This proof is credited to Don Zagier. It goes like this:
Consider the finite set $S=\{(x,y,z) \in \mathbb{N}^3 : x^2+4yz=p\}$ where p is a prime. Then consider the following involution:
$$(x,y,z) \mapsto \begin{cases} (x + 2z, z, y - x + z) & x< y-z \\ (2y - x, y, x - y + z) & y-z<x<2y \\ (x - 2y, x - y + z, y) & x > 2y \end{cases}$$ It's not difficult to see that this function is an involution and that it has at least one fixed point of the form $(x,x,k)$. But Zagier says that there is exactly one fixed point, $(1,1,k)$, which shows that the cardinality of $S$ is odd and $$(x,y,z)\mapsto (x,z,y)$$ Also has exactly one fixed point, which concludes the proof.
The part that I'm uncomfortable with is $(1,1,k)$ being the only fixed point. I assume its based on the definition of $S$. In the defining equation for $S$, $x$ must be odd because otherwise $x^2$ would be a multiple of 4 and the left side of the equation would not be prime. I'm not sure how all other possible values for $x$ are excluded though. Clarification or hints would be appreciated.
This is different from this question because I grasp the strategy of the proof I am just wondering about this particular detail.
Suppose that $(x,x,k)∈S$. Then $$x^2+4xk=p$$ In particular this implies that $x$ divides $p$ (and it is strictly smaller than $p$). Thus, since $p$ is prime, it is only possible that $x=1$.