Existence of root of a polynomial over $\mathbb F_p$. I came accross the following question and I can't find an easy proof of this fact : 
Let $p\geq 17$ be a prime number such that $p\equiv 1 \pmod 4$. 
Show that for any $z\in \mathbb F_p\backslash\{0\}$, there exist $x,y\in \mathbb F_p\backslash\{0\}$ such that $$(x^2+1)(y^2+1)=xyz.$$
I tried to use combinatorial facts (such that the number of square in $\mathbb F_p$ but it is not enough) or using the existence of roots of a polynomial with one variable over $\mathbb F_p$ but it doesn't seem to be conclusive.
Any idea ?
 A: Searching the bag of tools for a big hammer... finding this.
Fix $z$. Rewrite the equation, treating $y$ as the unknown, in the form
$$
y^2-\frac{zx}{x^2+1}y+1=0.
$$
This has a solution $y\in\Bbb{F}_p$, iff the discriminant
$$
\Delta=\Delta(x):=\frac{z^2x^2-4(x^2+1)^2}{(x^2+1)^2}
$$
is a square in $\Bbb{F}_p$. Let $\eta:\Bbb{F}_p\to\{0,\pm1\}$ be the Legendre character, i.e.
$\eta(a)=+1$, if $a$ is a non-zero square, $\eta(0)=0$ and $\eta(a)=-1$, if $a$ is a non-square.
Assuming that the quartic in the numerator $p(x)=z^2x^2-4(x^2+1)^2\in\Bbb{F}_p[x]$ is not of the form constant times a square, we have the Weil bound
$$
\left\vert\sum_{x\in\Bbb{F}_p}\eta(p(x))\right\vert\le (\deg p -1)\sqrt p.
$$
Here $\deg p=4$, so if $p>16$ we get that $p-4\sqrt p>0$, and thus
$$
\sum_{x\in\Bbb{F}_p}\eta(p(x))\ge-3\sqrt p> -p+\sqrt p.
$$
Therefore there are at least $\sqrt p>4$ choices for $x$ such that $\eta(p(x))\neq-1$.
Two of those choices may satisfy $x^2+1=0$, and one may be $x=0$, and these must be discarded, but at least one remains.
So for $p\ge17$ we have that for some $x\in\Bbb{F}_p^*$ we have that $x^2+1\neq0$ and $\Delta(x)$ is a square. For that $x$ we have a solution $y\in\Bbb{F}_p$. Clearly the conditions on $x$ mean that $y\neq0$.

We need to deal with the possibility that for some $K,a,b,c\in\Bbb{F}_p$ we have in the ring $\Bbb{F}_p[x]$
$$
p(x)= K(ax^2+bx+c)^2.
$$
Clearly then $K\neq0$. Looking at the constant terms implies that $c\neq0$. Looking at the linear terms then implie $b=0$. We are left with the possibility
$$
-4x^4+(z^2-8)x^2-4=K(ax^2+c)^2.
$$
Then $Ka^2=-4=Kc^2$, so $c=\pm a$. Therefore
$$
p(x)=K'(x^2\pm1)^2
$$
for $K'=-4a^2$. This implies that
$$
\Delta(x)=-4a^2\left(\frac{x^2\pm1}{x^2+1}\right)^2.
$$
As we assume that $p\equiv1\pmod4$, we know that $\eta(-1)=1$. Thus this implies that $\eta(\Delta(x))=+1$ for all $x$ such that $x^2+1\neq0$. Again the existence of a solution follows.
