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I almost see a brute-force attack on this problem, but before messing with the details I wonder there is some theory here, or at least a nice way to group the terms so I can see the cancellation.

Let $\omega$ be a primitive cube root of unity and $\zeta$ be a primitive $m^\text{th}$ root of unity where $m\not\equiv 2 \bmod 3$.

I claim that: $$ \sum_{a,b=0}^m \omega^{a-b}\zeta^{a+b}=1$$

This might fail for small $m$ but I have good reason to believe that it works for, say, $m>10$.

Remarks: In general I expect that replacing $\zeta$ by $\zeta^k$ and similarly for $\omega$, there is at most one $k$ value for which the above sum fails, roughly at $k\approx m/3$ or $2m/3$ (for those values I expect the sum is $m$ instead). It's possible that there's nothing special about cube roots, in which case the general condition for $\ell^\text{th}$ roots should be that $\gcd(m+1,\ell)=1$. On the other hand, there is a wallpaper group in the background for my application, so it's not unreasonable that it only works for $\ell=3,4,6$.

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  • $\begingroup$ Apart from the exception that you already noted ($m \not\equiv 2 \mod 3$), it dosen't fail for "small" $m$, except $m = 3$ (it's obvious from the final equation in the solution how that fails; and obvious from the original expression why it fails). $\endgroup$
    – M. Vinay
    Mar 23, 2019 at 3:55

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$$\sum_a(\zeta\omega)^a\sum_b(\zeta/\omega)^b=\frac{1-(\zeta\omega)^{m+1}}{1-\zeta\omega}\frac{1-(\zeta/\omega)^{m+1}}{1-\zeta/\omega}.$$Feel free to tidy that up using $1-\exp i\theta=-2i\sin\frac{\theta}{2}\exp\frac{i\theta}{2}$.

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