 # diophantine equations – Finding Pythagorean quadruples on a given airplane? Answer

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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 to rework Pythagorean triples from the 2D airplane.

 All Pythagorean quadruples have (at the least two) perpendiculars owing to the figure of their parameterization — https://en.wikipedia.org/wiki/Pythagorean_quadruple

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