$\{\sqrt n|n\in\Bbb N\; \text{is Square-free integer}\}$

wiki 给出的定义是: 不被不是 $$1$$ 的完全平方整除的整数称为无平方因子整数. 因此, $$1$$ 算无平方因子正整数. 鉴于此, 我们认为: 无平方因子整数定义为”不被质数的平方整除的整数”更为恰当.

$$n_1$$, $$n_2$$, $$\dotsc$$, $$n_k$$ 是互不相同的无平方因子正整数; $$a_1$$, $$a_2$$, $$\dotsc$$, $$a_k$$ 都是整数. 令

$S=a_1\sqrt{n_1}+a_2\sqrt{n_2}+\dotsb+a_k\sqrt{n_k},$

$S^\prime=b_1\sqrt{m_1}+b_2\sqrt{m_2}+\dotsb+b_l\sqrt{m_l},$

(这里 $$b_1$$, $$b_2$$, $$\dotsc$$, $$b_l$$ 都是整数; $$m_1$$, $$m_2$$, $$\dotsc$$, $$m_l$$ 都是无平方因子正整数, 并且都没有 $$p_1$$, $$p_2$$, $$\dotsc$$, $$p_N$$ 以外的质因子), 使得 $$SS^\prime\ne0$$ 是整数. 进而, 顺水推舟, $$S\ne0$$.

$$\alpha$$ 是无理数, 则 $$\{n^k\alpha\}$$$$(k=1, 2, \dotsc)$$ 在区间 $$(0, 1)$$ 稠密．

{nkα}An Elementary Proof on dense of

Sam Northshield 用一句话就说明了质数的无穷性. 这个很精彩的证明是这样的:

Proof.  If the set of primes is finite, then

$0\lt \prod_p \sin\left(\frac \pi p\right)= \prod_p \sin\left(\frac{\pi\Big(1+2\prod\limits_{p’}p’\Big)}p\right)=0. \qquad \Box$

$\sin\left(\frac{\pi\Big(1+2\prod\limits_{p’}p’\Big)}q\right)=\sin k\pi=0.$

### References

1. Sam Northshield, A One-Line Proof of the Infinitude of Primes, The American Mathematical Monthly, Vol. 122, No. 5 (May 2015), p. 466
2. Michael Hardy and Catherine Woodgold, Prime Simplicity,  the Mathematical Intelligencer, Volume 31, Issue 4, December 2009,  44-52
3. Harold Edwards, Contradict or Construct?,  the Mathematical Intelligencer, Volume 32, Issue 1, March 2010, p.3
4. Harold Edwards, Essays in Constructive Mathematics,  Springer, 2010

$$p$$ is a odd prime,  then

1. there are infinite primes $$q$$ such that $$q$$ is a quadratic residue modulo $$p$$;
2. there are infinite primes $$q$$ such that $$q$$ is a quadratic nonresidue modulo $$p$$.

$a_0=q, a_1=q+p,a_2=q+pa_1,a_3=q+pa_1a_2,\dotsc, a_n=q+pa_1a_2\dotsb a_{n-1},\dotsc$

$\bigg(\dfrac{kq_1q_2q_3\dotsm q_n+1}p\bigg)=-1.$

$\bigg(\frac qp\bigg) = \bigg(\frac pq\bigg).$

$a_1 = 4^k – p,$
$a_2 = 4^ka_1^2 -p,$
$a_3 = 4^ka_1^2a_2^2 -p,$
$\dotsc$
$a_n = 4^ka_1^2a_2^2\dotsm a_{n-1}^2-p,$
$\dotsc$

$\bigg(\frac qp\bigg)=\bigg(\frac {-p}q\bigg).$

$a_1 = 4+p,$
$a_2 = 4a_1^2 + p,$
$a_3 = 4a_1^2a_2^2 + p,$
$a_4 = 4 a_1^2a_2^2 a_3^2 + p,$
$\dotsc$
$a_n = 4 a_1^2a_2^2\dotsm a_{n-1}^2+ p,$
$\dotsc$

Which integers can be expressed as $$a^3+b^3+c^3-3abc$$? $$a$$, $$b$$, $$c\in\Bbb Z$$.

$(a\pm1)^3+a^3+a^3-3(a\pm1)a^2=3a\pm1$

$(a-1)^3+a^3+(a+1)^3-3a(a+1)(a-1)=9a$

$2(a^3+b^3+c^3-3abc)=3(a+b+c)(a^2+b^2+c^2)-(a+b+c)^3$

If $$3\mid(a^3+b^3+c^3-3abc)$$, then $$3\mid(a+b+c)^3$$, $$3\mid(a+b+c)$$. so $$9\mid(a^3+b^3+c^3-3abc)$$.

All $$n$$ such that $$3\nmid n$$ or $$9\mid n$$.

Let $$f(x)=x^n+a_{n-1}x^{n-1} +\dotsb+a_1x+a_0$$ be a polynomial with integer coefficients, and let $$d_1$$,$$\dotsc$$, $$d_n$$ be pairwise distinct integers. Suppose that for infinitely many prime numbers $$p$$ there exists an integer $$k_p$$ for which

$$f\left(k_p+d_1\right)\equiv f\left(k_p+d_2\right)\equiv\dotsb\equiv f\left(k_p+d_n\right)\equiv0\pmod p.$$

Prove that there exists an integer $$k_0$$ such that

$f\left(k_0+d_1\right)=f\left(k_0+d_2\right)=\dotsb= f\left(k_0+d_n\right)=0.$

$F(x)=n^nf\Big(\frac xn\Big)=x^n+na_{n-1}x^{n-1}+n^2a_{n-2}x^{n-2}\dotsb+n^{n-1}a_1x+n^na_0.$

$F\big(K_p+D_i\big)=n^nf\bigg(\frac{K_p+D_i}n\bigg)=n^nf\big(k_p+d_i\big)\equiv0\pmod p.$

$\big(K_p+D_1\big)+\big(K_p+D_2\big)+\dotsb+\big(K_p+D_n\big)\equiv-na_{n-1}\pmod p,$

$\big(K_p+nd_1-u\big)+\big(K_p+nd_2-u\big)+\dotsb+\big(K_p+nd_n-u\big)\equiv-na_{n-1}\pmod p,$

$nK_p\equiv-n\big(a_{n-1}+d_1+d_2+\dotsb+d_n-u\big)=0\pmod p.$

$(-1)^ln^la_{n-l}\equiv\prod_{1\leq i_1\lt\dotsb\lt i_l\leq n}\big(K_p+D_{i_1}\big)\dotsm\big(K_p+D_{i_l}\big)\pmod p,$

$$(-1)^ln^la_{n-l}\equiv\prod_{1\leq i_1\lt\dotsb\lt i_l\leq n}D_{i_1}\dotsm D_{i_l}\pmod p.$$

$(-1)^ln^la_{n-l}=\prod_{1\leq i_1\lt\dotsb\lt i_l\leq n}D_{i_1}\dotsm D_{i_l}.$

$F(x)=\big(x-D_1\big)\big(x-D_2\big)\dotsm\big(x-D_n\big).$

$f(x)=\Big(x-d_1-\frac un\Big)\Big(x-d_2-\frac un\Big)\dotsm\Big(x-d_n-\frac un\Big).$

$$f$$ 是首一多项式, 其有理根必是整数. 故 $$\dfrac un$$ 是整数.

## 单墫 余红兵 冯志刚 刘培杰

1. 设 $$p$$ 为大于 $$3$$ 的素数, 证明 $$\dfrac{p^p-1}{p-1}$$ 和 $$\dfrac{p^p+1}{p+1}$$ 不能都是素数幂;
2. 设 $$n\gt5$$, 证明 $$n!$$ 不能整除它的正约数之和;
3. 设 $$A$$, $$B$$ 划分正整数集, 如果$$A+A$$ 和 $$B+B$$ 都只含有有限个素数, 证明$$A$$ 或 $$B$$ 是全体奇数的集合;
4. 设 $$M$$ 是给定正整数, 证明对每个充分大的素数 $$p$$, 存在$$M$$个连续的 $$\bmod p$$ 的二次非剩余;
5. 设 $$q$$ 是一个不大于$$\dfrac{\pi^2}6 -1$$ 的正有理数, 证明 $$q$$ 可写为若干互异单位分数的平方和;
6. 对每个充分大的正整数 $$k$$, 存在若干互异正整数, 其和为 $$k$$, 其倒数和为 $$1$$;
7. 在 $$n^2$$ 和 $$(n+1)^2$$ 间总有一些正整数的积是一个平方数的两倍;
8. 若一些单位根之和在单位圆上, 则必亦为单位根;
9. 设 $$f(x)=a_0+a_1x+a_2x^2+\dotsb$$ 是一个整系数的形式幂级数, 假定 $$\dfrac{f^\prime(x)}{f(x)}$$ 也是一个整系数的形式幂级数, 证明对任意下标 $$k$$, $$a_k$$ 能被 $$a_0$$ 整除.

$$\dfrac pq$$ can expressed as the finite sum of reciprocals of distinct squares if and only if

$\frac pq\in[0, \frac{\pi^2}6-1)\cup[1,\frac{\pi^2}6).$

If  $$n$$ is an integer exceeding $$77$$ then there exist positive integers $$k$$, $$a_1$$, $$a_2$$, $$\dotsc$$, $$a_k$$ such that:

1. $$1\lt a_1\lt a_2\lt \dotsc \lt a_k;$$
2.  $$a_1+ a_2+ \dotsb + a_k=n;$$
3.  $$\frac1{a_1}+ \frac1{a_2}+ \dotsb + \frac1{a_k}=1.$$

His proof is constructive and fairly short, but it does require a long table of decompositions for relatively small values of $$n$$. It would be interesting to see a non-constructive proof that doesn’t require such a long list.

Granville and Selfridge, Product of integers in an interval, modulo squares: “We prove a conjecture of Irving Kaplansky which asserts that between any pair of consecutive positive squares there is a set of distinct integers whose product is twice a square.”

The details are Electronic Journal of Combinatorics, Volume 8(1), 2001.

$|\sum_{i=1}^k n_i\zeta_i|= 1,$

where $$n_i\in\mathbb Z$$, then $$\sum\limits_{i=1}^k n_i \zeta_i$$ is also an $$n$$-th root of unit.