<s>[INST] 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.
 [/INST] 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, $$127^2+236^2=1^2+268^2=257^2+76^2=191^2+188^2=1105^2.$$ 
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