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I'm in a class on proofs and problem-solving, and my final paper is on Zagier's "one sentence" proof regarding primes and their relations to the sum of two squares. One portion of the explanation of Zagier's proof requires me to prove that $S = A\cup B\cup C$ Definitions can be found here, to sum up the sets: $S = \{ (x,y,z) \in \mathbb N^3 : x^2 + 4yz = p \}$ $(x,y,z) \mapsto \begin{cases} (x+2z, z, y-x-z) & \text{ if } x < y-z \\ (2y-x, y, x-y+z) & \text{ if } y-z < x < 2y \\ (x-2y, x-y+z, y) & \text{ if } x > 2y. \end{cases}$

Once again, I'd just like to know how to prove $S = A\cup B\cup C$. Thanks in advance.

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What you’ve written in the question is inadequate to make the question comprehensible, so I’m going to expand it. Let $p\equiv1\pmod4$ be prime, set $S=\big\{\langle x,y,z\rangle\in\Bbb N^3:x^2+4yz=p\big\}\;,$ and let

$\varphi:S\to\Bbb N^3:\langle x,y,z\rangle\mapsto\begin{cases} \langle x+2z,z,y-x-z\rangle,&\text{if }x2y\;; \end{cases}\tag{1}$

then $\varphi$ is an involution on $S$ with a unique fixed point.

I’m guessing that your sets $A,B$, and $C$ correspond to the cases of the definition of $\varphi$:

$\begin{align*} A&=\big\{\langle x,y,z\rangle\in S:x2y\big\}\;. \end{align*}$

By definition $A\cup B\cup C\subseteq S$, so in order to show that $A\cup B\cup C$ is all of $S$, we need only prove that if $\langle x,y,z\rangle\in S$, then $x\ne y-z$ and $x\ne 2y$. Suppose, then, that $x,y,z\in\Bbb N$ and $x^2+4yz=p$.

  1. If $x=y-z$, then $p=(y-z)^2+4yz=y^2+2yz+z^2=(y+z)^2$, which is impossible, since $p$ is prime.
  2. If $x=2y$, then $p=4y^2+4yz=4y(y+z)$, which is again impossible because $p$ is prime.

The three cases of $(1)$ are therefore exhaustive $-$ no element of $S$ falls through the cracks $-$ and $S=A\cup B\cup C$.

Do you also need help showing that $\varphi$ is an involution and that it has a unique fixed point?

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    @Nick: No worries: it happens.2012-12-05
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Don Zagier took this involution because it's easy to show, that this involution has exactly one fixed point. Hence it appears that involution $(x,y,z) \mapsto (x,z,y)$ also has exactly one fixed point, implying the theorem.