Greatest Common Divisor Proof: $\gcd(m^2-n^2, m^2+n^2) = 1$ Prove that if $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$ then $\gcd(m^2-n^2, m^2+n^2) = 1$
Workings:
Suppose that $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$
$\gcd(m^2-n^2, m^2+n^2)$
$= gcd((m-n)(m+n), (m-n)(m+n)+2n^2)$
Now I know that $m+n=1 (\text{mod} ~2)$ means that one of $m$ or $n$ is odd or even
But now I'm not too sure on what to do.
Any help will be appreciated.

We can make use of the fact that $\gcd(a, b) = \gcd(a, b-ka)$ for any integer $k$. Note that 
$\gcd(m^2 - n^2, m^2 + n^2) = \gcd(m^2 - n^2, 2n^2).$ You noted that one of $m, n$ is odd and one is even, so $2 \nmid m^2 -n ^2$. So $$\gcd(m^2 - n^2, 2n^2) = \gcd(m^2 - n^2, n^2) (\text{why?}) = \gcd(m^2, n^2).$$
From there, you can argue that this gcd must be 1.
