Finding Pythagorean quadruples on a given plane? 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
 A: 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 $\mathbb{Z}[i]$
$$(a+ix)^2+(b+iy)^2+(c+iz)^2=0.$$
A: 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)=$(-8,12,-9,17)$
A: 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.
