Can we find two non-congruent right triangles with whole-number lengths and congruent hypotenuses? I know some ways to find some Pythagorean triples.
And I understand that if $a^2 + b^2 = c^2$ then $(a-b)^2 + (a+b)^2 = 2c^2$.
I feel like that suggests a way forward, but I cannot find that way.
Is there an algorithm to pick four whole numbers to serve as the legs of two right triangles such that the triangles are not congruent but the hypotenuses are the same whole number?

Let $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ be arbitrary non-proportional pythagotrean triples, and let $d:={\rm gcd}(c_1,c_2)$. Then
$${c_2\over d}\left(a_1,b_1,c_1\right),\qquad{c_1\over d}\left(a_2,b_2,c_2\right)$$
are pythagorean triples to noncongruent right triangles with the same hypotenuse $c={c_1c_2\over d}$.
