If \(x\) and \(y\) are coprime and \(gcd(x^2+y^2,x^2-y^2)=d\), then \(d\) is equal to either 2 or 1.
Proof:
By our hypothesis, we have: \[x^2+y^2=d\cdot a\] \[x^2-y^2=d\cdot b\]
For integers \(a\), \(b\in\mathbb{Z}\). We can combine these equations through addition. Doing so gives us: \[\phantom{+\text{ }}x^2+y^2=d\cdot a\] \[+\text{ }(x^2-y^2=d\cdot b)\] \[\line(200,0){100}\] \[2x^2=da+db\]
By factoring out like terms, we see: \[2x^2=d(a+b)\] This implies \(d\mid 2x^2\). From this we can conclude \(d\mid 2\) or \(d\mid x\).
We can also subtract these equations to get: \[\phantom{-\text{ }}x^2+y^2=d\cdot a\] \[-\text{ }(x^2-y^2=d\cdot b)\] \[\line(200,0){100}\] \[2y^2=da-db\]
By factoring out like terms again, we see: \[2y^2=d(a-b)\] This implies \(d\mid 2y^2\). From this we can conclude \(d\mid 2\) or \(d\mid y\).
Therefore we see that: \[d\mid 2\text{ or }d\mid x\quad \text{and}\quad d\mid 2 \text{ or } d\mid y\]
Since it is possible for both cases that \(d\mid 2\), it is possible \(d=2\). However, since it is possible \(d\mid x\) or \(d\mid y\), it follows that \(d=1\) since \(d\) can not divide both \(x\) and \(y\) simultaneously.
Hence, the \(gcd(x^2+y^2,x^2-y^2)\) is either 2 or 1. Q.E.D.