My proof extends to still a more general case that includes Pythagorean Triples
Theorem
For [math]A,B,x,y,m \in Z | (A,B = Constants)[/math]
Let [math] N = F_{x,y} = x^{2} + Bxy -y^{2}[/math] where gcd(x,y) = 1
Then there exists a coprime pair [math](x_{1},y_{1})[/math] such that
[math]F_{x_{1},y_{1}} = N^{2^{m}}[/math]
I have a recursive formula that gives [math](x_{1},y_{1})[/math]
It even gives values if [math] N = F_{x,y} = Ax^{2} + Bxy -y^{2}[/math] e.g. [math](x_{1},y_{1})[/math] such that [math]F = N^{2}[/math]
If A is a perfect square and B = 0 then my formula relates to Pythagorean triples
e.g.
[math]2^{2} - 1^{2} = 3[/math]
[math]5^{2} - 4^{2} = 3^{2}[/math]
[math]41^{2}-40^{2}=3^{4}[/math]
[math]3281^{2}-3280^{2}=3^{8}[/math]
[math]3^{2}-2^{2}=5[/math]
[math]13^{2}-12^{2}=5^{2}[/math]
[math]313^{2}-312^{2}=5^{4}[/math]
[math]195313^{2}-195312^{2}=5^{8}[/math]