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diophantine equations – Finding Pythagorean quadruples on a given airplane?

In 2D one can’t assemble Pythagorean triples $x^2+y^2=m^2$ ($x,y,minmathbb{Z}$) that equivocate on each line via the inception (e.g., a Pythagorean triple with $x=y$ would require $sqrt{2}$ to breathe rational).

What occurs when stirring to planes in 3D?

**Given $a,b,cinmathbb{Z}$ can one discover $x,y,z,minmathbb{Z}$ such that $mne 0$, $x^2+y^2+z^2=m^2$, and $ax + by + cz = 0$?**

I’d breathe fortunate with a counterexample (as within the 2D illustration) however happier with a building, since it could result in a nifty algorithm for approximating a 3D mannequin with one which has solely rational-coordinate unit-length normals.

## What I’ve celebrated up to now:

If $a,b,c,||(a,b,c)||$ is itself a Pythagorean quadruple the respond is clearly “yes”, and the development entails utilizing $(a,b,c)$‘s perpendiculars[1] to rework Pythagorean triples from the 2D airplane.

[1] All Pythagorean quadruples have (at the least two) perpendiculars owing to the figure of their parameterization — https://en.wikipedia.org/wiki/Pythagorean_quadruplewe are going to proffer you the answer to diophantine equations – Finding Pythagorean quadruples on a given airplane? query by way of our community which brings all of the solutions from a number of dependable sources.

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