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Given $m,n\in\Bbb N$ with $m<n$, how many pythagorean triples $p^2+r^2=q^2$ satisfy $$m\leq p<r\leq n?$$

Is there a way to give a sharp estimate?

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2 Answers 2

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See my preprint. Page 12 (this has since been published, but preprint is easiest to access).

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For each type of Pythagorean Triples $p=2xy$, $r=x^2-y^2$ and $p=x^2-y^2$, $r=2xy$ your inequalities describe a simple region on $Oxy$ plane. Boundary consits from the straight line $2xy=x^2-y^2$ and hyperbolas $2xy=p, x^2-y^2=r$ and $x^2-y^2=p, 2xy=r$.

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  • $\begingroup$ Yes. But how big or small is this as a function of $m,n$? $\endgroup$
    – user76479
    Sep 26, 2015 at 12:20
  • $\begingroup$ Find the area of this region. $\endgroup$ Sep 26, 2015 at 12:21
  • $\begingroup$ is that sufficient? will it be upper or lower bound and what is error term? $\endgroup$
    – user76479
    Sep 26, 2015 at 12:25
  • $\begingroup$ Yes. If the region has perimeter $P$ then the area is $\approx P^2$. The simplest error term for the number of point is $O(P)$ and for the number of primitive points is $O(P\log P)$. $\endgroup$ Sep 26, 2015 at 13:14
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    $\begingroup$ Arul: as you know, I also gave this answer (cf. my comment above), and gave an explicit integral for the area. I now feel as if I've been wasting my breath. $\endgroup$
    – Todd Trimble
    Sep 26, 2015 at 13:40