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.

Introduction to Modular Curves

## 目录

ISBN: 978-7-301-23438-9

## 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.

Richard Taylor(就是协助 Andrew Wiles 完成了Fermat’s Last Theorem 的证明的那位) 写了一篇很有趣的文章 Modular Arithmetic: Driven by Inherent Beauty and Human Curiosity(The Institute Letter, 2012, Summer, 6-8). 这文章指出: Euclid 在他的几何原本 已经得到方程

$$x^2+y^2=z^2$$

$$x^2+y^2=2z^2$$

$$x^2+y^2=nz^2,$$

Taylor 就说了这么多. 那么, 我们来尝试找出这方程的所有有理解, 以及所有整数解.

$$x^2+y^2=(a^2+b^2)z^2,$$

For which positive integers $$a, b, c, d$$, any natural number $$n$$ can be represented as

$n=ax^2+by^2+cz^2+dw^2,$

where $$x, y,z,w$$ are integers?

Lagrange’s four-square theorem states that $$(a,b,c,d)=(1,1,1,1)$$ works. Ramanujan proved that there are exactly $$54$$ possible choices for $$a, b, c, d$$.

For which positive integers $$a, b, c, d$$,

$n=ax^2+by^2+cz^2+dw^2,$

is solvable in integers $$x, y,z,w$$ for all positive integers $$n$$ except one number? For example, $$n=x^2+y^2+2z^2+29w^2$$ is solvable for all natural number $$n$$ except $$14$$, $$n=x^2+2y^2+7z^2+11w^2$$ and $$n=x^2+2y^2+7z^2+13w^2$$ except $$5$$.

P.R.Halmos proved that there are exactly $$88$$ possible choices for $$a, b, c, d$$.

What integers are not in the range of $$a^2+b^2+c^2-x^2$$? Ramanujan also thought about that.

#### 证明 1

• a=1. 此时 $$1+(d+2)d=(d+1)^2$$ 是合数;
• $$a\geqslant2$$. 此时 $$a+ad=a(d+1)$$ 是合数.

#### 证明 2

$$(m+1)!+2,(m+1)!+3,\dotsc,(m+1)!+m+1$$ 是 $$m$$ 个连续合数.

#### 证明 3

$a, a+d, a+2d, \dotsc, a+(m-1)d$

#### 证明 4

$f(x)=\sum\limits_{i=0}^ma_ix^i,$

#### 证明 6

$m_i|\left(a+i\right), i=1, 2, \dotsc, n.$

[证明 6 更新于 北京时间 2015 年 6 月 24 日]

1.21 唯一因式分解定理的证明

2.5  五边形与五角数

$p_m(k)=\dfrac{mk(k-1)}2+k.$

2.8  一个不平凡的结论

2.9 什么数恰好有 $$60$$ 个因数?

$$kn = x^2+y^2+1$$

$$n$$ is a odd number, then there exists positive integer $$k\gt0$$ such that $$kn = x^2+y^2+1$$ for some integers $$x,y$$.

with use of the Chinese remainder theorem we have to solve this problem only for power of primes:

suppose that $$n=p_1^{a_1}p_2^{a_2}\dotsm p_k^{a_k}$$, then we know that for each $$i$$, there exist $$x_i, y_i$$ such that $$p_i^{a_i}$$ divides  $$x_i^2+y_i^2+1$$. Now consider these equations:

$X\equiv x_i\pmod {p_i^{a_i}}, i= 1,2,\dotsc,k.$

these equations have solution because of  Chinese remainder theorem.

similarly these equation have solution:

$Y\equiv y_i\pmod {p_i^{a_i}}, i= 1,2,\dotsc,k.$

now $$n$$ divides  $$X^2+Y^2+1.$$

then we can apply hansel’s lemma. Actually we want to show that if for some $$\alpha$$, there exist $$x,y$$ such that $$p^\alpha$$ divides  $$x^2+y^2+1$$, then for  $$\alpha +1$$ such $$x$$  and $$y$$ exist. For this because in case $$\alpha$$, $$p$$ cannot divide both $$x$$  and $$y$$, then we can use hansel for improve $$\alpha$$ to $$\alpha+1.$$

#### References

1. 华罗庚, 数论导引.
2. Hardy, An introducton to the theory of numbers. 有中文本
3. Tom M. Apostol, Introduction to analytic number therory. 有中文本