$(x^2+xy+y^2)(z^2+zw+w^2)=(xz-yw)^2+(xz-yw)[wx+y(z+w)]+[wx+y(z+w)]^2$

\begin{vmatrix}
x& y\cr
-y & x+y
\end{vmatrix}

Let $$f(x_1,x_2,\dotsc,x_n)$$ be a homogeneous polynomial. Let

$S=\{f(a_1,a_2,\dotsc,a_n)\mid a_1,a_2,\dotsc,a_n \in\Bbb Z\}.$

If $$S$$ satisfies the following condition: for all $$m,n\in S$$, we have $$mn\in S$$. Can we determine all the homogeneous polynomials $$f$$?

For example, $$x^n(n\in\Bbb N),x^2+n y^2(n\in\Bbb Z), x^2+xy+y^2,x^3+y^3+z^3-3xyz$$, and $$x^2+y^2+z^2+w^2$$ are all appropriate examples.

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.

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

#### 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

Number theory will be understood, not as a collection of tricks and isolated results, but as a coherent and interconnected theory.

Green(与 Tao 合作 Green–Tao theorem 的那位)的(博士)导师 Gowers, 写了两篇关于此定理的文章: Why isn’t the fundamental theorem of arithmetic obvious?  和 Proving the fundamental theorem of arithmetic.

$H=\{1,5,9,13,17,\dotsc\}$

Sey Y.Kim 的证明也收录在 Biscuits of Number Theory 一书.

If $$p,q$$ are distinct odd primes,  then

$\left( \frac pq\right) \left( \frac qp\right) = (-1)^{\frac{(p-1)(q-1)}4},$

where $$\left( \frac{}{}\right)$$ is the Legendre symbol. 这就是被 Gauss 称为”数论酵母” 的二次互反律.

• 简单是首要条件. 基于简单的想法, 能揭示问题的本质, 加深对事物的理解.
• 美. 构思巧妙, 论证精妙方能展现出深刻. 当然, 简单其实也是一种美.
• 自然. 方法能用在更广泛的地方, 解决更多的问题.

For any odd $$n \in\Bbb N$$, denote by $$N_n$$ the number of solutions in $$\left( \Bbb Z / (q)\right)^n$$ to the equation

$x_1^2-x_2^2+x_3^2-\dotsb + x_n^2 = 1.$

If we substitute $$x_1 \gets x_1+x_2$$, then we get

$x_1^2+x_3^2-\dotsb + x_n^2-1 = -2x_1x_2.$

For any non-zero $$x_1$$-value and any values of $$x_3,\dotsc,x_n$$, there is a unique corresponding $$x_2$$-value. If $$x_1=0$$, there are no solutions, except if $$x_3^2-\dotsb + x_n^2 = 1$$ (which happens in $$N_{n-2}$$ cases): then all possible values of $$x_2$$ do the job. We find that

$N_n = q^{n-2}(q-1)+qN_{n-2}.$

and hence $$N_n = q^{n-1}+q^{\frac{n-1}{2}}(N_1-1) = q^{n-1}+q^{\frac{n-1}{2}}.$$ In particular,

$$N_p\equiv1+\left(\frac qp\right) \pmod p.$$

Next, $$N_p$$ can be classically determined as

$\sum_{t_1+\dotsb + t_p = 1}N(x_1^2 = t_1)N(x_2^2 = -t_2)N(x_3^2=t_3)\dotsm N (x_p^2=t_p),$

where the $$t_i$$ are in $$\Bbb Z/(q)$$ and $$N(\cdots)$$ denotes the number of solutions to the corresponding univariate equation. This can be rewritten as

$\sum_{t_1+\dotsb+t_p=1}\left(1+\left(\frac{t_1}q\right)\right)\left( 1+\left( \frac{-t_2}q\right)\right)\left( 1+\left( \frac{t_3}q\right)\right)\dotsm \left( 1+\left( \frac{t_p}q\right)\right),$

When expanding out the product, only the terms $$1\cdot 1 \cdot 1 \cdots 1$$ and $$\left(\frac{t_1}q\right)\cdot\left(\frac{-t_2}q\right)\cdot\left( \frac{t_3}q\right)\cdots\left( \frac{t_p}q\right)$$ should be taken into consideration; the other terms disappear because Legendre symbols sum up to zero: $$\sum\limits_{t\in\Bbb Z/(q)}\left(\frac tq\right) = 0.$$ Therefore, the above expression simplifies to

$q^{p-1}+\left( \frac{(-1)^{\frac{p-1}2}}q\right)\sum_{t_1+\dotsb+t_p = 1}\left(\frac{t_1t_2t_3\dotsm t_p}q\right).$

Modulo $$p$$, the latter sum almost completely vanishes, since the tuples $$(t_1,\dotsc,t_p)$$ satisfying $$t_1+\dotsb + t_p = 1$$ with not all $$t_i$$ equal to $$p^{-1}$$ can be collected in groups of size $$p$$ by cyclic permutation. Note that $$p$$ is indeed a multiplicative unit in $$\Bbb Z/(q)$$. We thus obtain

$$N_p\equiv1+\left( \frac{(-1)^{\frac{p-1}2}}q\right) \left( \frac{p^{-p}}q\right) \equiv1+(-1)^{\frac{p-1}2\frac{q-1}2}\left( \frac pq\right) \pmod p.$$

The last congruence follows from Euler’s criterion $$\left(\frac aq\right) \equiv a^{\frac{q-1}2} \pmod q$$ and the observation that $$p^{-p}$$ is a square in $$\Bbb Z/(q)$$ if and only if $$p$$ is a square in $$\Bbb Z/(q)$$.

Comparing $$(1)$$ and $$(2)$$, the reciprocitylaw follows. $$\Box$$

Let $$p = 2m+1$$ and $$q$$ be distinct odd primes and let $$N$$ denote the number of solutions of the equation

$x_1^2 + \ldots + x_p^2 = 1$

in the finite field $$\Bbb F_q$$.

The group $$\Bbb Z/p\Bbb Z$$ acts on the solution space $$X$$ by shifting indices: if $$(x_1, \ldots, x_p) \in X$$, then so is $$(x_a,x_{a+1}, \dotsc)$$ for each $$a \in {\mathbb Z}/p{\mathbb Z}$$, where the indices have to be read modulo $$p$$. Each orbit has exactly $$p$$ elements except if there is an $$x$$ with $$(x,x,\ldots,x) \in X$$: the orbit of this element has $$1$$ element. Now $$(x,x,\ldots,x) \in X$$ if and only if $$px^2 = 1$$ is solvable in $$\Bbb F_q$$, hence

$$N \equiv \Big( \frac pq \Big) + 1 \pmod p.$$

We make a change of variables to transform the diagonal equation into an equation where counting the number of solutions is easier. To this end, consider the matrix

$A = \left( \begin{matrix} 0 & 1 & & & & & & \\ 1 & 0 & & & & & & \\ & & 0 & 1 & & & & \\ & & 1 & 0 & & & & \\ & & & & \ddots & & & \\ & & & & & 0 & 1 & \\& & & & & 1 & 0 & \\& & & & & & & a\end{matrix} \right)$

with $$a = (-1)^{\frac{p-1}2}$$. Since $$\det A = 1$$, this matrix is congruent to the unit matrix, hence $$X$$ and the solution spaces $$X^\prime$$ of the equation $$x^T A x = 1$$,i.e.,recall that $$p=2m+1$$, of

$2(y_1z_1+\dotsb+y_mz_m)+ax_p^2=1$

are isomorphic.

For counting the number of solutions of $$X^\prime$$, observe that if $$(y_1, \dotsc, y_m) = 0$$, we have $$q^m(1+a^{\frac{q-1}2})$$ possibilities for choosing $$z_1, \dotsc, z_m$$ and $$x_p$$.

If $$y = (y_1, \dotsc, y_m) \ne 0$$, on the other hand, then for each choice of $$y$$ and $$x_p$$ we have to count the number of points on a hyperplane of dimension $$m$$; there are $$q^{m-1}$$ points on such a hyperplane, and the number of overall possibilities in this case is $$(q^m-1) \cdot q \cdot q^{m-1} = q^m(q^m-1)$$.

Thus we find

$$\begin{split} N & = q^m (1+a^{\frac{q-1}2})+q^m(q^m-1) \\&=q^m(q^m+(-1)^{\frac{p-1}2\frac{q-1}2})\\ & \equiv\Big(\frac qp\Big)\bigg(\Big(\frac qp\Big)+(-1)^{\frac{p-1}2\frac{q-1}2}\bigg)\\&\equiv1+(-1)^{\frac{p-1}2\frac{q-1}2}\Big(\frac qp\Big)\pmod p.\end{split}$$

Comparing $$(3)$$ and $$(4)$$, gives the quadratic reciprocity law. $$\Box$$

book on the Riemann hypothesis

ISBN: 978-7-302-29324-8