} + s_{i}$ is divisible by $p_{i}$ but not by $p_{i}^{2}$, thus for each $i = 1, \cdots, m$, the number $a_{n+1} + s_{i}$ is not a power. This recursively constructs an infinite sequence $a_{1}, a_{2}, \cdots$ that meets the aforementioned requirements. Q.E.D.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Given a positive integer $n$, let $f(n)$ be the smallest positive integer such that $\sum_{k=1}^{f(n)} k$ is divisible by $n$. Prove that $f(n)=2n-1$ if and only if $n$ is a power of 2.
|
The first half of the problem is quite easy to prove. If $n=2^{m}$, then on the one hand,
$$\sum_{k=1}^{2 n-1} k=\frac{(2 n-1) \times 2 n}{2}=\left(2^{m+1}-1\right) \cdot 2^{m}$$
is divisible by $2^{m}=n$. On the other hand, if $r \leqslant 2 n-2$, then
$$\sum_{k=1}^{r} k=\frac{r(r+1)}{2}$$
is not divisible by $2^{m}$, because one of $r$ and $r+1$ is odd, and the other does not exceed $(2 n-2)+1=2^{m+1}-1$, and thus is not divisible by $2^{m+1}$. Combining these two aspects, we know that $f\left(2^{m}\right)=2^{m+1}-1$.
Now suppose $n$ is not a power of 2, i.e., $n=2^{m} a$, where $m \geqslant 0, a>1$ is an odd number. We will prove that there exists a positive integer $r<2 n-1$ such that $2^{m+1} \mid r$ and $a \mid (r+1)$, so that
$$\sum_{k=1}^{r} k=\frac{r(r+1)}{2}$$
is divisible by $2^{m} a=n$, and thus $f(n)<2 n-1$.
To prove the above claim, we consider
$$x \equiv 0\left(\bmod 2^{m+1}\right), x \equiv -1(\bmod a)$$
Since $\left(2^{m+1}, a\right)=1$, by the Chinese Remainder Theorem, the system of congruences (1) must have a solution $x_{0}$, and all solutions are $x \equiv x_{0}\left(\bmod 2^{m+1} a\right)$, i.e., $x \equiv x_{0}(\bmod 2 n)$. Therefore, we can determine an $r$ that satisfies (1) and $0<r \leqslant 2 n$. Furthermore, by the second congruence in (1), we know that $r \neq 2 n$. And by the first congruence, we see that $r \neq 2 n-1$, so in fact $r<2 n-1$. This proves the existence of an $r$ that meets the requirements.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 Let $f(x)$ be an integer-coefficient polynomial, and $a_{1}, \cdots, a_{m}$ be given non-zero integers, with the following property: for any integer $n$, the number $f(n)$ is divisible by one of the integers $a_{1}, \cdots, a_{m}$. Prove: there exists an $a_{i}(1 \leqslant i \leqslant m)$, such that for all integers $n, f(n)$ is divisible by $a_{i}$.
|
If one of $a_{1}, \cdots, a_{m}$ is $\pm 1$, then the conclusion is obviously true. Hereafter, assume each $a_{i} \neq \pm 1$. Suppose the conclusion is not true, then for each $a_{i}$, there is an integer $x_{i}$ such that $a_{i} \nmid f\left(x_{i}\right)$. $i=1, \cdots, m$. We will construct an integer $n$ such that all $a_{i}$ do not divide $f(n)$, which will contradict the given condition.
For $i=1, \cdots, m$, since $a_{i} \nmid f\left(x_{i}\right)$, there is a prime power $p_{i}^{a_{i}}$ such that $p_{i}^{a_{i}} \mid a_{i}$, but $p_{i}^{a_{i}} \nmid f\left(x_{i}\right)$. If $p_{1}^{q_{1}}, \cdots, p_{n m^{\prime}}^{a_{1}}$ contain the same prime power, we retain only the one with the lowest exponent and remove the higher (and equal) powers. After this process, assume the remaining prime powers are $p_{1}^{p_{1}}, \cdots, p_{t}^{p_{t}}(1 \leqslant t \leqslant m)$, which are pairwise coprime. By the Chinese Remainder Theorem, the system of congruences
$$n \equiv x_{i}\left(\bmod p_{i}^{a_{i}}\right), i=1, \cdots, t$$
has an integer solution $n$.
Since $f(x)$ is a polynomial with integer coefficients, it follows from (1) that
$$f(n) \equiv f\left(x_{i}\right)\left(\bmod p_{i}^{a_{i}}\right), i=1, \cdots, t .$$
For $i=1, \cdots, t$, since $p_{i}^{a_{i}} \nmid f\left(x_{i}\right)$, it follows from the above that $p_{i}^{a_{i}} \nmid f(n)$, and thus $a_{i} \nmid f(n)$. For $j=t+1, \cdots, m$, by the previous process and assumption, each $p_{j}$ equals some $p_{i}(1 \leqslant i \leqslant t)$, and $p_{i}^{a_{i}} \mid p_{j}^{a_{j}}$. Therefore, from $p_{i}^{a_{i}} \nmid f(n)$, we deduce $p_{j}^{a_{j}} \nmid f(n)$, and hence $a_{j} \nmid f(n)$. Thus, $f(n)$ is not divisible by any of $a_{1}, \cdots, a_{m}$, which contradicts the given condition. Therefore, the conclusion of the problem is true. Q.E.D.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
1 Let $p$ be an odd prime, $n=\frac{2^{2 p}-1}{3}$. Prove: $2^{n-1} \equiv 1(\bmod n)$.
|
1. From the condition, we have
$$3(n-1)=4\left(2^{n-1}+1\right)\left(2^{n-1}-1\right) .$$
Given a prime $p>3$, by Fermat's Little Theorem, we have $p \mid\left(2^{p-1}-1\right)$. Combining this with (1), we deduce $2 p \mid(n-1)$, thus $\left(2^{2 p}-1\right) \mid\left(2^{n-1}-1\right)$. From the condition, we know $n \mid\left(2^{2 p}-1\right)$, so $n \mid\left(2^{n-1}-1\right)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2 Let $m \geqslant 2, a_{1}, a_{2}, \cdots, a_{m}$ be given positive integers. Prove: there are infinitely many positive integers $n$, such that the number $a_{1} \cdot 1^{n}+a_{2} \cdot 2^{n}+\cdots+a_{m} \cdot m^{n}$ is composite.
|
2. Clearly $a_{1}+2 a_{2}+\cdots+m a_{m}>1$, so there exists a prime $p$ that divides $a_{1}+2 a_{2}+\cdots+m a_{m}$. Take $n=k(p-1)+1, k=1,2, \cdots$, then for $1 \leqslant i \leqslant m$, if $p \nmid i$, by Fermat's Little Theorem we have
$$i^{n}=i \cdot\left(i^{k}\right)^{p-1} \equiv i(\bmod p)$$
If $p \mid i$, the above equation is obviously also true. Therefore,
$$a_{1} \cdot 1^{n}+a_{2} \cdot 2^{n}+\cdots+a_{m} \cdot m^{n} \equiv a_{1}+2 a_{2}+\cdots+m a_{m} \equiv 0(\bmod p)$$
Moreover, $a_{1} \cdot 1^{n}+a_{2} \cdot 2^{n}+\cdots+a_{m} \cdot m^{n}$ is clearly greater than $p$, so it is a composite number. Therefore, the $n$ chosen above satisfies the requirement, and there are clearly infinitely many such $n$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
3 Let $m, n$ be positive integers, and $m \geqslant n$, with the property: the equation
$$(11 k-1, m)=(11 k-1, n)$$
holds for all positive integers $k$. Prove: $m=11^{r} n, r$ is a non-negative integer.
|
3. Let $m=11^{i} u, n=11^{j} v$, where $i, j$ are non-negative integers, and $u, v$ are positive integers not divisible by 11. We prove that there must be $u=v$, from which it follows that $m=11^{i-j} n$. If $u \neq v$, without loss of generality, assume $u>v$. Since $(u, 11)=1$, by the Chinese Remainder Theorem, there exists a positive integer $x$ such that
$$x \equiv 0(\bmod u), x \equiv-1(\bmod 11)$$
i.e., $x=11 k-1$ (where $k$ is some positive integer). From (1), it is easy to see that $(11 k-1, m)=\left(x, 11^{i} u\right)=u$, but $(11 k-1, n)=\left(x, 11^{j} v\right) \leqslant v<u$, which contradicts the condition $(11 k-1, m)=(11 k-1, n)$. Therefore, it must be that $u=v$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
The smallest positive integer $r$ that satisfies $a^{r} \equiv 1(\bmod n)$ is called the order of $a$ modulo $n$. From the above argument, we know that $1 \leqslant r \leqslant n-1$. The following (1) indicates that the order of $a$ modulo $n$ has a very sharp property:
(1) Suppose $(a, n)=1$, and the order of $a$ modulo $n$ is $r$. If a positive integer $N$ makes $a^{N} \equiv 1(\bmod n)$, then $r \mid N$.
This is because, let $N=r q+k(0 \leqslant k<r)$, then
$$1 \equiv a^{N} \equiv\left(a^{r}\right)^{q} \cdot a^{k} \equiv a^{k}(\bmod n)$$
Since $0 \leqslant k<r$, it follows from the above equation and the definition of $r$ that $k$ must be $0$, hence $r \mid N$.
Property (1) combined with Euler's theorem (from (2) in Unit 7) can lead to
(2) Suppose $(a, n)=1$, then the order $r$ of $a$ modulo $n$ divides $\varphi(n)$. In particular, if $n$ is a prime $p$, then the order of $a$ modulo $p$ divides $p-1$.
|
None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". Here is the formatted output as requested:
None
|
not found
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Let $n>1, n \mid\left(2^{n}+1\right)$, prove $3 \mid n$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Prove that $n$ is obviously odd. Let $p$ be the smallest prime divisor of $n$. We will prove that $p=3$, hence $3 \mid n$. Let the order of 2 modulo $p$ be $r$. From $2^{n} \equiv -1 \pmod{n}$, we know
$$2^{2n} \equiv 1 \pmod{p}$$
Since $p \geqslant 3$, Fermat's Little Theorem gives
$$2^{p-1} \equiv 1 \pmod{p}$$
From (1) and (2), and the properties of the order, we deduce that $r \mid 2n$ and $r \mid (p-1)$, hence $r \mid (2n, p-1)$. It is not difficult to prove that $(2n, p-1) = 2$. This is because, by $2 \nmid n$, we have $2 \mid (2n, p-1)$, but $2^2 \nmid (2n, p-1)$; on the other hand, if there is an odd prime $q \mid (2n, p-1)$, then $q \mid (p-1)$, and $q \mid n$, but the former indicates $q < p$, which contradicts the fact that $p$ is the smallest prime divisor of $n$. Therefore, $(2n, p-1) = 2$, hence $r = 2$, so $p = 3$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4. Let $a, b$ be two coprime positive integers. Prove: There must exist positive integers $m, n$, such that $a b \mid a^{m}+b^{n}-1$.
|
4. Hint: Take $m=\varphi(b), n=\varphi(a)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5. Let $m, n \in \mathbf{N}_{+}$, and for any $k \in \mathbf{N}_{+}$, we have $(17 k-1, m)=(17 k-1, n)$. Prove: There exists $l \in \mathbf{Z}$, such that $m=17^{l} \cdot n$.
|
5. Proof: Let $m=17^{i} p, n=17^{j} q$, where $i, j \in \mathbf{N}, p, q \in \mathbf{N}_{+},(p, 17)=(q, 17)=1$.
Below we prove: $p=q$.
Assume $p>q$ (the case $p<q$ can be discussed similarly), since $(p, 17)=1$, by the Chinese Remainder Theorem, there exists $a \in \mathbf{N}_{+}$, such that $a \equiv 0(\bmod p), a \equiv-1(\bmod 17)$.
Thus, $a=17 k-1\left(k \in \mathbf{N}_{+}\right)$ and $(17 k-1, m)=\left(a, 17^{i} p\right)=p$.
On the other hand, since $(a, 17)=1$, we have $(17 k-1, n)=\left(17 k-1,17^{j} q\right) \leqslant q<p$. This contradicts $(17 k-1, m)=(17 k-1, n)$! Therefore, $p=q$.
Hence, there exists some $l \in \mathbf{Z}$, such that $m=17^{l} \cdot n$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
1. (Euler's Theorem) If $(a, m)=1$, then: $a^{\phi(m)} \equiv 1(\bmod m)$.
|
Proof: Let $r_{1}, r_{2}, r_{3}, \cdots, r_{\alpha(m)}$ be a reduced residue system modulo $m$. Since $(a, m)=1$, it follows that $a r_{1}, a r_{2}, \cdots, a r_{\phi(m)}$ is also a reduced residue system modulo $m$.
Thus, $a r_{1} \cdot a r_{2} \cdots a r_{q(m)} \equiv r_{1} r_{2} \cdots r_{\rho^{(m)}}(\bmod m)$,
which means $m \mid\left(a^{q(m)}-1\right) r_{1} r_{2} \cdots r_{q(m)}$,
and since $\left(m, r_{1} r_{2} \cdots r_{\varphi^{(m)}}\right)=1$, it follows that $m \mid a^{q(m)}-1$.
Therefore, $a^{\varphi(m)} \equiv 1(\bmod m)$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
From the definition of the order of $a$ modulo $m$ (denoted as $k$), it is easy to see that $k$ has the following properties:
(1) Suppose $(a, m)=1, k$ is the order of $a$ modulo $m$, and $u, v$ are any integers, then $a^{u} \equiv a^{v}(\bmod m)$ if and only if $u \equiv v(\bmod k)$. In particular, $a^{u} \equiv 1(\bmod k)$ if and only if $k \mid u$.
Simple proof: The sufficiency is obvious.
Necessity: Suppose $u>v$, let $l=u-v$, then from $a^{u} \equiv a^{v}(\bmod m)$ it is easy to see that $a^{l} \equiv 1(\bmod m)$. Let $l=k q+r$, where $0 \leqslant r<k$, so $a^{k_{q}} \cdot a^{r} \equiv 1(\bmod m)$, hence $a^{r} \equiv 1(\bmod m)$.
By $0 \leqslant r<k$ and the definition of $k$, it must be that $r=0$.
Therefore, $u \equiv v(\bmod m)$.
(2) Suppose $(a, m)=1$, and the order of $a$ modulo $m$ is $k$, then the sequence $a, a^{2}, \cdots, a^{k}, a^{k+1}, \cdots$ is a periodic sequence modulo $m$, with the smallest positive period being $k$, and the $k$ numbers $a, a^{2}, \cdots, a^{k}$ are pairwise incongruent modulo $m$.
(3) Suppose $(a, m)=1$, then the order of $a$ modulo $m$ divides the Euler's totient function $\varphi(m)$.
|
None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". Here is the formatted output as requested:
None
|
not found
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
3. (Fermat's Theorem) If $p$ is a prime, then $a^{p} \equiv a(\bmod p)$.
|
Proof: If $(a, p) \neq 1$, then $p \mid a$, in which case the conclusion is obviously true. If $(a, p)=1$, then by Euler's theorem $a^{\varphi(p)} \equiv 1(\bmod p)$, and since $\varphi(p)=p-1$, we have $a^{p-1} \equiv 1(\bmod p)$, so $a^{p} \equiv a(\bmod p)$.
Theorem proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4. (Wilson's Theorem) If $p$ is a prime, then $(p-1)!\equiv-1(\bmod p)$.
|
Proof: For any integer $a, 1 \leqslant a \leqslant p-1$, by Bézout's theorem, there exists an integer $a^{\prime}$ such that $a a^{\prime} \equiv 1$ $(\bmod p)$. Without loss of generality, let $1 \leqslant a^{\prime} \leqslant p-1$, then $a^{\prime}$ is called the modular inverse of $a$.
If there is an integer $b$ such that $b a^{\prime} \equiv 1(\bmod p)$, then $b a^{\prime} a \equiv a(\bmod p)$,
so $b \equiv a(\bmod p)$.
This shows that for different integers $a, 1 \leqslant a \leqslant p-1$, there correspond different modular inverses $a^{\prime}$.
Furthermore, if the modular inverse of an integer $a$ is itself, then $a^{2} \equiv 1(\bmod p)$.
Thus, $a \equiv 1(\bmod p)$ or $a \equiv-1(\bmod p)$.
Therefore, when $p>2$, $2,3,4, \cdots,(p-2)$ can be paired into $\frac{p-3}{2}$ pairs of mutual modular inverses, so their product $(p-2)!\equiv 1^{\frac{p-3}{2}} \quad(\bmod p) \equiv 1 \quad(\bmod p)$,
Thus, $(p-1)!\equiv p-1 \quad(\bmod p) \equiv-1 \quad(\bmod p)$.
When $p=2$, $(p-1)!\equiv-1(\bmod p)$.
Hence, the proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Proof of a lemma: For a prime $p>3$ and $p \equiv 2(\bmod 3)$, if $x^{3} \equiv y^{3}(\bmod p)$, then $x \equiv y(\bmod p)$.
|
Lemma proof: (1) If one of $x, y$ is a multiple of $p$, then it is easy to see that the lemma holds in this case. (2) If $p \nmid x, p \nmid y$, then since $p$ is a prime, we have $(x, p)=(y, p)=1$.
By Fermat's Little Theorem:
$$x^{p-1} \equiv y^{p-1} \equiv 1(\bmod p)$$
Also, $x^{3} \equiv y^{3}(\bmod p), p \equiv 2(\bmod 3)$,
so $x^{p-2} \equiv y^{p-2}(\bmod p)$.
Since $p \mid x^{p-1}-y^{p-2} \cdot y$, it follows that $p \mid x^{p-1}-x^{p-2} \cdot y$.
Thus, $p \mid x^{p-2}(x-y)$, and since $\left(p, x^{p-2}\right)=1$,
we have $p \mid x-y$. Therefore, $x \equiv y(\bmod p)$.
Hence, the lemma is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 9 (2006 National Training Team Problem) Prove: For any positive integers $m, n$, there always exists a positive integer $k$, such that $2^{k} - m$ has at least $n$ distinct prime factors.
|
Prove that for a fixed $m$, assuming $m$ is odd. We will prove that for any positive integer $n$, there exists $k_{n}$ such that $2^{k_{n}}-m$ has at least $n$ distinct prime factors.
(1)
We will prove (1) using mathematical induction:
(1) When $n=1$, $2^{3m}-m$ clearly has at least one prime factor;
(2) Assume $2^{k_{n}}-m$ has at least $n$ distinct prime factors, let $A_{n}=2^{k_{m}}-m$, then $\left(A_{n}, 2\right)=1$, and $2^{k_{n}+\varphi\left(A_{n}^{2}\right)}-m \equiv 2^{k_{n}}-m \equiv A_{n}\left(\bmod A_{n}^{2}\right)$, so $A_{n} \mid 2^{k_{n}+\varphi\left(A_{n}^{2}\right)}-m$,
Take a prime $p \left\lvert\, \frac{2^{k_{n}+\varphi\left(A_{n}^{2}\right)}-m}{A_{n}}\right.$, by $\frac{2^{k_{n}}+\epsilon\left(A_{n}^{2}\right)-m}{A_{n}} \equiv 1\left(\bmod A_{n}\right)$ we know $p \nmid A_{n}$.
Therefore, $2^{p\left(A_{n}^{2}\right)+k_{n}}-m$ has at least $n+1$ distinct prime factors.
By mathematical induction, the conclusion holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
7. Let $p, q$ be two odd primes, $p>q$. Prove: For any positive integer $m, p q \nmid m^{p-q}+1$.
|
7. Proof: (by contradiction) Suppose there exists $m \in \mathbf{N}_{+}$, such that $p q \mid m^{p-q}+1$, then $(m, p)=(m, q)=1$. From the condition, we have: $p\left|m^{p-q}+1 \Rightarrow p\right| m^{p-1}+m^{q-1}$.
By Fermat's Little Theorem, $m^{p-1} \equiv 1(\bmod p)$, so $p \mid m^{q-1}+1$.
Let $q-1=2^{k} \cdot \alpha\left(k \in \mathbf{N}_{+},(2, \alpha)=1\right)$,
then $p\left|\left(m^{\alpha}\right)^{2^{k}}+1 \Rightarrow p\right|\left(m^{\alpha}\right)^{2^{k+1}}-1$. Let $l$ be the order of $m$ modulo $p$, then $l \mid 2^{k+1}$. If $l \neq 2^{k+1}$, then $l \mid 2^{k}$,
so $p \mid m^{\alpha}+1$, a contradiction! Therefore, $l=2^{k+1}$.
Also, $p \mid\left(m^{\alpha}\right)^{p-1}-1$ (by Fermat's Little Theorem),
so $2^{k+1} \mid p-1$. Let $p-1=2^{t} \cdot \beta$, then $t>k$.
Also, $q\left|m^{p-1}+m^{q-1} \Rightarrow q\right| m^{p-1}+1$,
similarly, $2^{t+1} \mid q-1 \Rightarrow k>t$ which contradicts $t>k$!
Thus, the original proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2. Let $a, b$ be positive integers, $a^{2}+a b+1$ is divisible by $b^{2}+a b+1$, prove: $a=b$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
2. Proof: According to the problem, \(a^{2}+a b+1\) is divisible by \(b^{2}+a b+1\), then \(\left(a^{2}+a b+1, b^{2}+a b+1\right)=b^{2}+a b+1\),
Also, since \(b\left(a^{2}+a b+1\right)-a\left(b^{2}+a b+1\right)=b-a\), if \(a \neq b\), by Bézout's theorem, \(\left(b^{2}+a b+1\right) \mid (b-a)\),
i.e., \(b^{2}+a b+1 \leqslant b-a \leqslant b\),
which is a contradiction!
Thus, \(a \neq b\) does not hold, i.e., \(a=b\).
| null |
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
12. Given that $p$ is an odd prime greater than 3, and $a>b>1$. Prove: $C_{a}^{b p} \equiv C_{a}^{b}\left(\bmod p^{3}\right)$.
|
12. Proof: Let $f(x)=(x p-1)(x p-2) \cdots(x p-p+1)\left(x \in \mathbf{N}_{+}\right)$,
Then $C_{a p}^{b p}=\frac{a p \cdot(a p-p) \cdots(a p-b p+p)}{b p(b p-p) \cdots 2 p \cdot p} \cdot \frac{f(a) f(a-1) \cdots f(a-b+1)}{f(b) f(b-1) \cdots f(1)}$
$$=C_{a}^{b} \frac{f(a) f(a-1) \cdots f(a-b+1)}{f(b) f(b-1) \cdots f(2) f(1)}$$
Therefore, it suffices to prove: $f(a) f(a-1) \cdots f(a-b+1)=f(b) f(b-1) \cdots f(2) f(1)\left(\bmod p^{3}\right)$
We will prove the following lemma: $f(x) \equiv(p-1)!\left(\bmod p^{3}\right) . \forall x \in \mathbf{Z}$.
In fact, let $\delta_{0}=1$, and for $k \geqslant 1$, $\delta_{k}=\sum_{1 \leqslant i_{1}<i_{2}<\cdots<i_{k} \leqslant p-1} i_{1} \cdot i_{2} \cdots i_{k}$.
By Vieta's formulas:
$$f(x p)=\sum_{k=0}^{p-1}(x p)^{k}(-1)^{p-1-k} \delta_{p-1-k} \equiv \delta_{p-1}-\delta_{p-2} x p+x^{2} p^{2} \cdot \delta_{p-3}\left(\bmod p^{3}\right)$$
We will prove: $p\left|\delta_{p-3}, p^{2}\right| \delta_{p-2}$.
If $s=t r(\bmod m)$, and $(m, t)=1$, then we denote $\frac{s}{t} \equiv r(\bmod m)$.
Let $\frac{s_{1}}{t_{1}} \equiv r_{1}(\bmod m), \frac{s_{2}}{t_{2}} \equiv r_{2}(\bmod m)$.
It is not difficult to prove: $\frac{s_{1}}{t_{1}}+\frac{s_{2}}{t_{2}} \equiv r_{1}+r_{2}(\bmod m)$.
And if $s_{1} \equiv s_{2}(\bmod m), t_{1} \equiv t_{2}(\bmod m)$, then $r_{1} \equiv r_{2}(\bmod m)$.
Under the above definition, we will prove (2)
$$\begin{array}{l}
\delta_{p-3}=(p-1)!\sum_{1 \leqslant i<j \leqslant p-1} \frac{1}{i j} \equiv(p-1)!\sum_{1 \leqslant i<j \leqslant p-1} i^{-1} j^{-1}(\bmod p) \\
\equiv \frac{(p-1)!}{2} \sum_{i \neq j} i^{-1} j^{-1}(\bmod p) \equiv \frac{(p-1)!}{2} \sum_{s \neq t} s t(\bmod p) \\
\equiv \frac{(p-1)!}{2}\left\{\left[\frac{p(p-1)}{2}\right]^{2}-\frac{(p-1) p(2 p-1)}{6}\right\}(\bmod p) \\
\equiv 0(\bmod p) \\
\delta_{p-2}=(p-1)!\sum_{i=1}^{p-1} \frac{1}{i}=p!\sum_{i=1}^{\frac{2-1}{2}} \frac{1}{i(p-i)} \\
\text { and } \sum_{i=1}^{\frac{p-1}{2}} \frac{1}{i(p-i)} \equiv \sum_{i=1}^{\frac{p-1}{2}}-\left(i^{-1}\right)^{2}(\bmod p)
\end{array}$$
It is easy to prove that for each $1 \leqslant i \leqslant \frac{p-1}{2}$, there exists a unique $1 \leqslant j \leqslant \frac{p-1}{2}$ such that $\left(i^{-1}\right)^{2} \equiv j^{2}(\bmod p)$, so $\sum_{i=1}^{\frac{p-1}{2}} \frac{1}{i(p-i)} \equiv \sum_{j=1}^{\frac{p-1}{2}}\left(-j^{2}\right)(\bmod p) \equiv-\frac{p\left(p^{2}-1\right)}{24}(\bmod p)$,
Substituting into (3) we know $p^{2} \mid \delta_{p-2}$, thus (2) is proved.
From the lemma, it is immediately clear that both the left and right sides are congruent to $[(p-1)!]^{b}$ modulo $p^{3}$, hence (1) holds.
Thus, the original proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
14. Let $S$ be the set of all prime numbers $p$ that satisfy the following condition: the number of digits in the smallest repeating block of the decimal part of $\frac{1}{p}$ is a multiple of 3, i.e., for each $p \in S$, there exists the smallest positive integer $r=r(p)$, such that $\frac{1}{p}=0 . a_{1} a_{2} \cdots a_{3 r} a_{1} a_{2} \cdots a_{3 r} \cdots$
For each $p \in S$ and any integer $k \geqslant 1$, define $f(k, p)=a_{k}+a_{k+r(p)}+a_{k+2 r(p)}$.
(1) Prove that the set $S$ contains infinitely many primes;
(2) For $k \geqslant 1$ and $p \in S$, find the maximum value of $f(k, p)$.
|
14. Prove: (1) The length of the smallest repeating cycle of $\frac{1}{p}$ is the smallest integer $d (d \geqslant 1)$ such that $10^{d}-1$ is divisible by $p$.
Let $q$ be a prime number, and $N_{q}=10^{2 q}+10^{q}+1$. Then $N_{q} \equiv 3(\bmod q)$. Let $p_{q}$ be a prime factor of $\frac{N_{q}}{3}$. Then $p_{q}$ cannot be divisible by 3, because $N_{q}$ is a factor of $10^{3 q}-1$. The repeating cycle length of $\frac{1}{p_{q}}$ is $3 q$, so the length of the smallest repeating cycle of $\frac{1}{p_{q}}$ is a divisor of $3 q$. If the smallest repeating cycle length is $q$, then from $10^{q} \equiv 1\left(\bmod p_{q}\right)$ we get $N_{q}=10^{2 q}+10^{q}+1 \left(\bmod p_{q}\right)$, which is a contradiction! If the smallest repeating cycle length is 3, there is only one case, i.e., $p_{q}$ is a factor of $10^{3}-1=3^{3} \times 37$, i.e., $p_{q}=37$. In this case, $N_{q}=3 \times 37 \equiv 3(\bmod 4)$, while $N_{q}=10^{2 q}+10^{q}+1 \equiv 1(\bmod 4)$, which is a contradiction! Thus, for each prime $q$, we can find a prime $p_{q}$ such that the length of the smallest repeating cycle of the decimal part of $\frac{1}{p_{q}}$ is $3 q$.
(2) Let the prime $p \in S$, and $3 r(p)$ be the length of the smallest repeating cycle of $\frac{1}{p}$. Then $p$ is a factor of $10^{3 r(p)}-1$ but not a factor of $10^{r(p)}-1$, so $p$ is a factor of $N_{r}=10^{2 r(p)}+10^{r(p)}+1$.
Let $\frac{1}{p}=0 . \overline{a_{1} a_{2} \cdots}, x_{j}=\frac{10^{j-1}}{p}, y_{j}=\left\{x_{j}\right\}=0 . \overline{a_{j} a_{j+1} \cdots}$, where $\{x\}$ denotes the fractional part of $x$, then $a_{j}<10 y_{j}$. Thus, $f(k, p)=a_{k}+a_{k+r(p)}+a_{k+2 r(p)}<10\left(y_{k}+y_{k+r(p)}+y_{k+2 r(p)}\right)$. Since $x_{k}+x_{k+r(p)}+x_{k+2 r(p)}=\frac{10^{k-1} \cdot N_{r(p)}}{p}$ is an integer, $y_{k}+y_{k+r(p)}+y_{k+2 r(p)}$ is also an integer and is a number less than 3, so $y_{k}+y_{k+r(p)}+y_{k+2 r(p)} \leqslant 2$, thus $f(k, p)<20$, so $f(k, p) \leqslant 19$. Since $f(2,7)=4+8+7=19$, the maximum value sought is 19.
|
19
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Theorem 1 Let $m=m_{1} m_{2}, x_{i}^{(1)}$ be a complete residue system modulo $m_{1}$, and $x_{j}^{(2)}$ be a complete residue system modulo $m_{2}$. Then $x_{i j}$ $=x_{i}^{(1)}+m_{1} x_{j}^{(2)}$ is a complete residue system modulo $m$. That is, as $x^{(1)}, x^{(2)}$ run through the complete residue systems modulo $m_{1}$, modulo $m_{2}$ respectively, $x=x^{(1)}+m_{1} x^{(2)}$ runs through the complete residue system modulo $m=m_{1} m_{2}$.
|
Prove that at this time, $x_{ij}$ has a total of $m=m_{1} m_{2}$ numbers, so it is only necessary to prove that they are pairwise distinct modulo $m$.
If $x_{i_{1}}^{(1)}+m_{1} x_{j_{1}}^{(2)} \equiv x_{i_{1} j_{1}} \equiv x_{i_{2} j_{2}} \equiv x_{i_{2}}^{(1)}+m_{1} x_{j_{2}}^{(2)}\left(\bmod m_{1} m_{2}\right)$,
then it must be that $x_{i_{1}}^{(1)} \equiv x_{i_{2}}^{(1)}\left(\bmod m_{1}\right)$,
thus it must be that $i_{1}=i_{2}, x_{i_{1}}^{(1)}=x_{i_{2}}^{(1)}$ (because $x_{i}^{(1)}$ takes values in the same complete residue system modulo $m$).
Furthermore, we get $m_{1} x_{j_{1}}^{(2)} \equiv m_{1} x_{j_{2}}^{(2)}\left(\bmod m_{1} m_{2}\right)$,
which means $x_{j_{2}}^{(2)} \equiv x_{1_{1}}^{(2)}\left(\bmod m_{2}\right)$.
Similarly, we have $j_{1}=j_{2}, x_{j_{1}}^{(2)}=x_{j_{2}}^{(2)}$, and the theorem is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Theorem 2 Let $m=m_{1} m_{2},\left(m_{1}, m_{2}\right)=1, x^{(1)}, x^{(2)}$ traverse the complete residue systems modulo $m_{1}, m_{2}$, respectively, then $x_{i j}=m_{2} x_{i}^{(1)}+m_{1} x_{j}^{(2)}$ traverses the complete residue system modulo $m_{1} m_{2}$.
|
Prove Theorem 2 for $k=2$.
Assume the theorem holds for $k=n (n \geqslant 2)$.
When $k=n+1$,
let $\bar{x}^{(n)}=\frac{m}{m_{1} m_{n+1}} x^{(1)}+\cdots+\frac{m}{m_{n} m_{n+1}} x^{(n)}$,
we have $x=m_{n+1} \bar{x}^{(n)}+\frac{m}{m_{n+1}} x^{(n+1)}$.
From the above two equations and the fact that $k=2$ holds, the conclusion is proved. Thus, we clearly demonstrate how the "large modulus" $m=m_{1} m_{2} \cdots m_{k}$ is characterized by the smaller moduli $m_{1}, m_{2}, \cdots$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4. For any positive integer $n$, there exist $n$ consecutive positive integers such that in the standard factorization of each positive integer, the exponent of each prime is less than or equal to 1.
|
4. We only need to prove that there exist $n$ consecutive positive integers, each of which has at least one prime factor that appears exactly once in its standard factorization. We select $n$ distinct primes $p_{1}, p_{2}, \cdots, p_{n}$, and consider the system of congruences $x \equiv -i + p_{i} \left(\bmod p_{i}^{2}\right), i=1,2, \cdots, n$.
By the Chinese Remainder Theorem, the above system of congruences has a solution. For $1 \leqslant i \leqslant n$, by $p_{i}^{2} \mid p_{i} - (x + i) \Rightarrow p_{i} \mid x + i$. Also, $p_{i}^{2} \nmid x + i$, meaning $p_{i}$ appears exactly once in the standard factorization of $x + i$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5. Let $n$ be any positive integer, prove: there must exist $n$ consecutive integers, none of which are prime numbers.
.
|
5. Take $2 n$ distinct prime numbers $p_{1}, p_{2}, \cdots, p_{2 n}$. By the Chinese Remainder Theorem, the system of congruences $x \equiv -k$ $\left(\bmod p_{2 k-1} p_{2 k}\right), k=1,2, \cdots, n$ has a solution. Since for any $k, x+k$ has at least two different prime factors, it cannot be a prime number.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
6. Given a positive integer $n$, let $f(n)$ be the smallest positive integer such that $\sum_{k=1}^{(n)} k$ is divisible by $n$. Prove: $f(n)=2n-1$ if and only if $n$ is a power of 2.
|
6. (1) When $n=2^{m}$, $\sum_{k=1}^{2 n-1} k=\frac{(2 n-1) \cdot 2 n}{2}=\left(2^{m+1}-1\right) \cdot 2^{m}$. On one hand, $2^{m}$ divides the above expression. On the other hand, if $r \leqslant 2 n-2$, then $\sum_{k=1}^{r} k=\frac{r(r+1)}{2}$ cannot be divided by $2^{m}$, because one of $r$ and $r+1$ is odd, and the other does not exceed $(2 n-2)+1=2^{m+1}-1$. From these two aspects, we know that $f\left(2^{m}\right)=2^{m+1}-1$.
(2) When $n$ is not a power of 2, let $n=2^{m} a$, where $m \geqslant 0, a>1$ is an odd number. We prove that there exists a positive integer $r<2 n-1$ such that $2^{m+1} \mid r$ and $a \mid r+1$, thus $\sum_{k=1}^{r} k=\frac{r(r+1)}{2}$ is divisible by $2^{m} \cdot a=n$, and therefore $f(n)<2 n-1$.
To prove the above statement, consider $x \equiv 0\left(\bmod 2^{m+1}\right)$, $x \equiv -1 (\bmod a)$. Since $\left(a, 2^{m+1}\right)=1$, the above system of congruences must have a solution $x_{0}$, and all solutions are $x \equiv x_{0}\left(\bmod 2^{m+1} a\right)$. Furthermore, the second equation implies $r \neq 2 n$, and the first equation implies $0 \leqslant r<2 n-1$, which proves the existence of the required $r$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4. (32nd United States of America Mathematical Olympiad) Prove: For each positive integer $n$, there exists an $n$-digit positive integer that is divisible by $5^{n}$ and has only odd digits.
|
4. Prove by induction: When $n=1$, $5 \mid 5$.
Assume when $n=m$, $5^{m} \mid \overline{a_{1} a_{2} \cdots a_{m}}$, where $a_{i}(i=1,2, \cdots, m)$ are single-digit odd numbers.
When $n=m+1$, consider $\overline{1 a_{1} a_{2} \cdots a_{m}}, \overline{3 a_{1} a_{2} \cdots a_{m}}, \cdots, \overline{9 a_{1} a_{2} \cdots a_{m}}$.
Since their differences $2 \times 10^{m}, 4 \times 10^{m}, 6 \times 10^{m}, 8 \times 10^{m}$ cannot be divided by $5^{m+1}$. Therefore, the remainders of these 5 numbers when divided by $5^{m+1}$ are all different.
Thus, among these five numbers, there exists one that can be divided by $5^{m+1}$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Theorem 2 (Euler's Criterion)
Let prime $p>2, p>d$, then $d$ is a quadratic residue modulo $p$ if and only if
$$d^{\frac{p-1}{2}} \equiv 1(\bmod p)$$
$d$ is a non-quadratic residue modulo $p$ if and only if
$$d^{\frac{p-1}{2}} \equiv-1(\bmod p)$$
|
First, we prove that for any $d, p \times d$, either equation (4) or (5) holds, but not both. By Fermat's Little Theorem, we know $d^{p-1} \equiv 1 \pmod{p}$, thus $\left(d^{\frac{p-1}{2}}-1\right)\left(d^{\frac{p-1}{2}}+1\right) \equiv 0 \pmod{p}$. Since the prime $p > 2$ and $\left(d^{\frac{p-1}{2}}-1, d^{\frac{p-1}{2}}+1\right) \mid 2$, it follows that either (4) or (5) holds, but not both.
Next, we prove that $d$ is a quadratic residue modulo $p$ if and only if (4) holds. First, we prove the necessity.
If $d$ is a quadratic residue modulo $p$, then there must be an $x_{0}$ such that
$d \equiv x_{0}^{2} \pmod{p}$, and thus $x_{0}^{p-1} \equiv d^{\frac{p-1}{2}} \pmod{p}$.
Since $p \nmid d$, it follows that $p \nmid x_{0}$. By Fermat's Little Theorem,
$d^{\frac{p-1}{2}} \equiv x_{0}^{p-1} \equiv 1 \pmod{p}$, which proves the necessity.
Next, we prove the sufficiency. Suppose (4) holds, then $p \nmid d$. Consider the linear congruence $a x \equiv b \pmod{p}$. Since $p \nmid d$, for each $j$ in the reduced residue system (2) modulo $p$, when $a = j$, there must be a unique $x = x_{j}$ in the reduced residue system (2) such that $a x \equiv b \pmod{p}$ holds. If $d$ is not a quadratic residue modulo $p$, then $j \neq x_{j}$, and thus the $p-1$ numbers in the reduced residue system (2) can be paired as $(j, x_{j})$, leading to
$d^{\frac{p-1}{2}} \equiv (p-1)! \equiv -1 \pmod{p}$ (using Wilson's Theorem).
This contradicts (4), so there must be some $j_{0}$ such that $j_{0} = x_{j_{0}}$, and thus $d$ is a quadratic residue modulo $p$. The sufficiency is proved.
This completes the proof of Euler's criterion.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2. Let $p$ be an odd prime. Prove: the sum $S$ of all quadratic residues modulo $p$ in $1,2, \cdots, p-1$ is $\frac{p\left(p^{2}-1\right)}{24}-$ $p \sum_{j=1}^{\frac{p-1}{2}}\left[\frac{j^{2}}{p}\right]$
|
2. Let $j^{2}=p q_{j}+r_{j}, 1 \leqslant r_{j}<p, 1 \leqslant j \leqslant \frac{p-1}{2}$, from this and $q_{j}=\left[\frac{j^{2}}{p}\right]$ we get $S=\sum r_{j}=\sum j^{2}-$ $p \Sigma\left[\frac{j^{2}}{p}\right]$
Using (3) from the previous problem, we get
$$S=\frac{p\left(p^{2}-1\right)}{24}-p \sum_{j=1}^{\frac{p-1}{2}}\left[\frac{j^{2}}{p}\right]$$
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5. Let $k$ be a positive integer. Prove: there exist infinitely many perfect squares of the form $n \cdot 2^{k}-7$, where $n$ is a positive integer.
|
5. First prove that for any given $k$, there exists a positive integer $a_{k}$, such that $a_{k}^{2} \equiv -7 \pmod{2^{k}}$, and prove it by mathematical induction on $k$.
Direct observation shows that when $k \leqslant 3$, taking $a_{k}=1$ satisfies the condition. Suppose for some $k>3$, we have $a_{k}^{2} \equiv -7 \pmod{2^{k}}$.
Next, consider the value of $a_{k}^{2}$ modulo $2^{k+1}$, which is either $a_{k}^{2} \equiv -7 \pmod{2^{k+1}}$ or $a_{k}^{2} \equiv 2^{k}-7 \pmod{2^{k+1}}$. For the former, we can take $a_{k+1}=a_{k}$, and for the latter, we can take $a_{k+1}=a_{k}+2^{k-1}$. This is because $k \geqslant 3$, and $a_{k}$ is odd, so $a_{k+1}^{2} \equiv a_{k}^{2} + 2^{k} a_{k} + 2^{2k-2} \equiv a_{k}^{2} + 2^{k} \equiv -7 \pmod{2^{k+1}}$.
The above derivation uses the induction hypothesis.
Finally, it is easy to see that the sequence $\{a_{k}\}$ has no maximum element, and thus we can require that for any positive integer $k$, $a_{k}^{2} \geqslant 2^{k}-7$. Therefore, $\{a_{k}\}$ contains infinitely many different values, and the proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
9. Prove: There are infinitely many primes of the form $4k+1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, since the text is already in English, here is the same text with the requested format preserved:
9. Prove: There are infinitely many primes of the form $4k+1$.
|
9. Suppose there are only finitely many such primes, and let them be $p_{1}, p_{2}, \cdots, p_{k}$.
We consider $\left(2 p_{1} \cdots p_{k}\right)^{2}+1=p$, by the assumption and $p \equiv 1(\bmod 4)$, so $p$ is not a prime, let $p_{0}$ be a prime factor of $p$, $p_{0}$ is of course odd, so -1 is a quadratic residue modulo $p_{0}$, i.e., $\left(\frac{-1}{p_{0}}\right)=1$, thus $p_{0} \equiv 1(\bmod 4)$, but $p_{0}$ is clearly not $p_{1}, p_{2}, \cdots, p_{k}$, which contradicts the assumption.
Therefore, there are infinitely many primes of the form $4 k+1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Theorem 2 If $(a, b)=1$, and $x_{0}, y_{0}$ is a solution of (1), then all solutions of equation (1) are $x=x_{0}+b t, y=y_{0}$
$$-a t(t \in \mathbf{Z})$$
|
Let $x_{0}, y_{0}$ be a solution of (1), then we have
$$a x_{0}+b y_{0}=c \text {. }$$
Let $x, y$ be any solution of (1), subtracting (3) from (1) yields
$$a\left(x-x_{0}\right)+b\left(y-y_{0}\right)=0 .$$
Since $(a, b)=1$, it follows that $b \mid\left(x-x_{0}\right)$, i.e., $x-x_{0}=b t$, thus $x=x_{0}+b t$. Substituting this result into the above equation gives $y=y_{0}-a t$, meaning that any solution of the equation can be expressed as $x=x_{0}+b t, y=y_{0}-a t(t \in \mathbf{Z})$.
Conversely, if $x_{0}, y_{0}$ is a solution of (1), it is easy to verify that $x=x_{0}+b t, y=y_{0}-a t$ are also solutions of (1), thus Theorem 2 is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
9. Prove: There are infinitely many positive integers $n$ such that $[\sqrt{2} n]$ is a perfect square, ( $[x]$ denotes the greatest integer not exceeding $x$).
untranslated text remains the same as requested.
|
9. Proof: Consider the Pell's equation $x^{2}-2 y^{2}=-1$ which has infinitely many positive integer solutions. Take any solution $u, v$, then $2 v^{2}=u^{2}+1$.
Multiply both sides of the above equation by $u^{2}$ to get $2(u v)^{2}=\left(u^{2}+1\right) u^{2}$, hence $u^{2}<\sqrt{2} u v<u^{2}+1$. Therefore, $[\sqrt{2} u v]=u^{2}$ is a perfect square, i.e., taking $n=u v$ yields infinitely many such $n$.
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proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 Proof: The indeterminate equation $x^{2}+y^{2}-8 z^{3}=6$ has no integer solutions.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Prove that $x, y$ have the same parity by modulo 2; modulo 4 shows that $x, y$ are both odd, thus $x^{2} \equiv 1(\bmod 8), y^{2} \equiv 1$ $(\bmod 8)$.
Taking both sides of the equation modulo 8, we have $6 \equiv x^{2}+y^{2}-8 z^{3} \equiv 2(\bmod 8)$, a contradiction. Therefore, the original equation has no integer solutions.
| null |
Number Theory
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proof
|
Yes
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Yes
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number_theory
| false |
Example 6 Proof: The only positive integer solution to the indeterminate equation $8^{x}+15^{y}=17^{z}$
is $x=y=z=2$.
|
First, we use congruence to prove that $y$ and $z$ are both even.
Taking equation (1) modulo 4, we get $(-1)^{y} \equiv 1(\bmod 4)$,
thus $y$ is even. Taking equation (1) modulo 16, we get $8^{x}+(-1)^{y} \equiv 1(\bmod 16)$, which simplifies to $8^{x} \equiv 0(\bmod 16)$, hence $x \geqslant 2$.
Note that $17^{2} \equiv 1,15^{2} \equiv 1(\bmod 32)$, so if $z$ is odd, then from (1) with $2 \mid y$ and $x \geqslant 2$, we can derive $1 \equiv 17(\bmod 32)$.
This is impossible, so $z$ must be even.
Let $y=2 y_{1}, z=2 z_{1}$, then equation (1) can be factored as
$\left(17^{z_{1}}-15^{y_{1}}\right)\left(17^{z_{1}}+15^{y_{1}}\right)=8^{x}$,
It is easy to see that the greatest common divisor of the two factors on the left side of (2) is 2, and the right side of (2) is a power of 2, so we must have
$$\left\{\begin{array}{l}
17^{z_{1}}-15^{y_{1}}=2 \\
17^{z_{1}}+15^{y_{1}}=2^{3 x-1}
\end{array}\right.$$
Taking (3) modulo 32, we know that $z_{1}$ and $y_{1}$ must both be odd (otherwise, the left side of (3) $\left.\equiv 0,-14,16(\bmod 32)\right)$, adding (3) and (4) gives $17^{z_{1}}=1+2^{3 x-2}$,
If $x \geqslant 3$, then the right side of (5) $\equiv 1(\bmod 32)$; and since $z_{1}$ is odd, the left side of (5) $\equiv 17(\bmod 32)$, which is impossible, so we must have $x=2$. From this and (5), we get $z_{1}=1$, i.e., $z=2$, and it is easy to see that $y_{1}=1$, i.e., $y=2$. Therefore, $x=y=z=2$.
|
x=y=z=2
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Number Theory
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proof
|
Yes
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Yes
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number_theory
| false |
1. (2004 China Mathematical Olympiad) Prove: there does not exist a pair of positive integers $x, y$ satisfying $3 y^{2}=x^{4}+x$.
|
1. Proof: By contradiction
If the equation $3 y^{2}=\left(x^{3}+1\right) x$ has a set of integer solutions $x, y$, since $\left(x, x^{3}+1\right)=1$, there exist $u, v \in \mathbf{N}_{+}$, such that $y=u v$, and $3 u^{2}=x^{3}+1, v^{2}=x$ or $u^{2}=x^{3}+1,3 v^{2}=x$.
The first case is obviously impossible, now assume the second case holds.
Notice, at this time, $u^{2}=(x+1)\left(x^{2}-x+1\right)$,
Since $x \equiv 0(\bmod 3)$, and $x^{2}-x+1=(x+1)(x-2)+3$.
Also, $x+1$ and $x^{2}-x+1$ are coprime, therefore, there exists a positive integer $t$ satisfying $x^{2}-x+1=t^{2}$.
It is easy to see that the equation $x^{2}-x+1=t^{2}$ has only one positive integer solution $x=1$, but this contradicts $x \equiv 0(\bmod 3)$. Therefore, the conclusion of the problem holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5. Prove that the indeterminate equation $5^{x}-3^{y}=2$ has only the positive integer solution $x=y=1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if you intended to have the proof or a detailed explanation of the solution, please let me know!
|
5. Proof: The equation clearly has a solution $x=y=1$. Taking the equation modulo 4, it is easy to see that $y$ is odd. If $y>1$, taking the equation modulo 9 yields
$$5^{x} \equiv 2(\bmod 9)$$
It is not difficult to find that for $x=1,2, \cdots, 5^{x}$ modulo 9, the cycle is $5,7,8,4,2,1$. Therefore, by (1), $x$ must be of the form $6k+5$. Taking the original equation modulo 7, it is easy to verify that for odd $y$, $3^{y} \equiv 3,5,6(\bmod 7)$.
When $x=6k+5$, by Fermat's Little Theorem, $5^{6} \equiv 1(\bmod 7)$, so
$$5^{x}=5^{6k+5} \equiv 5^{5} \equiv 3(\bmod 7)$$
Thus, the two sides of the original equation are not congruent modulo 7, so it has no solutions for $y>1$. Therefore, the only positive integer solution is $y=1, x=1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
11. (2005 USA Mathematical Olympiad Problem) Prove: The system of equations $\left\{\begin{array}{l}x^{6}+x^{3}+x^{3} y+y=147^{157} \\ x^{3}+x^{3} y+y^{2}+y+z^{9}=157^{147}\end{array}\right.$ has no integer solutions $x, y, z$.
|
11. Proof: Adding both sides of the equation and then adding 1, we get
$$\left(x^{3}+y+1\right)^{2}+z^{9}=147^{157}+157^{147}+1$$
We will now prove that both sides of equation (1) are not congruent modulo 19.
The choice of modulo 19 is because the least common multiple of 2 and 9 is 18, and by Fermat's Little Theorem, when $a$ is not a multiple of 19, $a^{18} \equiv 1(\bmod 19)$.
In particular, $\left(z^{9}\right)^{2} \equiv 0$ or $1(\bmod 19)$, so $z^{9} \equiv-1,0,1(\bmod 19)$.
By calculation, we have $n^{2} \equiv-8,-3,-2,0,1,4,5,6,7,9(\bmod 19)$,
By Fermat's Little Theorem, $147^{157}+157^{147}+1 \equiv 147^{13}+157^{3}+1(\bmod 19)$
$$\equiv-5^{13}+5^{3}+1(\bmod 19) \equiv 14(\bmod 19)$$
Since $z^{9}+n^{2} \equiv 14(\bmod 19)$, equation (1) has no integer solutions.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
10. Write the first $2 n$ natural numbers in any order in a row, and number them sequentially as $1,2, \cdots, 2 n$. Then, find the sum of each number and its position number, resulting in $2 n$ sum numbers. Next, divide each sum number by $2 n$, obtaining $2 n$ quotients and remainders. Prove that at least two of the remainders are the same.
|
10. Proof: Let the first $2n$ natural numbers be written in any order as $a_{1}, a_{2}, a_{3}, \cdots, a_{2n-1}, a_{2n}$, where $1 \leqslant a_{i} \leqslant 2n$, $a_{i} \neq a_{j}(i \neq j),(i, j=1,2,3, \cdots, 2n)$, then the corresponding sum numbers are $1+a_{1}, 2+a_{2}, \cdots, 2n-1+a_{2n-1}$, $2n+a_{2n}$.
These sum numbers, when divided by $2n$, can only result in one of the following $2n$ remainders: $0,1,2, \cdots, 2n-2,2n-1$.
If we can prove that all $2n$ remainders cannot appear, the problem is solved. A direct approach is difficult, so we consider the opposite.
Suppose these $2n$ remainders all appear, then their sum is $r_{2n}=0+1+2+\cdots+(2n-1)=2n^2-n>0$.
This sum should be congruent modulo $2n$ to the sum of the $2n$ sum numbers divided by $2n$ (i.e., the remainders are the same), which indicates a remainder of zero. This is a contradiction.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 (2004 Belarusian Mathematical Olympiad Problem) Positive integers $a, b, c$ satisfy the equation $c(a c+1)^{2}=(5 c+2 b)(2 c+b)$,
(1) Prove that if $c$ is odd, then $c$ is a perfect square;
(2) For some $a, b$, does there exist an even $c$ that satisfies (1);
(3) Prove: Equation (1) has infinitely many positive integer solutions $(a, b, c)$.
|
Solve (2) first.
(2) Assume $c$ is even, let $c=2 c_{1}$. Then the known equation can be written as $c_{1}\left(2 a c_{1}+1\right)^{2}=\left(5 c_{1}+b\right)\left(4 c_{1}+b\right)$. Let $d=\left(c_{1}, b\right)$, then $c_{1}=d c_{0}, b=d b_{0}$, where $\left(c_{0}, b_{0}\right)=1$.
Thus, we have $c_{0}\left(2 a d c_{0}+1\right)^{2}=d\left(5 c_{0}+b_{0}\right)\left(4 c_{0}+b_{0}\right)$.
Obviously, $\left(c_{0}, 5 c_{0}+b_{0}\right)=\left(c_{0}, 4 c_{0}+b_{0}\right)=\left(d,\left(2 a d c_{0}+1\right)^{2}\right)=1$.
Therefore, $c_{0}=d$. Thus, $\left(2 a d^{2}+1\right)^{2}=\left(5 d+b_{0}\right)\left(4 d+b_{0}\right)$.
Notice that $\left(5 c_{0}+b_{0}, 4 c_{0}+b_{0}\right)=\left(5 c_{0}+b_{0}-4 c_{0}-b_{0}, 4 c_{0}+b_{0}\right)$
$$\begin{array}{l}
=\left(c_{0}, 4 c_{0}+b_{0}\right)=\left(c_{0}, 4 c_{0}+b_{0}-4 c_{0}\right) \\
=\left(c_{0}, b_{0}\right)=1
\end{array}$$
So, $5 d+b_{0}=m^{2}, 4 d+b_{0}=n^{2}, 2 a d^{2}+1=m n, m, n \in \mathbf{N}_{+}$.
Then, $d=m^{2}-n^{2}$ (obviously $m>n$ ), thus
$$m n=1+2 a d^{2}=1+2 a(m-n)^{2}(m+n)^{2}$$
$\geqslant 1+2 a(m+n)^{2} \geqslant 1+2 a \cdot 4 m n \geqslant 1+8 m n$, contradiction.
Therefore, $c$ is odd, i.e., for some $a, b$, there does not exist an even $c$ that satisfies equation (1).
(1) Similarly to (2), let $(c, b)=d, c=d c_{0}, b=d b_{0}$, and $\left(c_{0}, b_{0}\right)=1$.
Then the known equation can be rewritten as $c_{0}\left(a d c_{0}+1\right)^{2}=d\left(5 c_{0}+2 b_{0}\right)\left(2 c_{0}+b_{0}\right)$.
Notice that $\left(c_{0}, 5 c_{0}+2 b_{0}\right)=\left(c_{0}, 2 c_{0}+b_{0}\right)=\left(d,\left(a d c_{0}+1\right)^{2}\right)=1$.
Therefore, $c_{0}=d$, thus, $c=d c_{0}=d^{2}$.
(3) Let $c=1$, it suffices to prove that the equation $(a+1)^{2}=(5+2 b)(2+b)$ has infinitely many integer solutions $(a, b)$.
In fact, let $5+2 b=m^{2}, 2+b=n^{2}$, then $a=m n-1$.
Thus, it suffices to prove: there exist infinitely many $m, n \in \mathbf{N}_{+}$ satisfying $5+2 b=m^{2}, 2+b=n^{2}$, i.e., $m^{2}-2 n^{2}=1$. This is a Pell's equation, which has infinitely many solutions, thus the original proposition holds.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 (2003 Croatian National Mathematical Olympiad) For any integer $n(n>2)$, prove: $\left[\frac{n(n+1)}{4 n-2}\right]=$
$$\left[\frac{n+1}{4}\right]$$
|
Analysis When a fraction with a small positive integer as the denominator appears inside a Gaussian function, in many cases, it is necessary to discuss the remainder of the unknown divided by this denominator.
Proof Since $\frac{n(n+1)}{4 n-2}=\frac{n(n+1)-\frac{n+1}{4} \times(4 n-2)}{4 n-2}+\frac{n+1}{4}$
$$=\frac{\frac{1}{2}(n+1)}{4 n-2}+\frac{n+1}{4}=\frac{n+1}{4(2 n-1)}+\frac{n+1}{4},$$
and when $n>2$, $0<\frac{n+1}{4(2 n-1)}<\frac{1}{4}$, thus $\frac{n(n+1)}{4 n-2}<\frac{n+1}{4}+\frac{1}{4}$.
Hence $\left[\frac{n+1}{4}\right] \leqslant\left[\frac{n(n+1)}{4 n-2}\right]$.
Next, we prove
$$\frac{n(n+1)}{4 n-2}<\left[\frac{n+1}{4}\right]+1 \quad (n>2)$$
We discuss in two cases:
(1) Let $n=4 k+r, r=0,1,2$, then
$$\left[\frac{n+1}{4}\right]+1=k+1=\frac{n-r}{4}+1=\frac{n+(3-r)}{4}+\frac{1}{4} \geqslant \frac{n+1}{4}+\frac{1}{4} \text {, }$$
Combining $\frac{n(n+1)}{4 n-2}<\frac{n+1}{4}+\frac{1}{4}$, we know (2) holds.
In both cases, inequality (2) holds.
From (2), we know $\left[\frac{n(n+1)}{4 n-2}\right]<\left[\frac{n+1}{4}\right]+1=\left[\frac{n+1}{4}\right]+1$.
Thus $\left[\frac{n(n+1)}{4 n-2}\right] \leqslant\left[\frac{n+1}{4}\right]$.
Combining (1), we know $\left[\frac{n(n+1)}{4 n-2}\right]=\left[\frac{n+1}{4}\right]$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Prove the inequality $[\sqrt{\alpha}]+[\sqrt{\alpha+\beta}]+[\sqrt{\beta}] \geqslant[\sqrt{2 \alpha}]+[\sqrt{2 \beta}]$ for any real numbers $\alpha$ and $\beta$ not less than 1.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Analyzing the original inequality, which contains 5 floor functions, is quite complex and it is not easy to segment the values of $\alpha$ and $\beta$ directly. Therefore, we should first appropriately relax the inequality to remove some of the floor functions.
Proof: Let $a=\sqrt{\alpha}, b=\sqrt{\beta}$, then the original inequality can be written as
$[a]+\left[\sqrt{a^{2}+b^{2}}\right]+[b] \geqslant[\sqrt{2} a]+[\sqrt{2} b]$,
Here, by $\alpha \geqslant 1, \beta \geqslant 1$ we know $a \geqslant 1, b \geqslant 1$.
Using the mean inequality $\sqrt{\frac{a^{2}+b^{2}}{2}} \geqslant \frac{a+b}{2}$, we get
$$\left[\sqrt{a^{2}+b^{2}}\right] \geqslant\left[\frac{a+b}{\sqrt{2}}\right] \geqslant\left[\frac{\sqrt{2}}{2} a\right]+\left[\frac{\sqrt{2}}{2} b\right]$$
From this, we know that to prove (1), it is sufficient to prove the inequality
$$[x]+\left[\frac{\sqrt{2}}{2} x\right] \geqslant[\sqrt{2} x]$$
for any real number $x \geqslant 1$.
We will discuss this in 5 cases:
(1) When $x \geqslant 2+\sqrt{2}$, $\left(1-\frac{\sqrt{2}}{2}\right) x \geqslant 1$, so $\left[\left(1-\frac{\sqrt{2}}{2}\right) x\right] \geqslant 1$.
Therefore, $[x]+\left[\frac{\sqrt{2}}{2} x\right]>x-1+\frac{\sqrt{2}}{2} x-1=\sqrt{2} x+\left(1-\frac{\sqrt{2}}{2}\right) x-2$
$$\geqslant[\sqrt{2} x]+\left[\left(1-\frac{\sqrt{2}}{2}\right) x\right]-2 \geqslant[\sqrt{2} x]-1$$
That is, $[x]+\left[\frac{\sqrt{2}}{2} x\right]>[\sqrt{2} x]-1$.
Since both sides of the above inequality are integers, we have $[x]+\left[\frac{\sqrt{2}}{2} x\right] \geqslant[\sqrt{2} x]$.
(2) When $2 \sqrt{2} \leqslant x \leqslant 2+\sqrt{2}$, $2 \leqslant \frac{\sqrt{2}}{2} x<\sqrt{2}+1,4 \leqslant \sqrt{2} x<2+2 \sqrt{2}<5$.
Thus, $[x]+\left[\frac{\sqrt{2}}{2} x\right] \geqslant 2+2=4,[\sqrt{2} x]=4$.
(3) When $\frac{3}{2} \sqrt{2} \leqslant x<2 \sqrt{2}$, $\frac{3}{2} \leqslant \frac{\sqrt{2}}{2} x<2,3 \leqslant \sqrt{2} x<4$.
Thus, $[x]+\left[\frac{\sqrt{2}}{2} x\right]=2+1=3,[\sqrt{2} x]=3$.
(4) When $\sqrt{2} \leqslant x<\frac{3}{2} \sqrt{2}$, $1 \leqslant \frac{\sqrt{2}}{2} x<\frac{3}{2}, 2 \leqslant \sqrt{2} x<3$.
Thus, $[x]+\left[\frac{\sqrt{2}}{2} x\right] \geqslant 1+1=2,[\sqrt{2} x]=2$.
(5) When $1 \leqslant x<\sqrt{2}$, $\frac{\sqrt{2}}{2} \leqslant \frac{\sqrt{2}}{2} x<1, \sqrt{2} \leqslant \sqrt{2} x<2$.
Thus, $[x]+\left[\frac{\sqrt{2}}{2} x\right]=1+0=1,[\sqrt{2} x]=1$.
In summary, for any real number $x \geqslant 1$, (2) holds. Therefore, the inequality (1) holds, and the proof is complete.
|
proof
|
Inequalities
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 7 Let $[x]$ denote the greatest integer not exceeding $x$, and let $a$ be the positive root of the equation $x^{2}-m x-1=0$, where $m$ is a given positive integer. Prove that there exist infinitely many positive integers $n$ satisfying the equation
$$[a n+m a[a n]]=m n+\left(m^{2}+1\right)[a n] .$$
|
Given that $a$ satisfies $a^{2}-m a-1=0$, i.e., $a=m+\frac{1}{a}$, use this equation to simplify the expression inside the floor function, and finally obtain infinitely many positive integers $n$ that are related to $a$.
Proof: Since $a$ is the positive root of the equation $x^{2}-m x-1=0$,
we have $a^{2}-m a-1=0$, i.e., $a-m-\frac{1}{a}=0$.
Thus, $a=m+\frac{1}{a}>m$.
For every positive integer $n$ greater than $a$, we have
$$\begin{array}{l}
{[a n+m a[a n]]-m n-\left(m^{2}+1\right)[a n]} \\
=\left[\left(m+\frac{1}{a}\right) n+m\left(m+\frac{1}{a}\right)[a n]\right]-m n-\left(m^{2}+1\right)[a n] \\
=\left[m n+m^{2}[a n]+\frac{n}{a}+\frac{m}{a}[a n]\right]-m n-\left(m^{2}+1\right)[a n] \\
=\left[\frac{n}{a}+\frac{m}{a}[a n]\right]-[a n] \\
=\left[\frac{n}{a}+\frac{m}{a}(a n-\{a n\})\right]-[a n] \\
=\left[n\left(\frac{1}{a}+m\right)-\frac{m}{a}\{a n\}\right]-[a n]=\left[n a-\frac{m}{a}\{a n\}\right]-[a n] \\
=\left[n a-[n a]-\frac{m}{a}\{a n\}\right] \\
=\left[\{n a\}-\frac{m}{a}\{a n\}\right]=\left[\frac{a-m}{a} \cdot\{a n\}\right] .
\end{array}$$
Since $a>m>0$, we have $0<\frac{a-m}{a}<1$.
Also, $0 \leqslant\{n a\}<1$, so $0 \leqslant \frac{a-m}{a} \cdot\{n a\}<1$.
Thus, $\left[\frac{a-m}{a}\{n a\}\right]=0$, so when $n$ is a positive integer greater than $a$,
$$[a n+m a[a n]]-m n-\left(m^{2}+1\right)[a n]=0,$$
i.e., $[a n+m a[a n]]=m n+\left(m^{2}+1\right)[a n]$.
Since there are infinitely many positive integers greater than $a$, the original proposition is proved.
|
proof
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4. $n \in \mathbf{N}_{+}, n \geqslant 2$, then $\sum_{k=2}^{n}[\sqrt[k]{n}]=\sum_{k=2}^{n}\left[\log _{k} n\right]$.
|
Prove the construction of the planar region $D=\left\{(x, y) \mid y^{x} \leqslant n, x \geqslant 2, y \geqslant 2\right\}$, and consider the number of integer points in $D$.
If counted column by column, when $x=2$, there are $[\sqrt{n}]$ points, when $x=3$, there are $[\sqrt[3]{n}]$ points, $\cdots$, when $x=n$, there are $[\sqrt[n]{n}]$ points, so there are a total of $\sum_{k=2}^{n}[\sqrt[k]{n}]$ points.
If counted row by row, when $y=2$, there are $\left[\log _{2} n\right]$ points, when $y=3$, there are $\left[\log _{3} n\right]$ points, $\cdots$, when $y=n$, there are $\left[\log _{n} n\right]$ points, so there are a total of $\sum_{k=2}^{n}\left[\log _{k} n\right]$ points.
In summary, the original equation is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Let $p_{1}, p_{2}, p_{3}, \cdots$ be the prime numbers in increasing order, and let $x_{0}$ be a real number between 0 and 1. For a positive integer $k$, define
$$x_{k}=\left\{\begin{array}{ll}
0, & \text { if } x_{k-1}=0 \\
\left\{\frac{p_{k}}{x_{k-1}}\right\} . & \text { if } x_{k-1} \neq 0
\end{array}\right.$$
Here $\{x\}$ denotes the fractional part of $x$. Find all $x_{0}$ such that $0<x_{0}<1$ and the sequence $x_{0}, x_{1}, x_{2}, \cdots$ eventually contains 0, and prove your answer.
|
Analysis When $x$ is an irrational number, $\frac{p}{x}$ is also an irrational number $(p \in \mathbf{Q}$ and $p \neq 0)$, thus $\left\{\frac{p}{x}\right\}$ is also an irrational number. Therefore, we can conjecture that $x_{0}$ can take all rational numbers.
Solution We prove that the sequence eventually appears 0 if and only if $x_{0}$ is a rational number.
First, we prove that for $k \geqslant 1$, if $x_{k}$ is a rational number, then its previous term $x_{k-1}$ is also a rational number.
If $x_{k}=0$, then by the problem's condition, $x_{k-1}=0$ or $\frac{p_{k}}{x_{k-1}} \in \mathbf{N}_{+}$, thus $x_{k-1}$ is a rational number.
If $x_{k} \neq 0, x_{k} \in \mathbf{Q}$, then by
$x_{k}=\left\{\frac{p_{k}}{x_{k-1}}\right\}=\frac{p_{k}}{x_{k-1}}-\left[\frac{p_{k}}{x_{k-1}}\right]$, we get $x_{k-1}=\frac{p_{k}}{x_{k}+\left[\frac{p_{k}}{x_{k-1}}\right]}$,
thus $x_{k-1}$ is also a rational number.
In summary, for some $k$, if $x_{k}$ is a rational number, especially if the sequence eventually appears 0, then $x_{0}$ is also a rational number. Now, assume $x_{0}$ is a rational number, then the sequence $x_{0}, x_{1}, x_{2}, \cdots$ consists of rational numbers.
If $x_{k-1}=\frac{m}{n}, 0<m<n$, then $x_{k}=\left\{\frac{p_{k}}{\frac{m}{n}}\right\}=\left\{\frac{n \cdot p_{k}}{m}\right\}=\frac{r}{m}, 0<r<m$.
Here, $r$ is the remainder when $n p_{k}$ is divided by $m$.
Therefore, the denominator of each non-zero term in the sequence $\left\{x_{k}\right\}$ is strictly less than the denominator of the previous term, so the number of non-zero terms does not exceed the denominator of $x_{0}$.
Thus, the sequence $\left\{x_{n}\right\}$ must eventually appear 0.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 8 For real numbers $x, y$, let $s(x, y)=\{s \mid s=[n x+y], n \in \mathbf{N}\}$. Prove that if $r>1$ is a rational number, then there exist real numbers $u, v$, such that
$$s(r, 0) \cap s(u, v)=\varnothing, s(r, 0) \cup s(u, v)=\mathbf{N}$$
|
Proof: Let $r=\frac{p}{q}, p, q \in \mathbf{N}, p>q$, then
$u=\frac{p}{p-q}$ and satisfies $-\frac{1}{p-q} \leqslant vq$, then when $v(p-q)m$, $[n r]>[m r],[n u+v]>[m u+v]$.
Therefore, there are no identical elements in $s(u, v)$.
Thus, the number of elements in $s(u, v) \bigcap\{1,2, \cdots, k-1\}$ equals $[m u+v]-\frac{k-v}{u}$.
So, $m_{1}=-\left[\frac{u+v-k}{u}\right]$.
Similarly, the number of elements in $s(r, 0) \cap\{1,2, \cdots, k-1\}$, $n_{1}$, satisfies $n_{1}=-\left[\frac{r-k}{r}\right]$.
Thus, $-m_{1}-n_{1} \leqslant \frac{u+v-k}{u}+\frac{r-k}{r}=2-k+\frac{v(p-q)}{p}k-2$, i.e., $m_{1}+n_{1} \geqslant k-1$.
Since $s(u, v) \cap s(r, 0)=\varnothing$, we have
$$(s(u, v) \cup s(r, 0)) \cap\{1,2, \cdots, k-1\}=\{1,2, \cdots, k-1\}$$
This equation holds for all $k$.
Therefore, $s(u, v) \bigcup s(r, 0)=\mathbf{N}$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4. Let $m$ be a positive integer, and the number of factor 2s in $m!$ is denoted as $n(m)$. Prove that there exists a natural number $m > 2006^{2006}$, such that $m = 3^{2006} + n(m)$.
|
4. Proof: Let $m=2^{a}-1$, then the number of factor 2 in $m!$ is $\left[\frac{2^{a}-1}{2}\right]+\left[\frac{2^{a}-1}{2^{2}}\right]+\cdots+\left[\frac{2^{a}-1}{2^{a}}\right]$
$$\begin{array}{l}
=\left(2^{a-1}-1\right)+\left(2^{a-2}-1\right)+\cdots+(1-1) \\
=\left(2^{a-1}+2^{a-2}+\cdots+1\right)-a=2^{a}-1-a
\end{array}$$
That is, $n(m)=2^{a}-1-a$.
Take $a=3^{2006}$, then $m=2^{a}-1=\left(2^{a}-1-a\right)+a=3^{2006}+n(m)$, so it only needs to prove that $m>2006^{2006}$.
In fact, $m=2^{3^{2006}}-1>2^{3^{2005}}=2^{3^{2002} \cdot 3^{3}}=\left(2^{27}\right)^{3^{2002}}>2006^{3^{2002}}>2006^{2006}$. Hence the problem is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 1 (1) Let $n$ be a positive integer, prove: $(21 n+4,14 n+3)=1$; (2) Prove: $(n!+1,(n+1)!+$ $1)=1$; (3) Let $F_{k}=2^{2^{k}}+1, k \geqslant 0$, prove: for $m \neq n$, $(F_{m}, F_{n})=1$.
|
(1) and (2) can be solved using the Euclidean algorithm, while (3) requires the use of Bézout's identity.
Proof (1) By the Euclidean algorithm, we get
$$\begin{aligned}
(21 n+4,14 n+3) & =(21 n+4-14 n-3,14 n+3) \\
& =(7 n+1,14 n+3)=(7 n+1,14 n+3-2(7 n+1)) \\
& =(7 n+1,1)=1
\end{aligned}$$
(2) Since $(n+1,(n+1)!+1)=1$,
we have $(n!+1,(n+1)!+1)=((n!+1)(n+1),(n+1)!+1)$
$$\begin{array}{l}
=((n!+1)(n+1)-(n+1)!-1,(n+1)!+1) \\
=(n,(n+1)!+1) .
\end{array}$$
Let $d$ be the greatest common divisor of $n!+1$ and $(n+1)!+1$,
from the above, we know $d \mid n$, i.e., $d \mid n!$.
Since $d \mid(n!+1)$, it follows that $d=1$.
Thus, $(n!+1,(n+1)!+1)=1$.
(3) From the previous example 2, we know that for $m>n$,
$F_{n} \mid\left(F_{m}-2\right)$,
i.e., there exists an integer $x$ such that $F_{m}-2=x F_{n}$,
i.e., $F_{m}-x \cdot F_{n}=2$.
Let $d=\left(F_{m}, F_{n}\right)$, from $F_{m}-x F_{n}=2$ we deduce $d \mid 2$, so $d=1$ or 2, but $F_{n}$ is clearly odd.
Therefore, $d=1$, i.e., $\left(F_{m}, F_{n}\right)=1$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let $a$ be a given positive integer, and $A$ and $B$ be two real numbers. Determine the necessary and sufficient condition for the system of equations
$$\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}=(13 a)^{2} \\
x^{2}\left(A x^{2}+B y^{2}\right)+y^{2}\left(A y^{2}+B z^{2}\right)+z^{2}\left(A z^{2}+B x^{2}\right)=\frac{1}{4}(2 A+B)(13 a)^{4}
\end{array}\right.$$
to have positive integer solutions (express the condition in terms of a relationship between $A$ and $B$, and provide a proof).
|
The necessary and sufficient condition for the original system of equations to have positive integer solutions is $B=2A$. Below is the proof of this conclusion.
First, we prove the sufficiency, i.e., given $B=2A$, we prove that the original system of equations has positive integer solutions.
In this case, we can take $x=12a$, $y=3a$, $z=4a$, which is a set of positive integer solutions for the original system of equations.
Next, we prove the necessity: substituting (1) into (2) yields:
\[ x^{2}(A x^{2} + B y^{2}) + y^{2}(A y^{2} + B z^{2}) + z^{2}(A z^{2} + B x^{2}) = \frac{1}{4}(2A + B)(x^{2} + y^{2} + z^{2})^{2}, \]
which simplifies to:
\[ (2A - B)(x^{4} + y^{4} + z^{4} - 2x^{2}y^{2} - 2y^{2}z^{2} - 2z^{2}x^{2}) = 0. \]
If $2A \neq B$, then:
\[ x^{4} + y^{4} + z^{4} - 2x^{2}y^{2} - 2y^{2}z^{2} - 2z^{2}x^{2} = 0, \]
which can be rewritten as:
\[ (x + y + z)(x + y - z)(x + z - y)(y + z - x) = 0. \]
Assume $x + y - z = 0$, substituting into (1) gives:
\[ x^{2} + y^{2} + (x + y)^{2} = (13a)^{2}. \]
If there exists a set of $x_{0}, y_{0}, a_{0} (x_{0}, y_{0}, a_{0} \in \mathbf{N}_{+})$ that satisfies (3), it is clear that $a_{0}$ is even. Let $a_{0} = 2a'_{0} (a_{0} \in \mathbf{N}_{+})$, then:
\[ 4 \mid x_{0}^{2} + y_{0}^{2} + (x_{0} + y_{0})^{2}, \]
which implies that $x_{0}$ and $y_{0}$ are both even (otherwise, $x_{0}^{2} + y_{0}^{2} + (x_{0} + y_{0})^{2} \equiv 2 \pmod{4}$). Let $x_{0} = 2x'_{0}$, $y_{0} = 2y'_{0} (x_{0}, y_{0} \in \mathbf{N}_{+})$, then:
\[ x'_{0}^{2} + y'_{0}^{2} + (x'_{0} + y'_{0})^{2} = (13a'_{0})^{2}, \]
which means $\frac{x_{0}}{2}, \frac{y_{0}}{2}, \frac{a_{0}}{2}$ is also a set of positive integer solutions for (3). Repeating this process indefinitely, we find that for any $k \in \mathbf{N}_{+}$, $2^{k} \mid a_{0}$, but $a_{0} \in \mathbf{N}_{+}$, which is clearly impossible.
Therefore, if the original system of equations has positive integer solutions, then $2A = B$ must hold.
In summary, the necessary and sufficient condition for the original system of equations to have positive integer solutions is $B = 2A$.
|
B = 2A
|
Algebra
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2. Find the integer solutions of the equation $x^{2}+y^{2}+z^{2}=x^{2} y^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
2. Solution: $x_{0}=y_{0}=z_{0}$ is a solution to the equation. Suppose the equation has another solution $(x, y, z)$. If $x, y, z$ include three or two odd numbers, then $x^{2}+y^{2}+z^{2} \equiv 2$ or $3(\bmod 4)$, while $x^{2} y^{2} \equiv 0$ or $1(\bmod 4)$, which is obviously impossible; if $x, y, z$ include exactly one odd number, then at least one of $x, y$ is even, so $x^{2} y^{2} \equiv 0(\bmod 4)$, but $x^{2}+y^{2}+z^{2} \equiv 1$ ( $\bmod 4)$, a contradiction, therefore, $x, y, z$ must all be even.
Let $x_{1}=\frac{x}{2}, y_{1}=\frac{y}{2}, z_{1}=\frac{z}{2}\left(x_{1}, y_{1}, z_{1}\right.$ be integers), then we have $\left(2 x_{1}\right)^{2}+\left(2 y_{1}\right)^{2}+\left(2 z_{1}\right)^{2}=$ $\left(4 x_{1} y_{1}\right)^{2}$, i.e., $x_{1}^{2}+y_{1}^{2}+z_{1}^{2}=4 x_{1}^{2} y_{1}^{2}$. The right side of this equation is divisible by 4, so $x_{1}, y_{1}, z_{1}$ must all be even (otherwise $x_{1}^{2}+y_{1}^{2}+z_{1}^{2} \equiv 1$ or 2 or $3(\bmod 4)$, a contradiction). Thus, we can set $x_{2}=\frac{x_{1}}{2}, y_{2}=\frac{y_{1}}{2}, z_{2}=\frac{z_{1}}{2}$, then $x_{2}, y_{2}, z_{2}$ are all integers and $\left(2 x_{2}\right)^{2}+\left(2 y_{2}\right)^{2}+\left(2 z_{2}\right)^{2}=4 \cdot\left(4 x_{1} y_{1}\right)^{2}$, i.e., $x_{2}^{2}+y_{2}^{2}+z_{2}^{2}=16 x_{2}^{2} y_{2}^{2}$.
Continuing this discussion, we can obtain three infinite integer sequences $\left\{x_{n}\right\},\left\{y_{n}\right\},\left\{z_{n}\right\}$, where $z_{k}=\frac{1}{2} z_{k-1}, x_{k}=$ $\frac{1}{2} x_{k-1}, y_{k}=\frac{1}{2} y_{k-1}(k \geqslant 2), x_{1}=\frac{1}{2} x, y_{1}=\frac{1}{2} y, z_{1}=\frac{1}{2} z$. However, no non-zero integer can be divisible by all powers of 2, so $x, y, z$ must all be 0, which contradicts the assumption. Therefore, the original equation has only $x_{0}=y_{0}=z_{0}=0$ as the unique integer solution.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
4. Let positive integers $a, b, k$ satisfy $\frac{a^{2}+b^{2}}{a b-1}=k$, prove that: $k=5$.
|
4. Prove: If $a=b$, then $k=\frac{2 a^{2}}{a^{2}-1}=2+\frac{2}{a^{2}-1}$.
Then $\left(a^{2}-1\right) \mid 2$ and $a^{2}-1>0$.
So $a^{2}-1=1$ or 2, but this contradicts $a \in \mathbf{N}_{+}$.
Therefore, $a \neq b$, without loss of generality, let $a>b$.
When $b=1$, $k=\frac{a^{2}+1}{a-1}=a+1+\frac{2}{a-1}$, then $(a-1) \mid 2$ and $a \geqslant b+1=2$, so $a=2$ or 3, in either case, $k=5$.
Without loss of generality, assume $(a, b)$ is the ordered pair that minimizes $a+b$ among all ordered pairs satisfying $\frac{a^{2}+b^{2}}{a b-1}=k$.
From $\frac{a^{2}+b^{2}}{a b-1}=k$, we get $a^{2}-k b \cdot a+b^{2}+k=0$.
Viewing (1) as a quadratic equation in $a$, by Vieta's formulas, the roots are $a$ and $(k b-a)$, and $k b-a = \frac{b^{2}+k}{a} > 0$, and since $k b-a \in \mathbf{Z}$, we have $k b-a \in \mathbf{N}_{+}$, and $\frac{(k b-a)^{2}+b^{2}}{(k b-a) b-1}=k$. Thus, by the definition of $(a, b)$, we have $a+b \leqslant \frac{b^{2}+k}{a}+b\left(\frac{b^{2}+k}{a}=k b-a\right)$, so $a^{2}-b^{2} \leqslant k=\frac{a^{2}+b^{2}}{a b-1}$.
Therefore, $a^{2}-b^{2}a-b \geqslant 1$. That is, $(a-1)(b-1)b=2$, a contradiction. In conclusion, $k=5$.
|
5
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 On the coordinate plane, if the coordinates $x_{0}, y_{0}$ of a point $\left(x_{0}, y_{0}\right)$ are both integers, then the point is called an integer point. Try to prove: on the coordinate plane, there does not exist a regular $n$-gon $(n \geqslant 7)$, such that all its vertices are integer points.
|
Proof Assume there exists a regular $n$-sided polygon $A_{1} A_{2} A_{3} \cdots A_{n}(n \geqslant 7)$, whose vertices $A_{1}, A_{2}, \cdots, A_{n}$ are all integer points.
Take any integer point $M$ in the coordinate plane, and draw vectors $\overrightarrow{M B_{1}}, \overrightarrow{M B_{2}}, \overrightarrow{M B_{3}}, \cdots, \overrightarrow{M B_{n}}$ such that $\overrightarrow{M B_{1}}=\overrightarrow{A_{1} \vec{A}_{2}}, \overrightarrow{M B_{2}}=\overrightarrow{A_{2} A_{3}}, \overrightarrow{M B_{3}}=\overrightarrow{A_{3} A_{4}}, \cdots, \overrightarrow{M B_{n-1}}=\overrightarrow{A_{n-1} A_{n}}, \overrightarrow{M B_{n}}=\overrightarrow{A_{n} \overrightarrow{A_{1}}}$ (as shown in Figure 6-1).
Since $A_{1}, A_{2}, \cdots, A_{n}, M$ are all integer points, by vector coordinate operations, it can be known that $B_{1}, B_{2}, \cdots, B_{n}$ are all integer points.
It is also easy to see that $\left|\overrightarrow{M B_{i}}\right|=\left|\overrightarrow{M B_{i+1}}\right|\left(i=1,2, \cdots, n\right.$, define $\left.B_{n+1}=B_{1}\right)$, and the angle between $\overrightarrow{M B_{i}}$ and $\overrightarrow{M B_{i+1}}$ is $\frac{2 \pi}{n}$.
Therefore, the $n$-sided polygon $B_{1} B_{2} \cdots B_{n}$ is a regular $n$-sided polygon.
It is easy to see that the $n$-sided polygon $B_{1} B_{2} \cdots B_{n}$ is similar to the $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, with the similarity ratio being $\frac{\left|B_{1} B_{2}\right|}{\left|A_{1} A_{2}\right|}=\frac{\left|B_{1} B_{2}\right|}{\left|M B_{1}\right|}$, and since $\angle B_{1} M B_{2}=\frac{2 \pi}{n}$, then $\frac{\left|B_{1} B_{2}\right|}{\left|M B_{1}\right|}=2 \cos \frac{\pi}{n}$.
So $B_{1} B_{2}=2 \cos \frac{\pi}{n} \cdot\left|A_{1} A_{2}\right| \leqslant 2 \cos \frac{\pi}{7}\left|A_{1} A_{2}\right|$.
This process can be repeated indefinitely, i.e., from the original regular $n$-sided polygon, a series of regular $n$-sided polygons can be obtained, each with a side length no greater than the previous regular $n$-sided polygon's side length multiplied by $2 \cos \frac{\pi}{7}$, i.e., for any $m \in \mathbf{N}_{+}$, there exists a convex $n$-sided polygon with all vertices being integer points, whose side length is no greater than $\left(2 \cos \frac{\pi}{7}\right)^{m} \cdot\left|A_{1} A_{2}\right|$, but $2 \cos \frac{\pi}{7}<1$, so when $m \rightarrow +\infty$, $\left(2 \cos \frac{\pi}{7}\right)^{m} \cdot\left|A_{1} A_{2}\right| \rightarrow 0$, while the distance between any two different integer points is at least 1, i.e., the side length of the regular $n$-sided polygon is no less than 1, leading to a contradiction.
Therefore, there does not exist a regular $n$-sided polygon $(n \geqslant 7)$ whose vertices are all integer points.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Assume that $2 n+1$ positive integers $a_{1}, a_{2}, \cdots, a_{2 n+1}$ have the property: any $2 n$ of these numbers can be divided into two disjoint $n$-element subsets, and the sums of the $n$ numbers in the two subsets are equal, then these $2 n+1$ numbers must all be equal.
|
To prove that if there are $2n+1$ positive integers satisfying the property described in the problem, we call these $2n+1$ positive integers a "good array". Let's assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2 n+1}$, and $a_{1}=1$ (because reducing the smallest number to 1 and subtracting the same number from the other numbers still results in a "good array").
From the problem statement, we know that $2 \mid\left(\left(\sum_{i=1}^{2 n+1} a_{i}\right)-a_{j}\right)(j=1,2,3, \cdots, 2 n+1)$, which means $a_{1} \equiv a_{2} \equiv a_{3} \equiv a_{4} \equiv \cdots \equiv a_{2 n+1} \equiv \sum_{i=1}^{2 n+1} a_{i}(\bmod 2)$.
Since $a_{1}=1$, it follows that $a_{1}, a_{2}, a_{3}, \cdots, a_{2 n+1}$ are all positive odd numbers.
Let $b_{i}=\frac{a_{i}+1}{2}(i=1,2, \cdots, 2 n+1)$. Since $a_{1}, a_{2}, \cdots, a_{2 n+1}$ are all positive odd numbers, $b_{1}, b_{2}, b_{3}, \cdots, b_{2 n+1}$ are all positive integers and it is easy to see that $b_{1}, b_{2}, b_{3}, \cdots, b_{2 n+1}$ also form a "good array".
If $x > 1$, then $b_{2 n+1}=\frac{a_{2 n+1}+1}{2}<a_{2 n+1}$. Repeatedly performing this transformation, we will eventually obtain a "good array" whose maximum term is 1. Since in the above transformation, the smallest term in each array is always 1, the terms in this "good array" are all 1. Considering the way the transformation is performed (changing $x$ to $\frac{x+1}{2}$), it follows that the original $2n+1$ numbers $a_{1}, a_{2}, \cdots, a_{2 n+1}$ are all 1.
This completes the proof of the original proposition.
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Let $f(x)=x^{2}+x+p, p \in \mathbf{N}$, prove: If $f(0), f(1), f(2), \cdots, f\left(\left[\sqrt{\frac{p}{3}}\right]\right)$ are prime numbers, then the numbers $f(0), f(1), \cdots, f(p-2)$ are all prime numbers.
|
Analysis If there exists some $m \in\{0,1,2, \cdots, p-2\}$, such that $m^{2}+m+p$ is a composite number, and we can find a way to decrease $m$, i.e., find a smaller number $m^{\prime}\left(m^{\prime} \in\{0,1,2, \cdots, p-2\}\right)$, such that $f\left(m^{\prime}\right)$ is a composite number, then by the method of infinite descent, we can derive a contradiction.
Proof When $0 \leqslant x \leqslant p-2$, $p \leqslant x^{2}+x+p \leqslant(p-2)^{2}+(p-2)+p=(p-1)^{2}+1$.
Let $x_{0}$ be the smallest natural number $x$ such that $f(x)=x^{2}+x+p$ is a composite number.
If $0 \leqslant x_{0} \leqslant p-2$, then $x_{0}^{2}+x_{0}+p \leqslant(p-1)^{2}+1$,
thus the smallest prime factor $d_{0} \leqslant p-1$.
Let $x_{0}^{2}+x_{0}+p=d_{0} m\left(m \in \mathbf{N}_{+}\right)$.
(1) If $x_{0} \geqslant d_{0}$, let $x^{\prime}=x_{0}-d_{0}$,
then $x^{\prime 2}+x^{\prime}+p=\left(x_{0}-d_{0}\right)^{2}+\left(x_{0}-d_{0}\right)+p$
$$\begin{array}{l}
=x_{0}^{2}+x_{0}+p-d_{0}\left(2 x_{0}-d_{0}+1\right) \\
=d_{0}\left(m-2 x_{0}+d_{0}-1\right)
\end{array}$$
Since $x^{\prime 2}+x^{\prime}+p \geqslant p>p-1 \geqslant d_{0}$, it follows that $m-2 x_{0}+d_{0}-1>1$.
Thus, $d_{0}\left(m-2 x_{0}+d_{0}-1\right)$ is a composite number, i.e., $f\left(x^{\prime}\right)$ is a composite number, and $0 \leqslant x^{\prime}=x_{0}-d_{0} \leqslant p-2$, then by the definition of $x_{0}$, we know that $f(1), f(2), \cdots, f(p-2)$ are all prime numbers.
The original proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 8 (29th Russian Mathematical Olympiad) The sequence of positive integers $\left\{a_{n}\right\}$ is constructed as follows: $a_{0}$ is some positive integer, if $a_{n}$ is divisible by 5, then $a_{n+1}=\frac{a_{n}}{5}$; if $a_{n}$ is not divisible by 5, then $a_{n+1}=\left[\sqrt{5} a_{n}\right]$ (where $[x]$ denotes the greatest integer not exceeding $x$). Prove: The sequence $\left\{a_{n}\right\}$ is increasing from some term onwards.
|
Prove that for any positive integer $x$, we have $[\sqrt{5} x] \geqslant x+1$.
This is because when $x \in \mathbf{N}_{+}$, $\sqrt{5} x > 2 x \geqslant x+1$.
Therefore, $[\sqrt{5} x] \geqslant [x+1] = x+1$.
From this, we can see that the conclusion to be proven is equivalent to: there exists some positive integer $m$, such that for all $n \geqslant m, n \in \mathbf{N}_{+}, a_{n}$ is not a multiple of 5 (at this time $a_{n+1} = [\sqrt{5} a_{n}] > a_{n}$). Below, we will prove this.
To do this, we first prove that the sequence $\{a_{n}\}$ contains two consecutive terms that are not multiples of 5.
We use proof by contradiction to prove this. Assume that in the sequence $\{a_{n}\}$, any two consecutive terms have at least one that is a multiple of 5.
Consider $a_{1}, a_{2}, a_{3}$.
(1) If $5 \mid a_{1}$, then $a_{2} = \frac{a_{1}}{5}$, and by the recurrence relation of $\{a_{n}\}$, we know $a_{3} \leqslant [\sqrt{5} a_{2}]$, so $a_{3} \leqslant [\frac{\sqrt{5}}{5} a_{1}]$, i.e., $a_{3} \leqslant a_{1} - 1$.
(2) If $5 \nmid a_{1}$, then by the assumption in (1), $5 \mid a_{2}$, so $a_{2} = [\sqrt{5} a_{1}], a_{3} = \frac{a_{2}}{5}$, thus $a_{3} = \frac{[\sqrt{5} a_{1}]}{5} < a_{1}$, i.e., $a_{3} \leqslant a_{1} - 1$. $\square$
Combining (1) and (2), we see that in either case, $a_{3} \leqslant a_{1} - 1$, i.e., starting from $a_{1}$, we can find a term $a_{3}$ in the sequence $\{a_{n}\}$ such that $a_{3} \leqslant a_{1} - 1$. Similarly, starting from $a_{3}$, we can find a term $a_{5}$ in the sequence $\{a_{n}\}$ such that $a_{5} \leqslant a_{3} - 1, \cdots \cdots$, and this process can be repeated indefinitely, each new term being at least 1 less than the previous one. Since all terms in the sequence $\{a_{n}\}$ are positive integers, this process cannot continue indefinitely, leading to a contradiction.
Therefore, the assumption in (1) is false.
Thus, we have proven that the sequence $\{a_{n}\}$ must contain two consecutive terms that are not multiples of 5.
Hence, we can find two consecutive terms $a_{k}$ and $a_{k+1}$ that are not multiples of 5.
We now prove a lemma:
Lemma: If $a_{n}$ and $a_{n+1}$ are not multiples of 5, then $a_{n+2}$ is also not a multiple of 5.
Proof of the lemma: Since $a_{n}$ is not a multiple of 5, then by the problem statement,
$a_{n+1} = [\sqrt{5} a_{n}], a_{n+2} = [\sqrt{5} a_{n+1}]$.
Let $a_{n+1} = \sqrt{5} a_{n} - \alpha$, where $0 < \alpha < 1$.
Then $a_{n+2} = [\sqrt{5} a_{n+1}] = [\sqrt{5} (\sqrt{5} a_{n} - \alpha)] = 5 a_{n} + [-\sqrt{5} \alpha]$.
Since $0 < \alpha < 1$,
we have $-3 < -\sqrt{5} \alpha < 0$, so $[-\sqrt{5} \alpha] = -3$ or -2 or -1.
Thus, $a_{n+2} = [-\sqrt{5} \alpha] \equiv -3$ or -2 or $-1 \pmod{5}$.
Therefore, $a_{n+2}$ is not a multiple of 5, and the lemma is proven.
Using the lemma and the fact that $a_{k}$ and $a_{k+1}$ are not multiples of 5, we can conclude that for $n \geqslant k$, $a_{n}$ is not a multiple of 5. Combining this with the initial derivation, we have proven the original proposition.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2. Given $2 n$ points in the plane, no three of which are collinear, $n$ points represent farms $F=\left\{F_{1}, F_{2}, \cdots, F_{n}\right\}$, and the other $n$ points represent reservoirs $W=\left\{W_{1}, W_{2}, \cdots, W_{n}\right\}$. Each farm is connected to a reservoir by a straight road. Prove that there exists an assignment such that all roads do not intersect.
|
2. Proof: For any allocation method, if there exist two roads $F_{i} W_{m}$ and $F_{k} W_{j}$ that intersect (as shown in Figure 2 of Question 2), where $1 \leqslant i, k \leqslant n, 1 \leqslant j, m \leqslant n$ and $i \neq k, j \neq m$.
In this case, we make the following adjustment: remove roads $F_{i} W_{m}$ and $F_{k} W_{j}$, and replace them with $F_{k} W_{m}$ and $F_{i} W_{j}$. At this point, we have $F_{i} W_{m} + F_{k} W_{j} > F_{i} W_{j} + F_{k} W_{m}$ (since no three points among $F_{1}, F_{2}$, $\cdots, F_{n}, W_{1}, W_{2}, \cdots, W_{n}$ are collinear, the inequality must be strict). If there are still two roads intersecting after the adjustment, we make the same adjustment again. Each adjustment strictly reduces the total length of the $n$ roads, and there are $n!$ possible allocation methods (a finite number), so the decreasing process cannot continue indefinitely. It must stop after a certain number of adjustments, at which point the resulting allocation method will have no intersections between the $n$ roads.
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5. Let there be $2 n$ integers $a_{1}, a_{2}, \cdots, a_{2 n}$, and if any one of these numbers is taken out, the remaining $2 n-1$ numbers can always be divided into two disjoint subsets, such that the sum of the numbers in each subset is equal, then $a_{1}, a_{2}, \cdots, a_{2 n}$ must all be zero. (Here, "two disjoint subsets" means that for any remaining $a_{i}$, it belongs to exactly one subset).
|
5. Proof: According to the problem, we can know that $\sum_{i \neq k} a_{i} (k=1,2,3, \cdots, 2n)$ must be even. From this, we can conclude that for any $k \in \{1,2,3, \cdots, 2n\}, \sum_{i=1}^{2n} a_{i}$ and $a_{k}$ have the same parity.
Obviously, $a_{1}, a_{2}, \cdots, a_{2n}$ must all be even (otherwise, $a_{1}, a_{2}, \cdots, a_{2n}$ would all be odd, but in this case, $\sum_{i=1}^{2n} a_{i}$ would be even, which contradicts the previous conclusion).
In this way, $\frac{a_{1}}{2}, \frac{a_{2}}{2}, \cdots, \frac{a_{2n}}{2}$ are all integers, and they still satisfy the property described in the problem.
Using the same method, we can prove that $\frac{a_{1}}{2^{2}}, \frac{a_{2}}{2^{2}}, \cdots, \frac{a_{2n}}{2^{2}}$ are all integers; $\frac{a_{1}}{2^{3}}, \frac{a_{2}}{2^{3}}, \cdots, \frac{a_{2n}}{2^{3}}$ are all integers, and so on.
Repeating the above discussion, we can see that for any positive integer $m$, $\frac{a_{1}}{2^{m}}, \frac{a_{2}}{2^{m}}, \cdots, \frac{a_{2n}}{2^{m}}$ are all integers.
From this, we can conclude that $a_{1}=a_{2}=\cdots=a_{2n}=0$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
6. Prove: The area of a Pythagorean triangle cannot be a perfect square.
|
6. Proof: First, we provide a lemma.
Lemma: The system of equations $\left\{\begin{array}{l}x^{2}+y^{2}=z^{2}, \\ x^{2}-y^{2}=w^{2}\end{array}\right.$ has no positive integer solutions.
The proof of the lemma is omitted (refer to Chapter 6, Section 1, Exercise 5).
We now prove the original proposition:
By contradiction, assume there exists a right triangle with an area that is a perfect square.
Let the side lengths of such a triangle be $a, b, c$. (where $a, b$ are the lengths of the two legs, and $c$ is the length of the hypotenuse), then
$a^{2}+b^{2}=c^{2}$ and $\frac{1}{2} a b$ is a perfect square.
By the properties of Pythagorean triples, there exist $u, v \in \mathbf{N}_{+}$ such that $c=u^{2}+v^{2}$, and without loss of generality, assume $a=u^{2}-v^{2}, b=2 u v$ (we can assume $a, b, c$ are pairwise coprime, which leads to these expressions, and $(u, v)=1$ and $u$ and $v$ are coprime).
Then the area of the right triangle is $u v\left(u^{2}-v^{2}\right)$, i.e., $u v(u-v)(u+v)$ is a perfect square.
Since $u, v$ are coprime and one of $u, v$ is odd and the other is even, $u, v, u+v, u-v$ are pairwise coprime.
Thus, $u, v, u+v, u-v$ must all be perfect squares.
Let $u=s^{2}, v=t^{2}, u+v=p^{2}, u-v=q^{2}$ (where $s, t, p, q \in \mathbf{N}_{+}$).
Then we have the following system of equations $\left\{\begin{array}{l}s^{2}+t^{2}=p^{2}, \\ s^{2}-t^{2}=q^{2} .\end{array}\right.$
Using the lemma, we reach a contradiction. Therefore, there does not exist a right triangle with an area that is a perfect square.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 Please design a method to color all integer points, each point being colored one of white, red, or black, such that: (1) Points of each color appear on infinitely many lines parallel to the x-axis.
(2) For any white point $A$, red point $B$, and black point $C$, there always exists a red point $D$ such that $A B C D$ forms a parallelogram.
|
Analysis: Color the grid points according to their classification to satisfy condition (1). Label the points $A$, $B$, and $C$ with coordinates, and use the property that the diagonals of a parallelogram bisect each other and the midpoint formula to find the coordinates of $D$. This should yield a parallelogram $ABCD$ that satisfies condition (2).
Solution: Design the coloring method as follows: color (odd, odd) class grid points white, (even, even) class grid points black, and (odd, even) and (even, odd) class grid points red. It is easy to see that this coloring method satisfies condition (1).
Take any white point $A(2m+1, 2n+1)$, black point $C(2s, 2t)$, and red point $B(2p, 2q+1)$, where $m, n, s, t, p, q$ are all integers. Let the coordinates of point $D$ in parallelogram $ABCD$ be $(x, y)$.
By the property that the diagonals of a parallelogram bisect each other and the midpoint formula, we get
$$\left\{\begin{array}{l}
x + 2p = 2m + 1 + 2s, \\
y + 2q + 1 = 2n + 1 + 2t.
\end{array}\right.$$
Solving for $x$ and $y$, we get $x = 2m + 1 + 2s - 2p$, $y = 2(n + t - q)$.
Thus, $D(x, y)$ is an (odd, even) class grid point, and the color it is painted is red.
Similarly, if point $B$ is an (odd, even) class grid point, it can be deduced that $D$ is an (even, odd) class grid point, and the color it is painted is also red. Therefore, this design also satisfies (2).
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 Take any 6 lattice points $p_{i}\left(x_{i}, y_{i}\right)(i=1,2,3,4,5,6)$, satisfying
(1) $\left|x_{i}\right| \leqslant 2,\left|y_{i}\right| \leqslant 2(i=1,2,3,4,5,6)$;
(2) No three points are collinear.
Try to prove: Among all triangles with $p_{i}(i=1,2,3,4,5,6)$ as vertices, there must be one triangle whose area is no more than 2.
|
Proof Assume there exist 6 lattice points $p_{1}, p_{2}, \cdots, p_{6}$ within the region $S=\{(x, y)|| x|\leqslant 2| y \mid, \leqslant 2\}$, and any 3 of these points form a triangle with an area greater than 2. Let $p=\left\{p_{1}, p_{2}, \cdots, p_{6}\right\}$.
(1) If the number of points in $p$ on the $x$-axis is less than 2, then by the pigeonhole principle, the upper half-plane (or lower half-plane, not including the $x$-axis) must contain at least 3 points from $p$. The area of the triangle formed by these 3 points is no more than 2, leading to a contradiction. Therefore, the $x$-axis must have exactly 2 points from $p$ (since 3 points cannot be collinear). The remaining 4 points in $p$ cannot have any point on the line $y= \pm 1$, otherwise, a triangle with an area no more than 2 would be formed, which is a contradiction. This proves that on the lines $y=2$ and $y=-2$, there are exactly 2 points from $p$. Considering the symmetry of $S$, similarly, it can be proven that the lines $x=-2, x=0, x=2$ each have 2 points from $p$. Thus, on each line $y=2 j, x=2 j(j=0, \pm 1)$, there are exactly 2 points from $p$.
(2) $p$ cannot include the origin, otherwise, since all lattice points within $S$ with even coordinates lie on 4 lines passing through the origin, by the pigeonhole principle, the remaining 5 points in $p$ must have at least 2 points on one of these lines, leading to 3 collinear points, which is a contradiction.
Therefore, the two points of $p$ on the $x$-axis must be $(-2,0)$ and $(2,0)$. Similarly, the two points on the $y$-axis must be $(0,2)$ and $(0,-2)$. The remaining two points can only be $(-2,-2),(2,2)$ or $(-2,2),(2,-2)$. In either case, a triangle with an area no more than 2 is formed, leading to a contradiction.
|
proof
|
Combinatorics
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Theorem (1) Suppose the function $f(x)$ is continuous and non-negative on the closed interval $[a, b]$ (as shown in Figure 7-2), then the sum $\sum_{a<r \leqslant b}[f(t)]$ represents the number of lattice points in the planar region $a<x \leqslant b, 0<y \leqslant f(x)$, where $t$ denotes the integers in $(a, b]$, and $[f(t)]$ denotes the greatest integer function of $f(t)$.
(2) Suppose positive integers $p, q$ satisfy $(p, q)=1$, then
$$\sum_{0<x<\frac{p}{2}}\left[\frac{q}{p} x\right]+\sum_{0<y<\frac{p}{2}}\left[\frac{p}{q} y\right]=\left[\frac{p-1}{2}\right] \cdot\left[\frac{q-1}{2}\right]$$
|
The proof is as follows: As shown in Figure 7-3, because $(p, q)=1$, there are no lattice points on the line segment $O F$ except for $O$. Therefore, $\sum_{0<x<\frac{p}{2}}\left[\frac{q}{p} x\right]$ represents the number of lattice points inside $\triangle O D F$, and $\sum_{0<y<\frac{q}{2}}\left[\frac{p}{q} y\right]$ represents the number of lattice points inside $\triangle O E F$. Thus, $\sum_{0<x<\frac{p}{2}}\left[\frac{q}{p} x\right]+\sum_{0<y<\frac{q}{2}}\left[\frac{q}{p} y\right]$ represents the number of lattice points inside the rectangle ODFE, which numerically equals $\left[\frac{p-1}{2}\right] \cdot \left[\frac{q-1}{2}\right]$.
|
\left[\frac{p-1}{2}\right] \cdot \left[\frac{q-1}{2}\right]
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 5 For a lattice point $X$ on the coordinate plane, if the line segment $O X$ does not contain any other lattice points, then the point $X$ is said to be visible from the origin $O$. Prove that for any positive integer $n$, there exists a square $A B C D$ with area $n^{2}$ such that there are no lattice points visible from the origin inside this square.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Notice that a lattice point $(x, y)$ is visible from the origin if and only if $x, y$ are coprime. Thus, the problem is reduced to finding a square $A B C D$ such that every lattice point $(x, y)$ inside it satisfies that $x$ and $y$ are not coprime. Let $P_{1}, P_{2}, \cdots$ be a sequence of distinct primes. In an $n \times n$ matrix $M$, the elements of the first row are the first $n$ primes, the elements of the second row are the next $n$ primes, and so on. Let $m_{i}$ be the product of the elements in the $i$-th row of $M$, and $M_{j}$ be the product of the elements in the $j$-th column of $M$. Then $m_{1}$, $m_{2}, \cdots, m_{n}$ are pairwise coprime, and $M_{1}, M_{2}, \cdots, M_{n}$ are pairwise coprime. Next, consider the system of congruences
$$\left\{\begin{array}{l}
x \equiv-1\left(\bmod m_{1}\right) \\
x \equiv-2\left(\bmod m_{2}\right) \\
\vdots \\
x \equiv-n\left(\bmod m_{n}\right)
\end{array}\right.$$
By the Chinese Remainder Theorem, there is a unique solution $a\left(a \in \mathbf{N}_{+}\right)$ modulo $m_{1} m_{2} \cdots m_{n}$.
Similarly, the system of congruences
$$\left(\begin{array}{l}
y \equiv-1\left(\bmod M_{1}\right) \\
y \equiv-2\left(\bmod M_{2}\right) \\
\vdots \\
y \equiv-n\left(\bmod M_{n}\right)
\end{array} \text { has a unique solution } b\left(b \in \mathbf{N}_{+}\right) \text { modulo } M_{1} M_{2} \cdots M_{n} .\right.$$
In the square with vertices $(a, b)$ and $(a+n, b+n)$, the internal lattice points satisfy $(a+r, b+s)$, where $0<r<n, 0<s<n$.
We now prove that such points are not visible from the origin. Indeed, since $a \equiv-r\left(\bmod m_{r}\right), b \equiv-s\left(\bmod M_{s}\right)$, the prime in the $r$-th row and $s$-th column of matrix $M$ divides both $a+r$ and $b+s$, hence the lattice point $(a+r, b+s)$ is not visible from the origin.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 6 If $a, b$ are coprime positive integers, prove:
$$\left[\frac{a}{b}\right]+\left[\frac{2 a}{b}\right]+\cdots+\left[\frac{b-1}{b} a\right]=\frac{1}{2}(a-1)(b-1) .$$
|
Prove that, as shown in Figure 7-6, the four vertices of rectangle $O A B C$ are $O(0,0), A(b, 0), B(b, a)$, and $C(0, a)$. All integer points $(x, y)$ inside this rectangle satisfy $1 \leqslant x \leqslant b-1, 1 \leqslant y \leqslant a-1$, and there are a total of $(a-1) \cdot (b-1)$ such integer points.
The equation of the diagonal $O B$ of the rectangle is $y = \frac{a}{b} x$. Since $(a, b) = 1$, when $x = 1, 2, \cdots, b-1$, $y = \frac{a}{b} x$ cannot be an integer, meaning there are no integer points on $O B$. Thus, the number of integer points inside the triangle below $O B$ is equal to the number of integer points inside the triangle above $O B$, both equal to $\frac{1}{2}(b-1)(a-1)$. On the other hand, $\sum_{k=1}^{b-1}\left[\frac{k a}{b}\right]$ represents the number of integer points inside $\triangle O A B$. Therefore, $\sum_{k=1}^{b-1}\left[\frac{k a}{b}\right] = \frac{1}{2}(a-1)(b-1)$.
|
\frac{1}{2}(a-1)(b-1)
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
8. Prove: There exist 2008 lattice points in the plane, not all on a straight line, such that the distance between any two points is an integer.
|
8. Proof: Let $B(0, b)$, where $b=2 p_{1} p_{2} \cdots p_{2008}$. ($p_{i}$ are distinct factors, $i=1,2, \cdots, 2008$), and let $A_{j}\left(a_{j}, 0\right)$, where $a_{j}=\left|\left(p_{1} p_{2} \cdots p_{j}\right)^{2}-\left(p_{j+1} p_{j+2} \cdots p_{2008}\right)^{2}\right|, j=1,2, \cdots, 2007$.
By the Pythagorean triple formula, it is easy to see that $\left|A_{j} B\right|=\sqrt{b^{2}+a_{j}^{2}} \in \mathbf{N}_{+}$, and $\left|A_{i} A_{j}\right|=\left|a_{j}-a_{i}\right|(1 \leqslant i<j \leqslant 2007)$ is an integer, meaning that the 2008 lattice points $B, A_{j}(j=1,2, \cdots, 2007)$ are not all on the same line, and the distances between each pair are integers.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Minkowski's Theorem: Let $K$ be a convex figure in the $x O y$ plane, symmetric with respect to the origin, and with an area greater than 4. Prove: The figure $K$ must cover a lattice point other than the origin.
|
Prove that using lines parallel to the coordinate axes to divide the plane into some $2 \times 2$ squares, and then translating all the squares that cover the points of $K$ to the same square, since the area of $K$ is greater than 4, there must be a point in this square that is covered by the squares containing points of $K$ twice, i.e., there are two points $A \neq B$ in $K$, whose differences in both the x-coordinates and y-coordinates are multiples of 2. Since the figure $K$ is symmetric about the origin, the point symmetric to $A$ about the origin is $A^{\prime}$, then $A^{\prime} \in K$. Let $B\left(x_{1}, y_{1}\right), A\left(x_{2}, y_{2}\right)$, then $x_{1}-x_{2}=2 m, y_{1}-y_{2}=2 n, m, n \in \mathbf{Z}$. Therefore, $A^{\prime}$ is $\left(-x_{2},-y_{2}\right)$, so the coordinates of the midpoint of $A^{\prime} B^{\prime}$ are $\left(\frac{x_{1}-x_{2}}{2}, \frac{y_{1}-y_{2}}{2}\right)=(m, n) \neq(0,0)$. Since $K$ is a convex figure, $(m, n)$ is inside $K$.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let $L$ be a subset of the Cartesian plane, defined as follows:
$$L=\{(41 x+2 y, 59 x+15 y) \mid x, y \in \mathbf{Z}\}.$$
Prove that every parallelogram centered at the origin with an area of 2008 contains at least two points from $L$.
|
Prove that taking $(x, y)=(0,0),(0,1),(1,0),(1,1)$, we get four points in $L$: $(0,0),(2,15)$, $(41,59),(43,73)$.
Let these four integer points form the fundamental region $F$, which is a parallelogram containing no other integer points.
It is easy to see that the area of $F$ is 497.
Translating $F$ to form a network on the plane, its lattice points are all points in $L$.
Let the parallelogram $P$ be centered at the origin with an area of 2008.
With the origin as the center of homothety, perform a homothety with a ratio of 2:1, transforming $P$ into $\frac{1}{2}P$, at which point the area is $\frac{1}{4} \cdot 2008 = 502 > 497$.
Therefore, after translating each piece of $\frac{1}{2}P$ into $F$, there must be two points that coincide.
Let these two points be $D_{1}(x_{1}, y_{1}), D_{2}(x_{2}, y_{2})$, translated along $\overrightarrow{O A_{1}}, \overrightarrow{O A_{2}}$, where $A_{1}(a_{1}, b_{1})$, $A_{2}(a_{2}, b_{2})$ are points in $L$, then $(x_{1}, y_{1}) + (a_{1}, b_{1}) = (x_{2}, y_{2}) + (a_{2}, b_{2})$.
Thus, $(a_{1} - a_{2}, b_{1} - b_{2}) = (a_{1}, b_{1}) - (a_{2}, b_{2}) = (x_{2}, y_{2}) - (x_{1}, y_{1})$.
It is easy to see that the point $D(x_{2} - x_{1}, y_{2} - y_{1}) \in P$ (this is because $(x_{1}, y_{1}) \in \frac{P}{2}, (x_{2}, y_{2}) \in \frac{P}{2}$, and $\frac{P}{2}$ is a symmetric figure centered at the origin, so $(-x_{1}, -y_{1}) \in \frac{P}{2}$, then by the convexity of $\frac{P}{2}$, $(\frac{x_{2} - x_{1}}{2}, \frac{y_{2} - y_{1}}{2}) \in \frac{P}{2}$, so $D(x_{2} - x_{1}, y_{2} - y_{1}) \in P$).
And the point $D(a_{1} - a_{2}, b_{1} - b_{2}) \in L$, and $D$ is different from $(0,0)$.
Therefore, $P$ contains at least two points from $L$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
3. In a Cartesian coordinate system, construct a closed broken line such that each node of the broken line is a lattice point, and the lengths of each segment of the broken line are equal. Prove: the number of segments of such a broken line must be even.
|
3. Proof: Take a node on a grid paper as the origin to establish a Cartesian coordinate system, so that all nodes on the grid paper are lattice points.
Thus, the coordinates of each vertex $\left(x_{i}, y_{i}\right)(i=1,2, \cdots, n)$ of the closed polyline are integers.
Since it is a closed polyline, the projections of the closed polyline on the coordinate axes (including signs) satisfy $x_{1}+x_{2}+\cdots+x_{n}=0, y_{1}+y_{2}+\cdots+y_{n}=0, x_{i}^{2}+y_{i}^{2}=c(i=1,2, \cdots, n)$.
(1) If $c$ has the form $4 k+2$.
In this case, $x_{i}$ and $y_{i}$ are always odd, and since the sum of $n$ odd numbers is 0, it follows that $n$ is even.
(2) If $c$ has the form $4 k+1$.
In this case, $x_{i}$ and $y_{i}$ have different parities, i.e., $x_{i}+y_{i}$ is odd, and $\left(x_{1}+y_{1}\right)+\left(x_{2}+y_{2}\right)+\cdots+\left(x_{n}+\right.$ $\left.y_{n}\right)=0$. Therefore, $n$ is even.
(3) If $c$ has the form $4 k$.
In this case, $x_{i}$ and $y_{i}$ are even. Let $2^{k}$ be the highest power of 2 that divides all $x_{i}$ and $y_{i}$. At this point, dividing both sides of $x_{i}^{2}+y_{i}^{2}=c(i=1,2, \cdots, n)$ by $2^{k}$, we can reduce it to the case where $x_{i}$ and $y_{i}$ are either both odd or one is odd and the other is even, i.e., cases (1) and (2). In this case, $n$ is also even.
In summary, $n$ is even.
|
proof
|
Geometry
|
proof
|
Yes
|
Yes
|
number_theory
| false |
5. Let $n, k$ be positive integers, and $n>k$, prove: the greatest common divisor of $C_{n}^{k}, C_{n+1}^{k}, \cdots, C_{n+k}^{k}$ is 1.
|
5. Prove: Let $d \in \mathbf{N}$ be a common divisor of $C_{n}^{k}, C_{n+1}^{k}, \cdots, C_{n+k}^{k}$, then $d$ is also a common divisor of $C_{n}^{k-1}=C_{n+1}^{k}-C_{n}^{k}, C_{n+1}^{k-1}=$ $C_{n+2}^{k}-C_{n+1}^{k}, \cdots, C_{n+k-1}^{k-1}=C_{n+k}^{k}-C_{n+k-1}^{k}$.
Similarly, $d$ is also a common divisor of $C_{n}^{k-2}, C_{n+1}^{k-2}, \cdots, C_{n+k-2}^{k-2}, C_{n}^{k-3}, C_{n+1}^{k-3}, \cdots, C_{n+k-3}^{k-3}, \cdots$
Thus, $d$ is a divisor of $C_{n}^{0}$,
Therefore, the greatest common divisor of $C_{n}^{k}, C_{n+1}^{k}, \cdots, C_{n+k}^{k}$ is 1.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
1. Definition of prime numbers and some basic properties An integer $n$ greater than 1 has at least two distinct positive divisors: 1 and $n$. If $n$ has no divisor greater than 1 and less than $n$, then $n$ is called a prime number. If $n$ has a divisor greater than 1 and less than $n$, i.e., $n$ can be expressed in the form of $a \cdot b$, then $n$ is called a composite number.
(1) Any integer greater than 1 must have a prime factor.
(2) There is only one positive integer that is both even and prime, which is 2.
(3) Let $p$ be a prime number, and $n$ be any integer, then either $p$ divides $n$, or $p$ and $n$ are coprime.
(4) Let $p$ be a prime number, and $a, b$ be integers. If $p \mid a b$, then at least one of $a, b$ is divisible by $p$. Proof: If $p$ does not divide $a$ and $b$, then $p$ is coprime with $a$ and $b$, thus $p$ is coprime with $a b$, which is a contradiction!
(5) There are infinitely many prime numbers. Prove this proposition by contradiction, assume there are only finitely many prime numbers, let $p_{1}=2, p_{2}$ $=3, \cdots, p_{k}$ be all the prime numbers, consider $N=p_{1} p_{2} \cdots p_{k}+1$, clearly $N>1$. Therefore, $N$ has a prime factor $p$. Since $p_{1}, p_{2}$, $\cdots, p_{k}$ are all the prime numbers, $p$ must equal some $p_{i}(1 \leqslant i \leqslant k)$, thus $p$ divides $N-p_{1} p_{2} \cdots p_{k}=1$, which is impossible, therefore there are infinitely many prime numbers.
|
None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". Here is the formatted output as requested:
None
|
not found
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
2. The Fundamental Theorem of Arithmetic states that every positive integer greater than 1 can be decomposed into a product of a finite number of prime numbers. If the order of the prime factors in the product is not considered, then the decomposition is unique, i.e., \( n = p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \cdots \cdots \cdot p_{k}^{a_{k}} \), where \( p_{i} \) are prime numbers, \( \alpha_{i} \) are positive integers, and \( 1 \leqslant i \leqslant k \). Below, we will prove its uniqueness. Let \( n > 1 \) have two prime factorizations,
$$n = p_{1} p_{2} \cdots p_{k} = q_{1} q_{2} \cdots q_{t}$$
|
We need to prove that $k=l$ and the prime numbers $p_{1}, p_{2}, \cdots, p_{k}$ are a permutation of $q_{1}, q_{2}, \cdots, q_{l}$.
From (1), we see that $p_{1}$ divides $q_{1} q_{2} \cdots q_{l}$, so $p_{1}$ divides $q_{1}$ or $q_{2} \cdots q_{l}$. If $p_{1} \mid q_{1}$, since $p_{1}$ and $q_{1}$ are both primes, then $p_{1}=q_{1}$; if $p_{1} \mid q_{2} \cdots q_{l}$, repeating this argument shows that $p_{1}$ must be the same as one of $q_{2}, \cdots, q_{l}$. Therefore, we can cancel $p_{1}$ from both sides of (1), then consider $p_{2}$, and repeat the process. Eventually, we find that the two prime factorizations of $n$ are identical. Proof completed.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 (29th Russian Mathematical Olympiad) Prove: from any 5 pairwise coprime three-digit numbers, it is possible to select 4 numbers that are coprime.
untranslated text retained the original line breaks and formatting.
|
Analyzing, first set the greatest common divisor of any 4 numbers, then use proof by contradiction to prove that at least one of the 5 common divisors is 1.
Proof: Let the 5 numbers be \(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\).
If the proposition does not hold, then there exist
\[
\begin{array}{l}
\left(a_{1}, a_{2}, a_{3}, a_{5}\right)=d_{1},\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=d_{2},\left(a_{1}, a_{3}, a_{4}, a_{5}\right)=d_{3} \\
\left(a_{1}, a_{2}, a_{4}, a_{5}\right)=d_{4},\left(a_{2}, a_{3}, a_{4}, a_{5}\right)=d_{5}
\end{array}
\]
where \(d_{1}, d_{2}, d_{3}, d_{4}, d_{5}\) are pairwise coprime and none of them is 1. Without loss of generality, assume \(d_{1}>d_{2}>d_{3}>d_{4}>d_{5}\), then
\[
a_{1}=b_{1} d_{1} d_{2} d_{3} d_{4} \geqslant b_{1} \times 3 \times 5 \times 7 \times 11=1155 b_{1} \geqslant 1155>1000
\]
Contradiction!
Therefore, one of \(d_{1}, d_{2}, d_{3}, d_{4}, d_{5}\) must be 1.
That is, from any 5 coprime three-digit numbers, 4 coprime numbers can be found.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 4 (15th Asia Pacific Mathematical Olympiad) Let $k, k \geqslant 14, p_{k}$ be the largest prime less than $k$. If $p_{k} \geqslant \frac{3 k}{4}$, and $n$ is a composite number, prove:
(1) If $n=2 p_{k}$, then $n$ does not divide $(n-k)$ !.
(2) If $n>2 p_{k}$, then $n$ divides $(n-k)$ !.
|
Prove (1) When $n=2 p_{k}$,
Since $k>p_{k}$, then $p_{k}>2 p_{k}-k=n-k$, so, $p_{k} \nmid(n-k)!$, hence $n \nmid(n-k)!$.
(2) When $n>2 p_{k}$,
Since $n$ is a composite number, let $n=a b(2 \leqslant a \leqslant b)$, if $a \geqslant 3$.
( i ) $a \neq b$, then $n>2 p_{k} \geqslant \frac{3 k}{2}, b \leqslant \frac{n}{3}$.
Thus, $k\frac{n}{3} \geqslant b>a$.
So $n \mid(n-k)!$.
(ii) $a=b$, then $n=a^{2}, n-k>\frac{n}{3}=\frac{a^{2}}{3}$,
Since $k \geqslant 14$, then $p_{k} \geqslant 13, n \geqslant 26, a \geqslant 6$,
Thus $\frac{a^{2}}{3} \geqslant 2 a$. Hence $n-k>2 a$. So, $n \mid(n-k)!$.
If $a=2$, since $n \geqslant 26$, assume $b$ is not a prime, then $b=b_{1} b_{2}\left(b_{1} \leqslant b_{2}\right)$, since $b \geqslant 13$, then $b_{2} \geqslant 4$, thus $a b_{1} \geqslant 4$ falls into the case of $a \geqslant 3$,
Assume $b$ is a prime, then $b=\frac{n}{2}>p_{k}$,
Since $p_{k}$ is the largest prime less than $k$, then $b>k$,
Thus $n-k=2 b-k>b$. So, $n \mid(n-k)!$.
In summary, when $n>2 p_{k}$, $n \mid(n-k)!$.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
7. (2003 IMO Slovenia National Team Selection Test) Let $S$ be a finite non-empty set of positive integers greater than 1, and has the following property: there exists a number $s \in S$, such that for any positive integer $n$, either $(s, n)=1$ or $(s, n)=s$. Prove: there must exist two numbers $s, t \in S$ (not necessarily distinct), such that $(s, t)$ is a prime number.
|
7. Proof: Let $n$ be the smallest positive integer that is not coprime with any number in the set $S$ (such a number exists; in fact, since $1 \notin S$, the product of all numbers in $S$ is not coprime with any number in $S$).
Among all prime factors of $n$, each prime appears exactly once, otherwise, $n$ would not be the smallest number with the above property. Since the number $n$ is not coprime with any number in $S$, by the given condition, there exists $s \in S$ such that $(s, n)=s$, hence, $s \mid n$.
Let $p$ be any prime factor of the number $s$, then $p \mid n$. Furthermore, by the selection method of $n$, the number $\frac{n}{p}$ is coprime with some $t \in S$, from $\left(t, \frac{n}{p}\right)=1$ and $(t, n)>1$, we know $p \mid t$.
But the number $t$ cannot be divisible by any other prime factor of $n$ (since these primes all divide $\frac{n}{p}$), thus we can conclude that $(s, t)=p$, i.e., $(s, t)$ is a prime number.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 Let $p$ be a prime, $1 \leqslant k \leqslant p-1$, prove: $C_{p-1}^{k} \equiv(-1)^{k}(\bmod p)$.
|
$$\begin{array}{l}
A \equiv \frac{(p-1)(p-2) \cdots(p-k)}{k!}(\bmod p), \\
\text { so } k!\cdot A \equiv(p-1)(p-2) \cdots(p-k)(\bmod p) \text {, } \\
\equiv(-1) \cdot(-2) \cdots(-k) \quad(\bmod p) \\
\equiv(-1)^{k} k!\quad(\bmod p) . \\
\text { Also }(k!, p)=1,
\end{array}$$
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 2 (2003 Croatian Mathematical Olympiad) For all odd primes $p$ and positive integers $n(n \geqslant p)$, prove:
$$C_{n}^{p} \equiv\left[\frac{n}{p}\right](\bmod p)$$
|
Analysis and Solution Note that $C_{n}^{p}=\frac{n(n-1) \cdots(n-p+1)}{p!}$, and $n, n-1, \cdots, n-p+1$ form a complete residue system modulo $p$, so there exists $a \in\{n, n-1, \cdots, n-p+1\}$ such that $p \mid a$, clearly $a=p \cdot\left[\frac{n}{p}\right]$. Let $C_{n}^{p} \equiv c(\bmod p)$.
Then $C_{n}^{p}=\frac{n(n-1) \cdots(a+1)(a-1)(a-2) \cdots(n-p+1) \cdot a}{(p-1)!p}(\bmod p)$
So $p \left\lvert\, c \cdot(p-1)!-\left[\frac{n}{p}\right] n(n-1) \cdots(a+1)(a-1) \cdots(n-p+1)\right.$. Notice that $n(n-1) \cdots(a+1)(a-1)(a-2) \cdots(n-p+1) \equiv(p-1)!(\bmod p)$, so $p \left\lvert\,\left(c-\left[\frac{n}{p}\right]\right) \cdot(p-1)!\right.$. Also, $(p,(p-1)!)=1$, so $p \left\lvert\, c-\left[\frac{n}{p}\right]\right.$. Therefore, $c \equiv\left[\frac{n}{p}\right](\bmod p)$. That is, $C_{n}^{p} \equiv\left[\frac{n}{p}\right](\bmod p)$.
|
C_{n}^{p} \equiv\left[\frac{n}{p}\right](\bmod p)
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
Example 3 Let $n$ be a positive integer, integer $k$ be coprime with $n$, and $0<k<n$. Let $M=\{1,2, \cdots, n-1\}$, and color each number in $M$ with one of two colors, black or white, according to the following rules: (i) for each $i$ in $M, i$ and $n-i$ are the same color, (ii) for each $i$ in $M, i \neq k, i$ and $|k-i|$ are the same color. Prove: All numbers in $M$ must be the same color.
|
Analysis: From $(k, n)=1$, the numbers in $M$ can be reordered, and then it is only necessary to prove that adjacent numbers have the same color.
Proof: Since $(k, n)=1, 0, 1, 2, \cdots, n-1$ is a complete residue system modulo $n$,
so $k \cdot 0, k \cdot 1, k \cdot 2, \cdots, k(n-1)$ is also a complete residue system modulo $n$.
If we set $r_{j} \equiv j k(\bmod n)$ (where $1 \leqslant r_{j} \leqslant n-1, j=1,2, \cdots, n-1$),
then $M=\{1,2,3, \cdots, n-1\}=\left\{r_{1}, r_{2}, \cdots, r_{n-1}\right\}$,
it is only necessary to prove:
$r_{j}$ and $r_{j+1}$ have the same color $(1 \leqslant j \leqslant n-2)$.
Clearly, $r_{j+1} \equiv(j+1) k(\bmod n) \equiv r_{j}+k(\bmod n)$.
We discuss the following two cases:
(1) If $r_{j}+k \geq n$, then $r_{j+1}=r_{j}+k-n$, so by condition (i), $k-r_{j+1}=n-r_{j}$ has the same color as $n-(n-r_{j})=r_{j}$, and by condition (ii), $k-r_{j+1}$ has the same color as $\left|k-r_{j+1}-k\right|=r_{j+1}$, thus $r_{j+1}$ and $r_{j}$ have the same color. Therefore, (1) is proved.
In summary, the original proposition is proved.
|
proof
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false |
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