How to show if $m$ and $n$ are coprime and $m-n$ is odd, then $(m^2-n^2)^2$, $(2mn)^2$, $(m^2+n^2)^2$ have a common factor? We know that the Pythagorean triples can be generated by the Euclid's formula $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$ for any positive integers $m,n$ and $m>n$.
I am trying to prove the statement:
The triple generated by Euclid's formula is primitive if and only if $m$ and $n$ are coprime and $m-n$ is odd.
I could prove the $\Rightarrow$ direction. Now I am trying to show the $\Leftarrow$ proof.
I am trying to prove by contradiction. If $m$ and $n$ are coprime and $m-n$ is odd, let's assume the triple generated is not primitive. Then there is a factor $k$ in each term. But since $m$ and $n$ are coprime and $m-n$ is odd, we can deduce that $(m^2-n^2)^2$ is odd, $(2mn)^2$ is even and $(m^2+n^2)^2$ is odd. Then $k$ can only be odd. Let $k=2p+1$ where $p$ is an integer. But then I am not sure how to proceed to obtain a contradiction.
Any helps are greatly appreciated. Many thanks!

Informal outline Suppose for example that $(m^2-n^2)^2$ and $(2mn)^2$ are not relatively prime. Then there is a prime $p$ that divides $m^2-n^2$ and $2mn$. Since $m$ and $n$  have opposite parity, $p\ne 2$. 
Since $p$ divides $2mn$, and $p\ne 2$, it follows that $p$ divides one of $m$ or $n$, say $m$. Since $p$ divides $m^2-n^2$, it follows that $p$ divides $n^2$, and hence $n$. This contradicts the fact that $m$ and $n$ are relatively prime.  
Remark: We started, as you did, from the assumption that the numbers $(m^2-n^2)^2$, $(2mn)^2$, and $(m^2+n^2)^2$ are not pairwise relatively prime. However, a triple $(x,y,z)$ such that $x^2+y^2=z^2$ is usually called primitive if $x$, $y$, and $z$ are pairwise relatively prime. It makes no real difference, since $x^2$, $y^2$, and $z^2$ are pairwise relatively prime if and only if $x$, $y$, and $z$ are. 
