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.
Best Answer
Start from $1105 = 5\cdot 13\cdot 17$ whose prime factors all has the form $4k+1$.
Decompose the prime factors in $\mathbb{Z}$ over gaussian integers $\mathbb{Z}[i]$ as
$$5 = (2+i)(2-i),\quad 13 = (3+2i)(3-2i)\quad\text{ and }\quad17 = (4+i)(4-i)$$
Recombine the factors of $1105^2$ over $\mathbb{Z}[i]$ in different order and then turn them to sum of two squares. We get:
$$\begin{array}{rc:rr} 1105^2 = & 943^2 + 576^2 & ((2+i)(3+2i)(4+i))^2 = & -943 + 576i\\ = & 817^2 + 744^2 & ((2-i)(3+2i)(4+i))^2 = & 817 + 744i\\ = & 1073^2 + 264^2 & ((2+i)(3-2i)(4+i))^2 = & 1073 + 264i\\ = & 1104^2 + 47^2 & ((2-i)(3-2i)(4+i))^2 = & -47 - 1104i\\ \end{array} $$ A counter-example for the speculation that an integer can appear in at most two primitive Pythagorean triples.