# Small linear relations between primitive Pythagorean triples $\mathsf{II}$

WillJagy answered a linear relation question on Pythagorean Triples in Small linear relations between primitive Pythagorean triples $\mathsf I$.

Now let $$a^2+b^2=c^2$$ be a primitive Pythagorean triple and then consider the Linear Diophantine Equation $$ua+vb+zc=0$$ where $$(u,v,z)\in\mathbb Z^3$$ are variables. If $$(u,v,z)\neq(0,0,0)$$ then:

1. Is $$\|(u,v,z)\|_\infty$$ at least $$\sqrt{\max(|a|,|b|)}$$ up to constant factors or should the scale (disregarding constants) be smaller (perhaps $$\sqrt[3]{\max(|a|,|b|)}$$)?

2. What is the distribution of $$\|(u,v,z)\|_\infty$$?

Note if it were $$ua^2+vb^2+zc^2=0$$ then the answer is $$O(1)$$ since $$(u,v,z)=(1,1,-1)$$ suffices.

This is what I have $$a = m^2 - n^2$$ $$b = 2mn$$ $$c = m^2 + n^2$$ then $$n(m^2 - n^2 ) +(-m)(2mn) + n(m^2 + n^2) = 0$$ or triple $$(u,v,z)=(n,-m,n)$$ works and this gives morally $$\sqrt{\max(|a|,|b|)}$$ ($$(m,n,-m)$$ also works to give morally $$\sqrt{\max(|a|,|b|)}$$). Could there be something smaller?

1. Yes, the minimal $$\|(u,v,z)\|_\infty$$ is within a constant factor of $$\sqrt{|c|}$$ (equivalently, of $$\sqrt{\max(|a|,|b|)}$$.

The orthogonal complement of $$(a,b,c) = (m^2-n^2, 2mn, m^2+n^2)$$ contains the independent integer vectors $$v_1 := (n,-m,n)$$ (which you found) and $$v_2 := (m,n,-m)$$. Their $$\bf Z$$-span is the full integral complement of $$(a,b,c)$$, for example because that span has discriminant $$\|v_1\|_2^2 \|v_2\|_2^2 - \langle v_1, v_2 \rangle^2 = 2(m^2+n^2)^2$$ which equals $$a^2+b^2+c^2 = \|(a,b,c)\|_2^2$$. Moreover, $$(v_1,v_2)$$ is a reduced basis, because $$|\langle v_1, v_2 \rangle| = mn$$ is less than $$(m^2+n^2)/2$$, and thus less than both $$\frac12 \|v_1\|_2^2$$ and $$\frac12 \|v_2\|_2^2$$. Therefore the minimal $$\|(u,v,z)\|_2$$ exceeds $$\sqrt{m^2+n^2} = \sqrt{|c|}$$, whence the same is true of $$\|(u,v,z)\|_\infty$$ within a constant factor. QED

• @NoamElkies Fastest accept. However I am wondering if we take unbalanced diophantine equations $au+bv+c^2z=0$ or $a^2u+b^2v+cz=0$ or any $a^tu+b^{t'}v+c^{t''}z=0$ where $t,t',t''\in\{1,2\}$ and $1<tt't''<8$ also will we get similar $\sqrt{\max(|a|,|b|)}$ bounds? At lower limit $tt't''=1$ we get this bound and at upper limit $tt't''=8$ we get $O(1)$. Perhaps there is some interpolating rule? Or perhaps $\sqrt{\max(|a|,|b|)}$ until at $8$? Should I post a new post?
– VS.
Apr 15, 2020 at 0:49
• That's a different question (and I'm can't guess the motivation for those particular "unbalanced diophantines"). Once we have a reduced basis for the lattice, you can use it to start analyzing such equations where you require one or more of the coordinates to be square. Apr 15, 2020 at 0:56
• I see the point of analyzing that way however posted mathoverflow.net/questions/357507/…. Motivation is to see if it might give a smaller relations that cannot be ordinarily seen with just probabilistic analysis. I think only at $8$ it breaks down and at every value from $\{1,2,4\}$ it remains $\Omega(\sqrt{\max(|a|,|b|)})$ but I do not see a reason why it should fall only at $8$.
– VS.
Apr 15, 2020 at 0:59
• 1) Discriminant of ${\bf Z}$-span of any vectors $v_1,\ldots,v_m$ = determinant of Gram matrix $(\langle v_i, v_j\rangle)$. This is the formula for any quadratic form. 2) It is known that if $v \in {\bf Z}^n$ is primitive (not $cv'$ for some $c>1$ and integer vector $v'$) then its orthogonal complement has discriminant $\|v\|_2^2$. So if you've found $n-1$ vectors with the same discriminant you've got everything. (A superlattice of index $d>1$ would have discriminant $d^2$ times smaller.) 3) Not sure what exactly the first question means. For the second, [cont'd] Apr 16, 2020 at 18:54
• these days I'm even less able to dig up references than usual, but all this can surely be found in "SPLAG" = Conway and Sloane's Sphere Packings, Lattices, and Groups, not far from the beginning. Apr 16, 2020 at 18:55