# Vector Sum of Pythagorean Triples

Given any finite set of linearly independent Pythagorean Triples, show that the vector sum of this set is never a Pythagorean triple.

• Would this be better posted on the Math.SE instead of Puzzling? Sep 5, 2023 at 22:51
• Ok. Boundaries are fuzzy. Sep 5, 2023 at 23:11
• Math puzzles are welcome here as long as they are puzzles, not problems. See links for details. I don't know enough to determine which side of the line this falls on, but if it's on-topic here then it's on-topic here, regardless if there's another site which would also take it. Sep 5, 2023 at 23:16
• This is puzzle / problem that I made up. Thought it was cool / fun. Sep 5, 2023 at 23:33

The graph of real-valued Pythagorean triples $$x^2+y^2=z^2$$ forms an infinite cone if we restrict to $$z>0$$:
A sum of $$n$$ independent vectors on this cone is $$n$$ times their average, which lies within their convex hull and so is inside the cone, and so cannot be a Pythagorean triple.
By definition of (Euclidean) length, 2-dimensional vector $$v_j:=(a_j,b_j)$$ has length $$|v_j|=c_j$$ if and only if $$a_j^2+b_j^2=c_j^2$$ ($$c_j\ge 0$$). By repeated application of the
we have for any finite sum $$\left|\sum_j v_j\right| \le \sum_j c_j$$ with equality if and only if each $$v_j$$ is a nonnegative multiple of every other nonzero $$v_j$$. As the given vectors are assumed to be linearly independent the inequality must be strict. In particular, the sums do not form a Pythagorean triple.