Suppose the equation has a solution in positive integers, and choose a solution that minimizes $x^{2}+y^{2}$. Note that $x,y$, and $z$ are pairwise coprime, since otherwise we could divide out by their common divisor to get a smaller solution. Thus

so that $z,y^{2},x^{2}$ form a pythagorean triple. There are thus positive integers $p,q$ of opposite parity (and coprime since $x,y$, and $z$ are) such that $x^{2}=p^{2}+q^{2}$ and either $y^{2}=p^{2}-q^{2}$ or $y^{2}=2pq$.

Factoring the original equation, we get

If $y^{2}=p^{2}-q^{2}$, then $(xy)^{2}=p^{4}-q^{4}$, and clearly $p^{2}+q^{2}=x^{2}<x^{2}+y^{2}$, so we have found a solution smaller than the assumed minimal solution.

Assume therefore that $y^{2}=2pq$. Now, $x^{2}=p^{2}+q^{2}$; we may assume by relabeling if necessary that $q$ is even and $p$ odd. Then $p,q,x$ are pairwise coprime and form a pythagorean triple; thus there are $P>Q>0$ of opposite parity and coprime such that

Then

is a square; it follows that $P,Q$, and $P^{2}-Q^{2}$ are all (nonzero) squares since they are pairwise coprime. Write

for positive integers $R,S,T$. Then $T^{2}=R^{4}-S^{4}$, and

We have thus found a smaller solution in positive integers, contradicting the hypothesis. ∎

Suppose $x,y,z$ is a right triangle with $z$ the hypotenuse, and let $d=\gcd(x,y,z)$. Either $x/d$ or $y/d$ is even; by relabeling if necessary, assume $x/d$ is even. Then we can choose relatively prime integers $p,q$ with $p>q$ and of opposite parity such that

If the triangle’s area is to be a square, then

must be a square, and thus $pq(p^{2}-q^{2})$ must be a square. Since $p$ and $q$ are coprime, it follows that $p$, $q$, and $p^{2}-q^{2}$ are all squares, and thus that $p^{2}-q^{2}$ is the difference of two fourth powers. But then

must also be a square. Since both $p$ and $q$ are squares, this is impossible by the theorem. ∎