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I've had some difficulties finding answer to the two following questions:
1) Given one of natural numbers $a,b$ where $b$ is even and $a^2+b^2=c^2$ is there only one such a pythagorean triple?
2)How about a sum of $n$ squares of natural numbers that is equal to a square of a natural number? Given one of the summed squares is there only one such a pythagorean "n-let"?
I know that $a$ and $c$ are odd and I am aware of Euclid's formula but I don't know how to use it. I have no idea how to tackle 2). I only know that for exemple of $n=4$ there may be or not such two summed squares that would be lengths of the legs of a right triangle. Perhaps this is trivial but I don't know what to do.
Edit: 3) How about a given $c$ instead of $a,b$ in 1)?

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    $\begingroup$ $(5,12,13)$ and $(9,12,15)$ $\endgroup$ – lulu Nov 5 '18 at 11:09
  • $\begingroup$ If anyone were to object that $(9,12,15)$ is not primitive, consider in its stead $(35,12,37)$ $\endgroup$ – Keith Backman Nov 5 '18 at 15:07
  • $\begingroup$ Solve $2xy=2uv=N$ with $N$ having many divisors and then apply Pythagorean identity for $(x,y)$ and for $(u,v)$. $\endgroup$ – Piquito Nov 5 '18 at 16:07
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Question $1$ appears to follow from the two variable equivalent expression:

$$a^2+b^2=c^2, b\in 2k, a,b,c,k \in \Bbb Z $$ $$\to (p^2-q^2)^2 + (2pq)^2=(p^2+q^2)^2, p,q\in \Bbb Z $$

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HINT.- From $$(x_n^2+x_1^2+x_2^2+\cdots+x_{n-1}^2)^2-(x_n^2-x_1^2-x_2^2+\cdots-x_{n-1}^2)^2=2x_n^2(2x_1^2+\cdots+2x_{n-1}^2)$$ you do have the identity $$(x_n^2-x_1^2\cdots -x_{n-1}^2)^2+(2x_nx_1)^2+(2x_nx_2)^2\cdots+(2x_nx_{n-1})^2=(x_n^2+x_1^2+x_2^2+\cdots+x_{n-1}^2)^2$$ which is a kind of generalization of the identity giving the Pythagorean triples.

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