# Another Roots of Unity Sum

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 $$\ell=3,4,6$$.

• 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). – M. Vinay Mar 23 '19 at 3:55

## 1 Answer

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