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.

Euclid's algorithm gives:
\begin{align*}
\gcd(m^2 - n^2, m^2 + n^2)
&= \gcd(m^2 - n^2, (m^2 + n^2) - (m^2 - n^2)) \\
&= \gcd(m^2 - n^2, 2n^2)
\end{align*}
Since $m + n \equiv 1 \pmod{2}$, $m^2 - n^2 \equiv 1 \pmod{2}$, so we can forget about the $2$.
We get
\begin{align*}
&= \gcd(m^2 - n^2, n^2) \\
&= \gcd(m^2, n^2) \\
&= 1.
\end{align*}
