4
$\begingroup$

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

$\endgroup$
7
  • 4
    $\begingroup$ 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. $\endgroup$ Jan 11, 2021 at 17:27
  • $\begingroup$ Thank you! That's exactly the sort of straightforward counterexample I was hoping for. $\endgroup$
    – Jim McCann
    Jan 11, 2021 at 17:35
  • $\begingroup$ @Jim Please write up the answer so as not to leave the question unanswered. $\endgroup$
    – brainjam
    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
    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
    Jan 12, 2021 at 5:38

3 Answers 3

1
$\begingroup$

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 $\mathbb{Z}[i]$ $$(a+ix)^2+(b+iy)^2+(c+iz)^2=0.$$

$\endgroup$
0
$\begingroup$

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)=$(-8,12,-9,17)$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.