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.
 A: $16,  63,  65$
$25,  60,  65$ 
$33,  56 , 65$
Observed from: http://www.tsm-resources.com/alists/trip.html
Possible interesting reasearch question: Are there numbers that appears in infinitely many Pythagorean triples?
A: Any prime which is congruent to $1$ modulo $4$ is the largest member of a Pythagoren triplet. (That is not obvious.) 
If $P_1,..., P_n$ are $n$ distinct primes, each congruent to $1$ modulo 4, then $\prod_{j=1}^nP_j$ is the largest member of $2^{n-1}$ different primitive Pythagorean triplets. ( It is not obvious that the method, below, always yields that many.)
Use $(aa'+bb')^2+(ab'-ba')^2=(aa'-bb')^2+(ab'+ba')^2=(a^2+b^2)(a'^2+b'^2).$
Example : (For ease of typing let "$.$" denote "$\times$").   From $3^2+4^2=5^2$ and $5^2+12^2=13^2$  we have  $$(3.5\pm 4.12)^2+(3.12\mp 4.5)^2=5^2.13^2.$$ That is, $63^2+16^2=33^2+56^2=65^2.$...... Since $17=1^2+4^2$ we have $$(63.1\pm 16.4)^2+(63.4\mp 16.1)^2=17^2.65^2=$$ $$=  (33.1\pm 56.4)^2+(33.4\mp 56.1)^2=17^2.65^2.$$ That is, $$127^2+236^2=1^2+268^2=257^2+76^2=191^2+188^2=1105^2.$$ 
A: 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.
A: 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 $60^2 + 899^2 = 901^2$.
A: $120,160,200$ 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 $\frac{120}{5}$.
