Given any finite set of linearly independent Pythagorean Triples, show that the vector sum of this set is never a Pythagorean triple.
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$\begingroup$ Would this be better posted on the Math.SE instead of Puzzling? $\endgroup$– DanDan0101Sep 5 at 22:51
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$\begingroup$ Ok. Boundaries are fuzzy. $\endgroup$– gyanceySep 5 at 23:11
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$\begingroup$ 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. $\endgroup$– bobbleSep 5 at 23:16
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2$\begingroup$ This is puzzle / problem that I made up. Thought it was cool / fun. $\endgroup$– gyanceySep 5 at 23:33
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2 Answers
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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.
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1$\begingroup$ this is insanely neat - i agree now that it's well-suited for Puzzling! $\endgroup$ Sep 6 at 1:47
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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
triangle inequality
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.
