Is there a way of finding out the remaining two numbers of pythagorean triple if one of the side is given

Question

I am solving one question related to right triangle and I have to find out the remaining two numbers of the pythagorean triple if one of the number is given. I know there can be many triples possible , but I just need to find one triple. Thank you for your help :)

Answer 1

If the number is odd, its square can be represented as $2k+1$. Then one triple is $k, \sqrt{2k+1}, k+1$. If the number is even, it is $2k$ and you can use $k^2-1, 2k, k^2+1$

Answer 2

If your number $x$ is even, then any factorization $x/2=m\cdot n$, with $m>n$, gives a triple $(m^2+n^2, m^2-n^2,x)$.

It $x$ is odd, then again factor it as $x=m\cdot n$, with $m>n$, and a triple is $\left({m^2+n^2\over2},{m^2-n^2\over2},x\right)$.

If $x$ is not prime and $d$ is one of its divisors, then you can find other triples by considering the triples generated by $x/d$ and multiplying them by $d$

This works, however, as long as $x$ represents the length of a cathetus.