Proving divisibility involving Pythagorean triplet.

Question

Let $a$, $b$, $c$ be natural numbers such that $$ a^2 +b^2=c^2 $$ and $$c-b=1.$$

Prove that $$a^b + b^a $$ is divisible by $c$. Any hints??

Answer 1

$a^2=2b+1=2(c-1)+1\equiv-1\pmod c$

$b=c-1\equiv-1,b^a\equiv-1$ as $a$ is odd

If $a=2d+1,b=4\cdot\dfrac{d(d+1)}2$