「數學」好好做着幾何題，怎麼就變數論了呢？

Everything starts with an innocent-looking question from the last question of 2021 Euclid contest:

Question

Suppose that $c$ is a positive integer. Define $f(c)$ to be the number of pairs $(a,b)$ with $c < a < b$ for wich two circles of radius $a$ , two circles of radius $b$ , and one circle of radius $c$ can be drawn so that

each circle of radius $a$ is tangent to both circles of radius $b$ and to the circle of radius $c$ , and
each circle of radius $b$ is tangent to both circles of radius $a$ and to the circle of radius $c$ ,

as shown, Determine all positive interger $c$ for which $f(c)$ is even.

Solution (Part 1)

First it is easy to infer that the centres of the outside circles form the vertices of a rhombus. Hence, the centres of any three adjacent circles form a right-angled triangle, with three sides of $a+c, b+c$ , and $a+b$ . Pythagoras Theorem automatically results:

(a+c)^2+(b+c)^2=(a+b)^2\\(a-c)(b-c)=2c^2

The inequality between $a, b$ , and $c$ ensures that $a-c\ne b-c$ . The problem now becomes:

Find such $c$ that the number $2c^2$ can be expressed as the product of two integers in even number of ways.

[Detour] Ways of decompose an integer

For any natural number that is not a perfect square, its factors always come in pairs, since if $p|c$ then we must have $\frac cp|c$ . As a result of this, for a non-squared number $c$ , there are always $\frac n2$ ways to decompose it, where $n$ is the number of factors of $c$ .

[Detour] Number of factors of a natural number

Fundamental Theory of Arithmetic states that the prime factor decomposition is unique for every natural number. Hence, assume that:

c=\prod\limits_ip_i^{e_i}\in\mathbb{N}.

Consider the $i$ ^th prime factor, it can have $e_i+1$ states, namely: $\{0, 1, \ldots, e_i\}$ . Therefore, there will be $K$ factors of $c$ where

K=\prod_i(e_i+1).

Solution (Part 2)

Now that we know there should be even ways to decompose $2c^2$ , assume that $c=2^k\prod\limits_ip_i^{e_i}$ , then:

2c^2=2^{2k+1}\prod_ip_i^{2e_i},\\K_{2c^2}=(2k+2)\prod_i(2e_i+1)

That is, for any $c=2^k\prod\limits_ip_i^{e_i}$ , there are $f(c)=\frac{K_{2c^2}}2=(k+1)\prod\limits_i(2e_i+1)$ ways of decomposing $2c^2$ . Since $2e_i+1\equiv1\pmod2, f(c)\equiv0\pmod2\iff k\equiv1\pmod2$

Conclusion

To make $f(c)$ even, $c$ has to have the following prime factor decomposition:

c=2^{2k-1}\prod_ip_i^{e_i},\quad k\in\mathbb{Z}.

心路歷程

做完了題，來說說考慮問題時候的一些思路。首先看到三邊都是自然數的直角三角形，第一反應果然還是 Pythagorean Tuples 的公式，這個代換最初是人們爲了嘗試證明費馬大定理（Fermat's Last Theorem）時候寫下的：

Pythagorean Tuple

Given that the three sides of a right-angled triangle are all integers: $(a, b, c)\in\mathbb{Z}^3$ , then the tuple is referred to as Pythagorean tuple. Pythagorean tuples can be generated by two integers $u, v$ :

a=u^2-v^2,\\b=2uv,\\c=u^2+v^2.

然而這個思路最大的問題在於 $(u,v)\to(a,b,c)$ 是到上的，但不是滿的，於是並無法寫出一個合適的逆映射，也就是說無法討論 $c$ 的 $u,v$ 表出。

其次是考慮單純的 Pythagoras Theorem： $(a+b)^2=(a+c)^2+(b+c)^2$ 的化簡結果： $ab=ac+bc+c^2=cp$ ，其中 $p$ 是半週長。做到這裏發生的一個迷惑是顯然 $c|ab$ 然後就開始考慮三邊的分解質因數了。

解到分解 $c$ 的時候，一直在邊上看着的女友急得跺腳，總覺的女朋友戰勝我的一天又近了好多（捂臉