A very different property of primitive Pythagorean triplets: Can number be in more than two of them?

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.

Solution 1:

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.

Solution 2:

The "smallest" counterexample is \begin{gather*} 5^2 + 12^2 = 13^2 \\ 9^2 + 12^2 = 15^2 \\ 12^2 + 16^2 = 20^2 \end{gather*} ($12$ also satisfies $12^2 + 35^2 = 37^2$.)

Edit: now that the OP has stipulated that the elements be pairwise coprime, the smallest counterexample (in the sense that the repeated number is minimal) is

\begin{gather*} 11^2 + 60^2 = 61^2 \\ 60^2 + 91^2 = 109^2 \\ 60^2 + 221^2 = 229^2 \end{gather*}

$60$ also satisfies $60^2 + 899^2 = 901^2$.

Solution 3:

$120,160,200$ and $90,120,150$ and $72,96,120$

How I arrived at it: It is a common knowledge that if we scale the triplet $3,4,5$ by any constant, we get another triplet. So I found out a common multiple of $3,4,5$, which is $120$ and then scaled the triplet one by one with the constants $\frac{120}{3}$, $\frac{120}{4}$ and $\frac{120}{5}$.