Consider the identity $4AB=(A+B)^2 -(A-B)^2$ . Assume that $A,B$ are coprime. (This is no restriction: If $d$ is the greatest common divisor of $A,B$ , then one can write $A=dA^{'}, B=dB^{'}$ to get $4d^2(A^{'}B^{'})^2=d^2\left((A^{'}+B^{'})^2 -(A^{'} -B^{'})^2\right)$ and cancel $d^2$ .)

For $4AB,A+B, A-B$ to form a Pythagorean triple, each of $A, B$ must be squares. So $A=m^2, B=n^2$ where $m,n$ are coprime, $n<m$ . So we have \begin{equation} \label{eq1} a=2mn, b=m^2-n^2, c=m^2+n^2 \end{equation}and $\{a,b,c\}$ are a Pythagorean triple. But this needn't be primitive: If $m,n$ are odd, then $2 \mid m^2 \pm n^2$ , so not all of $a,b,c$ are relatively prime.

Suppose $2mn, m^2-n^2, m^2+n^2$ are pairwise coprime. Then $\gcd(2mn, m^2+n^2)=\gcd(2mn, (m+n)^2)=1=\gcd(2mn,(m-n)^2)$ , and it follows that $\gcd(2mn,m+n)=1, \gcd(2mn,m-n)=1$ . Thus $\gcd(2,m+n)=1$ , i.e. $m \pm n$ is odd. Furthermore, $\gcd(2mn,m+n)=1$ implies $\gcd(m,n)=1$ . And since the sum/difference of two integers is odd iff one is even, and the other is odd, only one of $m,n$ is odd. Thus, $\gcd(m^2-n^2,m^2+n^2)=\gcd(m^2+n^2,2n^2)=1$ . Conversely, if $m,n$ are coprime, and exactly one of $m,n$ is odd, then $\gcd(2,m \pm n)=1$ ; thus, $2mn$ ,$m^2-n^2$ are coprime. From the fact that $\gcd(a+b, a-b)=\gcd(a,b)$ if $a,b$ have opposite parity and $\gcd(m^2,n^2)=1$ it follows that $m^2-n^2$ ,$m^2+n^2$ are also coprime. And since $\gcd(2mn,m^2+n^2)=\gcd(2mn,(m+n)^2)$ and $\gcd(2mn,m+n)=1$ it follows that $2mn,m^2+n^2$ are coprime. So the conditions the Pythagorean triple $\{2mn, m^2-n^2,m^2+n^2\}$ is primitive, $\gcd(2mn,m+n)=1$ and $m,n$ are coprime and exactly one of them is odd are equivalent.

So if $a,b,c$ satisfy $a^2+b^2=c^2$ and $a,b,c$ are pairwise coprime, then $c$ is odd, and exactly one of $a,b$ are odd and the other is even.

Let $n,m$ be coprime positive integers of opposite parity, $n<m$ . Set \begin{equation} n^{'}=m+n, \; m^{'}=n-m \end{equation}in equation gives \begin{equation} b=m^{'}n^{'}, c=\frac{n^2+m^2}{2}=\frac{\left.m^{'}\right.^2 +\left.n^{'}\right.^2}{2}, \;a=\frac{\left.n^{'}\right.^2-\left.m^{'}\right.^2}{2} \end{equation}since $n=\frac{m^{'}+n^{'}}{2}$ , $m=\frac{n^{'}-m^{'}}{2}$ . Clearly, $\gcd(m^{'},n^{'})=1$ .

Now we prove that any primitive Pythagorean triple can be generated choosing odd coprime integers.

Proof. Since $\gcd(n^2, m^2)=1=\gcd(f_1+f_2, f_1-f_2)$ , the statement follows from the fact that $f_1, f_2$ have opposite parity since in this case $\gcd(f_1,f_2)=\gcd(f_1+f_2, f_1-f_2)$ . Since $4 \mid n^2-m^2$ , $f_2$ is odd, and since $f_1=f_2+n^2$ and $n$ is odd, $f_1,f_2$ have opposite parity.

Substituting $C=n^2$ , $B=m^2$ in $BC=\left(\frac{C+B}{2}\right)^2 -\left(\frac{C-B}{2}\right)^2$ yields that $mn,\, \frac{n^2-m^2}{2},\, \frac{n^2+m^2}{2}$ is a primitive Pythagorean triple.

To see that any primitive Pythagorean triple is of this form:

Proof. $a,b$ cannot both be even since $\gcd(a,b)=1$ . If both $a,b$ were odd we had $c^2 \equiv 2 \pmod{4}$ which is impossible since the square of any number is either congruent 0 or 1 modulo 4. Thus, $c$ must be odd. Now for any integers $a,b$ the congruence $a^2+b^2 \equiv (a+b)^2 \pmod{2}$ holds. Together with $c^2\equiv 1 \pmod{2}$ this gives $a+b\equiv 1 \pmod{2}$ , so $a,b$ have opposite parity.