Given hypotenuse, find the other two sides. Note that we are only interested in integral pythagorean triplets, we are given the hypotenuse $c$, how can I efficiently find the other two sides of the right angled triangle. I need something better than the bruteforce approach of iterating over all lengths $a$ below $c$, and checking perfect square for $b = \sqrt{c^2-a^2}$.
For multiple solutions, I need one with the smallest $a$ possible.

How about using the standard formula for generating Pythagorean triples?  Solve $c = m^2 + n^2$ for $m$ and $n$. Then you have $a = m^2 - n^2$ and $b = 2mn$. (If $m$ and $n$ are co-prime and of opposite parity, the triple is primitive, otherwise not.) This requires less brute force than the approach you wanted to avoid, since $c < c^2$.
