<s>[INST] Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square. Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square. [Hint: consider the triple $4n^2, n^4-4, n^4+4$, where $n$ is an arbitraty odd integer.]
What I got:
Using the hint. $4n^2, n^4-4, n^4+4$ is a Phytagorean triple if $x^2+y^2=z^2$. Replacing and solving the equation it is clear that, $(4n^2)^2+(n^4-4)^2=(n^4+4)^2$ where $n$ is odd is indeed a Pythagorean triple with $x=4n^2=(2n^2)^2$ a perfect square.
Now I have to prove that $gcd(4n^2, n^4-4, n^4+4)=1$.
But I'm stuck here. I tried $gcd(n^4-4, n^4+4)=1$ without success. Any ideas?
 [/INST] All Pythagorean triples can be generated by the formula:
$$A=(m^2-n^2)k\qquad B=2mnk\qquad C=(m^2+n^2)k$$
For a triple to be primitive, it is necessary but not sufficient that $k=1$. This leaves $B=2mn$. The smallest $m$ is $2$; $m,n$ must be of opposite parity; $m>n$, meaning 2mn must be a multiple of $4$
Every side-B is a multiple of $4$, e.g. $\quad (3,4,5)\quad (15,8,17)\quad (5,12,13)\quad (35,16,37)\quad ...$
Among these are all powers of $2$ that are perfect squares so, there being infinite combinations of $(m,n)$, there are an infinite number of triples where $\sqrt{B}=\lfloor\sqrt{B}\rfloor.$
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