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!
 A: 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. 
A: Suppose $k | (m^2-n^2)$ and $k | (m^2+n^2)$, then $k | 2m^2$ and $k | 2n^2$. But then $k$ divides $\gcd(2m^2,2n^2)=2\gcd(m^2,n^2)=2$. The last step follows from the fact that $\gcd(m,n)=1$. Thus $k=1$ or $k=2$. Ask yourself why can't $k$ be $2$?
A: I thought I had a proof that, if $m,n$ are mutually prime, $m,n$ would generate only primitive Pythagorean triples. The $proof$ is between the asterisk below.
$$\text{*************************}$$
$\text{We are given }\quad A=m^2n^2\quad B=2mn\quad C=m^2+n^2$
Let $x$ be the GCD of $m,n$ and let $p$ and $q$ be the cofactors of $m$ and $n$ respectively. Then we have
$$A=(xp)^2-(xq)^2\quad B=2xpxq\quad C=(xp)^2+(xq)^2$$
$$A=x^2(p^2-q^2)\quad B=2x^2(pq)\quad C=x^2(p^2+q^2)$$
If $GCD(m,n)=1$, then $GCD(A,B,C)=1$ and $(A,B,C)$ is a primitive triple. This means that $m$ and $n$ must be co-prime to generate a primitive.
$$\text{*************************}$$
However, a counter-example destroys to so-called proof: $\quad\text{Let }m,n=7,3$.
$$A=49-9=40\quad B=2*7*3=42\quad C=49+9=58\quad GCD(40,42,58)=2$$
$$\therefore GCD(m,n)=1\neg\implies GCD(A,B,C)=1$$
I believe but do not know that the proof may be valid if we insist that $m$ and $n$ be of opposite parity.
The only two formulas I know about that will generate only primitive triplets do not generate all of them but $C-B=1$ in the first one and $C-A=2$ in the second one.
$$A=2n^2+1\quad B=2n^2+2n\quad C=2n^2+2n+1$$
$$A=4n^2-1\quad B=4n\quad C=4n^2+1$$
$\mathbf{UPDATE:}$ I did some looking and found a variation of Euclid's formula that can let us prove when we have found a primitive with more confidence.
$$\text{Let}\quad A=(2m-1+n)^2-n^2\quad B=2(2m-1+n)n\quad C=(2m-1+n)^2+n^2\quad m,n\in\mathbb{N}$$
This formula generates only and all triples where $GCD(A,B,C)=(2k-1)^2,k\in\mathbb{N}$ which includes all primitives.
If we let $GCD((2m-1),n)=x$ and $p,q$ be the respective cofactors, then we have
 $$A=(xp+xq)^2-(xq)^2\quad B=2(xp+xq)xq\quad C=(xp+xq)^2+(xq)^2$$
$$A=x^2(p+q)^2-x^2q^2\quad B=2x^2(p+q)q\quad C=x^2(p+q)^2+x^2q^2$$
We can see by inspection that if $(2m-1)$ and $n$ have common factors, then GCD(A,B,C) an odd square and, if GCD((2m-1),n)=1, the triple is primitive.
