We say that $(a,b,c) \in \mathbb{N}^3$ is a Pythagorean triple if $a^2 + b^2 = c^2$. Is there a characterization of those Pythagorean triples $(a,b,c)$ for which $ab$ is a square-residue modulo $c^2$?

This is related to an open problem I've been working on.

  • $\begingroup$ Well done. I wish you success. I hope that we will solve this equation $$ax^2+bxy+cy^2=jz^2$$ $\endgroup$ – individ Apr 4 '15 at 4:32
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    $\begingroup$ Just a quick note: 2ab is a quadratic residual mod c^2. $\endgroup$ – Fan Zheng Apr 4 '15 at 6:23
  • $\begingroup$ In general the solutions of the pythagorean triples are $(m^2-n^2,2mn,m^2+n^2)$. We have $(a+b)^2\equiv 2ab (mod c^2)$. So $2ab$ is always a quadratic residue and $ab$ is a quadratic residue if and only if $2$ is a quadratic residue. From math.stackexchange.com/questions/180002/… it follows that the prime factors $p$ of $c^2$ satisfy that $p^2\equiv 1 mod (16)$. In johndcook.com/blog/quadratic_congruences it is described how to solve quadratic congruences. $\endgroup$ – user35593 Apr 5 '15 at 8:22

If $c$ is even $a$ and $b$ are also even. Hence we can first assume that $c$ is odd. By the comment above the answer is all triples $2^k(m^2-n^2,2mn,m^2+n^2)$ with $m\equiv 0 (mod 4)$ and $n\equiv 1,-1,7$ or $-7 (mod 16)$.


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