Problem 10.3. Solve in integers the equation

$$
z^{2}+1=x y(x y+2 y-2 x-4) \text {. }
$$

Ivan Landjev
10.3. We set $x=u-1, y=v+1$ and obtain the equation

$$
z^{2}+1=\left(u^{2}-1\right)\left(v^{2}-1\right)
$$

It is easy to see that $u, v$ and $z$ must be even. Hence, if $|u|>1$, then $u^{2}-1$ has a prime divisor $p$ such that $p \equiv 3(\bmod 4)$. Therefore $z^{2}+1 \equiv 0(\bmod p)$, which is impossible (it is well known that if $p$ is a prime such that $p \equiv 3$ $(\bmod 4)$ and $p$ divides $x^{2}+y^{2}$ then $p$ divides both $x$ and $\left.y\right)$. Thus $u=0$.

Analogously we get $v=0$ and therefore $z=0$. Hence $x=-1, y=1, z=0$.