Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0$

Question

Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0.$

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Let one of the $x,y,z$ be even number.Let $x=2p$

$x^2+y^2+z^2=x^2y^2$

This gives $y^2+z^2$ is also even,which means either $y,z$ are both even or $y,z$ are both odd.

So either $x=2p,y=2q,z=2r$ is the solution or $x=2p,y=2q+1,z=2r+1$ is the solution.I put $x=2p,y=2q+1,z=2r+1$ in the equation $x^2+y^2+z^2=x^2y^2$ and see that it does not satisfy both sides,so it is not the solution.

When I put $x=2p,y=2q,z=2r$,i get $p^2+q^2+r^2=4p^2q^2$

I am confused here how to prove that $p^2+q^2+r^2=4p^2q^2$ does not hold true.

Also i cannot solve when $x$ is odd which is the second case.

Is my method correct?Please help me complete the solution.

Answer 1

$$ z^2 + 1 = (x^2 - 1)(y^2 - 1) $$ From quadratic reciprocity, $z^2 + 1$ is not divisible by $4$ or by any prime $q \equiv 3 \pmod 4.$

If $x$ or $y$ is odd, then $z^2 + 1$ is divisible by $4.$ Nope.

If $x$ or $y$ is $0$ then all $x,y,z = 0.$

Otherwise, if $x$ or $y$ is even, say it is $x,$ then one of $x-1,x+1$ is $1 \pmod 4$ but the other is $3 \pmod 4,$ the latter therefore divisible by some prime $q \equiv 3 \pmod 4.$ Nope.