Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet Show that $(m^2 - n^2, 2mn, m^2 + n^2)$ is a primitive Pythagorean triplet
First, I showed that $(m^2 - n^2, 2mn, m^2 + n^2)$ is in fact a Pythagorean triplet.
$$\begin{align*} (m^2 - n^2)^2 + (2mn)^2 &= (m^2 + n^2)^2 \\
&= m^4 -2m^2n^2 + n^4 + 4m^2n^2 \\
&= m^4 + 2m^2n^2 + n^4 \\
&= 1\end{align*}$$ which shows that it respect $a^2+b^2 = c^2$
let p be a prime number, $ p|(m^2 + n^2) \text { and } p|(m^2 - n^2) $
if $gcd(m^2 + n^2, (m^2 - n^2)) = 1$
$p | (m^2 + n^2) , \text { so, } p |m^2 \text { and } p |n^2$
that means $ (m^2 + n^2) \text { and } (m^2 - n^2) $ are prime together
I'm kind of lost when I begin to show the gcd = 1... I think I know what to do, just not sure how to do it correctly. 
Thanks

The question is wrong.
It is not always primitive.
As when we put m=3, n=1
The triplet comes to be 8,6,10, which is not a primitive Pythagorean triplet.
