Basic questions about pythagorean triples and "n-lets" 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)?
 A: To construct n-lets, we can begin with this function to find values of $(m,n)$ for Euclid's formula:
$$n=\sqrt{m^2-A}\text{ where m varies from }\lceil\sqrt{A}\rceil\text{ to }\frac{A+1}{2}$$
This will let us find a triple with a matching side A, if it exists, for any $m$ that yields a positive integer $n$.
Let's begin with $(3,4,5)$ and find a triple to match the hypotenuse. In this case,
$$\lceil\sqrt{A}\rceil=\frac{A+1}{2}=3\text{ and } \sqrt{3^2-5}=2\text{ so we have }(3,2)$$
$$A=3^2-2^2=5\qquad B=2*3*2=12\qquad C=3^2+2^2=13$$
$\text{ The n-let that follows is }\\3^2+4^2+12^3=13^2\\\text{ and continuing with this process, we can get}$
$$3^2+4^2+12^2+84^2+3612^2=3613^2$$
or
$$3^2+4^2+12^2+84^2+132^2=157^2$$
Here we have two triples that match the hypotenuse of $(13,84,85)$ and here is how we found them
$$m =\lceil\sqrt{85}\rceil=10\text{ to }\frac{85+1}{2}=43$$
In the loop from $10$ to $43$, we find $(11,6)$ and $(43,42)$ with integer $n$.
$$A=11^2-6^2=85\qquad B=2*11*6=132\qquad C=11^2+6^2=157$$
$$A=43^2-42^2=85\qquad B=2*43*42=3612\qquad C=43^2+42^2=3613$$
A: Question $1$ appears to follow from the two variable equivalent expression:
$$a^2+b^2=c^2, b\in 2k, a,b,c,k \in \Bbb Z $$
$$\to (p^2-q^2)^2 + (2pq)^2=(p^2+q^2)^2, p,q\in \Bbb Z $$
A: 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.
