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In 2D one cannot construct Pythagorean triples $x^2+y^2=m^2$ ($x,y,m\in\mathbb{Z}$) that lie on every line through the origin (e.g., a Pythagorean triple with $x=y$ would require $\sqrt{2}$ to be rational).

What happens when moving to planes in 3D?

Given $a,b,c\in\mathbb{Z}$ can one find $x,y,z,m\in\mathbb{Z}$ such that $m\ne 0$, $x^2+y^2+z^2=m^2$, and $ax + by + cz = 0$?

I would be happy with a counterexample (as in the 2D case) but happier with a construction, since it would lead to a nifty algorithm for approximating a 3D model with one that has only rational-coordinate unit-length normals.

What I have noted so far:

If $a,b,c,||(a,b,c)||$ is itself a Pythagorean quadruple the answer is clearly "yes", and the construction involves using $(a,b,c)$'s perpendiculars[1] to transform Pythagorean triples from the 2D plane.

[1] All Pythagorean quadruples have (at least two) perpendiculars owing to the form of their parameterization -- https://en.wikipedia.org/wiki/Pythagorean_quadruple

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    $\begingroup$ If $a=b=-c=1$ then $z=x+y$, but $S=x^2+y^2+z^2=2x^2+2xy+2y^2$ cannot be a square. WLOG you can assume that $x, y$ are not both even, then you easily see that $S \equiv 2 \pmod{4}$ therefore not a square. $\endgroup$
    – Yaakov Baruch
    Commented Jan 11, 2021 at 17:27
  • $\begingroup$ Thank you! That's exactly the sort of straightforward counterexample I was hoping for. $\endgroup$
    – Jim McCann
    Commented Jan 11, 2021 at 17:35
  • $\begingroup$ @Jim Please write up the answer so as not to leave the question unanswered. $\endgroup$
    – brainjam
    Commented Jan 11, 2021 at 21:26
  • $\begingroup$ @Jim If we parametrize $(a,c,m)$ as a solution of $a^2+c^2=m^2$, we can get a parametric solution for $(x,y,z,m).$ Parametric solution(simple version) is $(a,b,c)=(2uv, b, u^2-v^2),(x,y,z,m)=((b-2uv)(2uv+b)(u-v)(u+v), -2(u^4+2u^2v^2+v^4+2uvb)(u-v)(u+v), 2u^4b+8u^3v^3+4v^2bu^2+2b^2uv+2v^4b, (u^2+v^2)(2u^4+2v^4+4uvb+b^2)).$ $b,u,v$ are arbitrary. For instance, one of numeric solution is $(a,b,c)=(4, 1, 3),(x,y,z,m)=(-45, -174, 118, 215)$. $\endgroup$
    – Tomita
    Commented Jan 12, 2021 at 4:24
  • $\begingroup$ First you need to solve a linear system of equations. These solutions are already substituted into the quadratic equation and examined for solvability. So, it's better to rephrase the question. Just find a solution in a general way. $\endgroup$
    – individ
    Commented Jan 12, 2021 at 5:38

3 Answers 3

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It is not the answer but some relevant information.

In the paper "Cubes in an Integer Lattice" Ivan Horozov gave parameriyation of all mutually perpendicular integer vectors $A_{1}=\left(x_{1}, y_{1}, z_{1}\right), A_{2}=\left(x_{2}, y_{2},\right.$ $\left.z_{2}\right), A_{3}=\left(x_{3}, y_{3}, z_{3}\right)$ of equal length (Mathematics and Informatics Quarterly, 1993, 3, 85-89). In partiqular this result allows to describe solutions of the system $$x^2+y^2+z^2=m,\quad ax + by + cz = 0,\quad a^2+b^2+c^2=m.$$ The idea is to replace this system by Pythagorean equation over $\mathbb{Z}[i]$ $$(a+ix)^2+(b+iy)^2+(c+iz)^2=0.$$

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Parametric solution is given below:

a= $k^2+12k-7$

b=$-3k^2+12k+5$

c=$4k^2-12$

x=$2k^2-6k-4$

y=$2k^2+6k-8$

z=$k^2-6k-1$

m=$3k^2-2k+9$

for k=2

(a,b,c)=$(21,17,4)$

(x,y,z,m)=$(-8,12,-9,17)$

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Conjecture: This system has a solution if and only if $a^2+b^2+c^2=\alpha^2+\beta^2$.

The lattice $\mathbb Z^4$ can be given a basis such that the corresponding Cartan matrix is diagonal with entries 1,-1,-1,-1. Then every automorphism in the Weyl group is a 4x4 matrix with the first row and first column a vector of square 1 with respect to the Cartan matrix and all the other rows and columns a vector of square -1.

Acting by such an automorphism on a vector $(r,p,q,s)\in\mathbb Z^4$ and asking if it is possible to obtain a vector $(\tilde r,\tilde p,\tilde q,\tilde s)$ where one of the differences $\tilde r-\tilde \cdot=0$ is equivalent to finding a solution to the Diophantine system $$ x^2+y^2+z^2=m^2\mbox{ and } rm=px+qy+zs. $$ So, if $(r,p,q,s)=(0,a,b,c)$, this problem is exactly the one above. However, the elements of the Weyl group preserve the quantity $r^2-p^2-q^2-s^2$.

Thus, if $a^2+b^2+c^2=\alpha^2+\beta^2$ and there exists (*) a $R$ in the Weyl group with $R(0,a,b,c)=(\tau,\alpha,\beta,\tau)$, then these equations have a solution. Conversely, if these equations have a solution, then there exists a Weyl element $R$ such that $R(0,a,b,c)=(\tau,\alpha,\beta,\tau)$ meaning that $-a^2-b^2-c^2=\tau^2-\alpha^2-\beta^2-\tau^2$ (with regard to product given by the Cartan matrix).

I cannot prove (*), even though I know it to be true in every example I have tried.

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