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Say $a^2+b^2=c^2$ is a primitive Pythagorean triple. Then consider the Linear Diophantine Equation $$ua^2+vb^2+xab+ybc+zca=0$$ where $(u,v,x, y, z)\in\mathbb Z^4$ are variables. If $(u,v,x, y, z)\neq(0,0,0,0,0)$ then can we say anything about $\|(u,v,x, y, z)\|_\infty$ or the probability distribution of $\|(u,v,x, y, z)\|_\infty$?

By this I mean can $\|(u,v,x, y, z)\|_\infty$ be much smaller than $\sqrt{\max(a^2,b^2)}$?

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  • $\begingroup$ Sorry, how do you want to get any small solution if $a|y,b|z$ and $c|x$ provided $(a,b)=1$? $\endgroup$ Apr 12, 2020 at 9:45
  • $\begingroup$ $ybc=-a(xb+zc)$, so $a|ybc$. If $a,b,c$ are pairwise coprime then $a|y$ is nesessary. $\endgroup$ Apr 12, 2020 at 10:58
  • $\begingroup$ @PavelKozlov I updated and for this I am pretty sure we can do better that $\sqrt{\max(a^2,b^2)}$. $\endgroup$
    – VS.
    Apr 12, 2020 at 11:32

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Yes, when $m>n>0$ and $$ a = m^2 - n^2 $$ $$ b = 2mn $$ $$ c = m^2 + n^2 $$ then $$ -n a^2 +(m-n)b^2 - n ab +(m-n)bc - n ca = 0 $$ or quintuple $$ -n, m-n, -n, m-n, -n $$

There is a second pattern that gives the same optimum when $n$ is small, quintuple $$ -2n, m-n, m-n, m-n, -2n $$

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  • $\begingroup$ Do we always such $m,n$? $\endgroup$
    – VS.
    Apr 13, 2020 at 0:42
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    $\begingroup$ @VS. of course. It's well-known that every primitive Pythagorean triple is given by those formula (with $\gcd(m,n)=1$, and $m,n$ not both odd). $\endgroup$ Apr 13, 2020 at 1:43
  • $\begingroup$ @WillJagy So $\sqrt{\max(|a|,|b|)}$ is the correct scale up to constants? $\endgroup$
    – VS.
    Apr 13, 2020 at 7:01
  • $\begingroup$ @WillJagy Now I am really curious. What about $ua+vb+zc=0$ with $(u,v,z)\neq(0,0,0)$? Is $\|(u,v,z)\|_\infty$ also at least $\sqrt{\max(|a|,|b|)}$ or should the scale (disregarding constants) be smaller (perhaps $\sqrt[3]{\max(|a|,|b|)}$)? $\endgroup$
    – VS.
    Apr 14, 2020 at 12:02
  • $\begingroup$ @WillJagy mathoverflow.net/questions/357444/… $\endgroup$
    – VS.
    Apr 14, 2020 at 12:17

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