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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?

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  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

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  • $\begingroup$ @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? $\endgroup$
    – VS.
    Commented Apr 15, 2020 at 0:49
  • $\begingroup$ 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. $\endgroup$
    – Noam D. Elkies
    Commented Apr 15, 2020 at 0:56
  • $\begingroup$ 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 $8$. $\endgroup$
    – VS.
    Commented Apr 15, 2020 at 0:59
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    $\begingroup$ 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] $\endgroup$
    – Noam D. Elkies
    Commented Apr 16, 2020 at 18:54
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    $\begingroup$ 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. $\endgroup$
    – Noam D. Elkies
    Commented Apr 16, 2020 at 18:55

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