$\text{Suppose a, b and c are positive integers such that a,c are odd.}$
$\text{We are given that:} \space a^2+b^2=c^2$
$\text{Prove that b+c is a perfect square.}$
..
$\text{What I got so far:}$
$odd+even=odd \to \text{b is even}$
$(2p+1)^2+(2q)^2=(2r+1)^2$
$4p^2+4p+1+4q^2=4r^2+4r+1$
$p(p+1)+q^2=r(r+1)$
$\text{For any positive integer x,} \space x(x+1) \equiv 0 \space (mod \space 2)$
$Therefore, \space q^2 \equiv 0 \space (mod \space 2) \to q \equiv 0 \space (mod \space 2)$
$\text{I don't know what to do anymore :(}$
Your claim is false: consider the $9$, $12$, $15$ right triangle. It's true for primitive triples (when $b$ and $c$ are relatively prime). Then write $$ c^2 - b^2 = a^2 $$ and factor to find a proof.
You can do this without Wikipedia's complete description of the primitive triples @labbhattacharjee points to in his comment.