Is it possible to modify Zagier's one-sentence proof of Fermat's sum of two squares theorem (see here) to prove certain non-trivial cases of Jacobi's four-square theorem (see here)?
Let $p$ be a prime number with $p\equiv1\pmod4$ and define the set
$$S=\{(x,y,z)\in\mathbb{N}^3\mid x^2+4yz=p\}$$
Zagier gave a very short proof of the fact $p$ is a sum of two squares of integers by constructing two involutions $\sigma$ and $\mu$ on this set $S$. The involution $\mu$ is a quite obvious one, sending $(x,y,z)$ to $(x,z,y)$; while $\sigma$ is less obvious: if $x<y-z$, $\sigma(x,y,z)=(x+2z,z,y-x-z)$; if $y-z<x<2y$, $\sigma(x,y,z)=(2y-x,y,x-y+z)$; if $x>2y$, $\sigma(x,y,z)=(x-2y,x-y+z,y)$.
Yet there is still a geometric interpretation of $\sigma$ (see here).
There have been some generalizations of this proof to the case $p\equiv3\pmod8$ expressed as $x^2+2y^2$(see for example here). Maybe there are other cases that I am not aware of.
My question is mainly about possible generalization of Zagier's proof to Jacobi's four-squares theorem. Let $p$ be an odd prime, then this theorem states that the number of solutions $(x,y,z,w)\in\mathbb{N}^4$ such that $x^2+y^2+z^2+w^2=p$ is exactly $(p+1)/2$ (see here). Suppose $p$ is NOT a sum of three squares of integers, then $p\equiv7\pmod8$. In the following, I will assume $p\equiv7\pmod{16}$ and define the following set
$$S'=\{(x_1,x_2,x_3,y,z)\in\mathbb{N}^5\mid x_1^2+x_2^2+x_3^2+4yz=p\}$$
There are involutions $\sigma'$ and $\mu'$ on $S'$:
$$\sigma'(x_1,x_2,x_3,y,z)=(x_2,x_1,\sigma(x_3,y,z)),\quad\mu'(x_1,x_2,x_3,y,z)=(x_1,x_2,x_3,z,y).$$
The set of fixed points of $\sigma'$ is just the set $S_1$ of $(x,y,z)\in\mathbb{N}^3$ such that $2x^2+y(y+4z)=p$, while the set of fixed points of $\mu'$ is just the set $S_2$ of $(x,y,z,w)\in\mathbb{N}^4$ such that $x^2+y^2+z^2+4w^2=p$ (if we apply Jacobi's theorem mentioned above, if $p\equiv7\pmod{16}$, then $S_2$ has cardinal $(p+1)/8$, which is odd).
So to show that $S_2$ has odd cardinal, it is enough to show that $S_1$ has odd cardinal (and vice versa). However, I am stuck here to show $S_2$ has odd cardinal (I have posted this as a question here, yet have not received any solutions so far). Has anyone every thought about this? Any help would be appreciated!
