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Let $L_n:$ the side length of a regular $n$-polygon inscribed in a unit fixed circle.

We have an interesting relationship:

$L_6^2+L_6^2=L_4^2$

$L_6^2+L_4^2=L_3^2$

$L_{10}^2+L_6^2=L_5^2$

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There are more solutions: $L_m^2+L_n^2=L_p^2$ ?

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I think the complete set of solutions is (3,6,2), (4,4,2), (4,6,3), (6,6,4), (6,10,5) but I don't know how to prove it. –  sperners lemma Nov 11 '12 at 19:13
    
What if you want $\sum\limits_{m\in A}L_m^2 = L_n^2$, where $|A|$ need not be $2$? Or $\sum\limits_{m\in A}L_m^2 = \sum\limits_{m\in B}L_m^2$? –  Michael Hardy Nov 11 '12 at 19:23
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I think we can consider it a problem about for each $n$ solving for $(r_1,r_2,r_3)$ (each $r|n$) the equation $\sin(\frac{\pi}{n} r_1)^2 + \sin(\frac{\pi}{n} r_2)^2 = \sin(\frac{\pi}{n} r_3)^2$. The idea is that the terms are all algebraic numbers in the same field ($\mathbb Q(\zeta_n)$). The key would be proving $n=2\cdot 3\cdot 5$ is the biggest $n$ this can ever happen. –  sperners lemma Nov 11 '12 at 19:54
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up vote -1 down vote accepted

This will not answer the question as phrase, but:

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