2015/031) if $a^2 + b^2 = c^2$ then show that $\dfrac{(c-a)(c-b)}{2}$ is a perfect square

We have
$2(c-a)(c-b) = 2 c^2- 2(a+b) c + 2ab$

= $(c^2+2ab) + (c^2- 2(a+b)c)$

= $(a^2+b^2 + 2ab) – 2(a+b)c + c^2$

= $(a+b)^2 – 2(a+b) c + c^2$

= $(a+b-c)^2$

so $\dfrac{(c-a)(c-b)}{2} = (\dfrac{(a+b-c)}{2})^2$

 

proved