While playing with numbers, I thought about squares of numbers, and then the first thing that came to mind was Pythagorean triplets.
I observed a very interesting fact that any $x\in\mathbb N$ can never be a member of more than two Pythagorean triplets of pairwise coprime numbers, like $(3,4,5)$ and $(8,15,17)$.
For example $$16^2+63^2=65^2$$$$33^2+56^2=65^2$$
are the possible triplets for $x=65$, and $65$ cannot exist in any other triplet of co-primes.
Now I need to prove this conjecture.
So I thought that in a Pythagorean triplet, the three numbers are of the form $(2mn, m^2-n^2, m^2+n^2)$.
Let $x$ be a number . Then I have to show that $$x=2mn$$$$x=a^2+b^2$$$$x=y^2-z^2$$ are not simultaneously possible.
But I am stuck and don't know where to go from here.
Please help me prove this or help me disprove it by giving a counter example.