# Small linear relations between primitive Pythagorean triples $\mathsf I$

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 $$\sqrt{\max(a^2,b^2)}$$?

• Sorry, how do you want to get any small solution if $a|y,b|z$ and $c|x$ provided $(a,b)=1$? – Pavel Kozlov Apr 12 '20 at 9:45
• $ybc=-a(xb+zc)$, so $a|ybc$. If $a,b,c$ are pairwise coprime then $a|y$ is nesessary. – Pavel Kozlov Apr 12 '20 at 10:58
• @PavelKozlov I updated and for this I am pretty sure we can do better that $\sqrt{\max(a^2,b^2)}$. – VS. Apr 12 '20 at 11:32

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 $$-2n, m-n, m-n, m-n, -2n$$

• Do we always such $m,n$? – VS. Apr 13 '20 at 0:42
• @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). – Gerry Myerson Apr 13 '20 at 1:43
• @WillJagy So $\sqrt{\max(|a|,|b|)}$ is the correct scale up to constants? – VS. Apr 13 '20 at 7:01
• @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 $\sqrt[3]{\max(|a|,|b|)}$)? – VS. Apr 14 '20 at 12:02
• @WillJagy mathoverflow.net/questions/357444/… – VS. Apr 14 '20 at 12:17