Show $a$ and $b$ must be relatively prime where $a= m^2 - n^2$ and $ b = 2mn$. From this show that $r$ and $s$ are relatively prime. Show this implies that $r$ and $s$ must be perfect squares as well. $r = n^2$ and $s = m^2$.
I don't really know where to start here. I'm new with proofs. My first inclination is to plug $r$ and $n$ into $a$ and $b$, but this doesn't get me anywhere. Suggestions?
Let $GCD(m,n)=f$ (a common factor) so that $(m,n)=(fx,fy)*emphasized text*$.
$$\text{ then }A=(m^2-n^2)= f^2x^2-f^2y^2=f^2(x^2-y^2)\quad \land \quad B=2mn=2fxfy=f^2(2xy)$$ $$\text{ then }(m,n)\text{ are co-prime if and only if } f=1 $$. If any natural numbers are co-prime, then their squares are co-prime because they have no common factors to square as in the example of $GCD(3^2, 2^2)-1$.
$$\therefore GCD(m,n)=1\land r=m^2\land s=n^2\implies GCD(r,s)=1 $$