Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.

Multiply by conjugate: $${ab\over a+b+c}={ab\over a+b+c}\cdot \frac{a+b-c}{a+b-c}=\frac{ab(a+b-c)}{2ab}=\frac{a+b-c}{2}\in \mathbb Z^+,$$ because: $$a+b>c$$ and there are two cases for $a^2+b^2=c^2$: 1) $a,b,c$ are even; 2) one is even, the other two are odd. And for each case, $a+b-c$ is even.

Alternatively, use Formulas for generating Pythagorean triples

WLOG $a=2pqk, b=(p^2-q^2)k,c=(p^2+q^2)k$

$c-a=k(p-q)^2$

$c-b=2kq^2$