$$\frac{1}{u^2}+\frac{1}{v^2}=\frac{1}{w^2}$$
I want to generate all primitive solutions up to $u \le N$. Is there a parametric solution?
By brute force, I got these solutions:
$(15, 20, 12),(20, 15, 12),(65, 156, 60),(136, 255, 120),(156, 65, 60),(175, 600, 168),(255, 136, 120),(580, 609, 420),(600, 175, 168),(609, 580, 420)$.
It seems that $c$ is a multiple of $12$.
Another observation is $u^2+v^2=t^4$ where $t$ is the hypotenuse of a triangle.
Best Answer
With $A$ being even, $r>s>0,$ $\gcd(r,s) = 1$ and $r+s$ odd,
$$ A = 2rs\left( r^2 + s^2 \right) \; , \; \; B = r^4 - s^4 \; , \; \; C = 2rs\left( r^2 - s^2 \right) \; . \; \; $$
The proof starts with $\gcd(A,B,C) = 1,$ goes on with $\gcd(A,B) = g, \; \;$ $\gcd(g,C) = 1,$ $A = ga, \; \; B = gb, \; \; $ $\gcd(a,b)=1,$ and goes on from there...