Primitive 18-th root of unity problem involving congruences.

Let $\omega$ be a primitive 18-th root of unity. Find $n \in \mathbb Z$ such that:

$\omega^n = i^{n(n+1)}$

I have been told to look at the RHS possible values $(i,-i , 1, -1)$ and from here try to arrive at the conclusion that the LHS has to necessarily be either 1 or -1.

What I did so far is trying to give it the following approach:

Since $\omega$ is a primitive 18-th root of unity, then $\omega$ is of the form: $\omega^n= e^{\theta i}$ where $\theta = \frac {2k\pi n}{18}$ and $GCF(k,18)=1$ , $k \in \mathbb Z$.

The RHS can be written as: $i^{n(n+1)}= e^{\phi i}$, where $\phi=\frac {\pi }{2} n(n+1)$

Now since neither side is equal to zero I can say that: $e^{\theta i}= e^{\phi i} \iff e^{(\theta - \phi) i} =1 \iff Re[e^{(\theta - \phi) i}]=1 \land Im[e^{(\theta - \phi) i}]=0 \iff (\theta - \phi)=2t\pi$ for some $t \in \mathbb Z$

and from $(\theta - \phi)=2t\pi$ , I have that $2k\pi n-9\pi n(n+1)=36t\pi \iff 2kn\equiv 9n(n+1) \pmod{36}$ and here I'm a bit stuck... I've tried to do some Chinese Remainder Theorem argument with the prime factors of 36 to see how n has to be, but still don’t know how to arrive to the answer

Thanks for the help!!

• Also, integral powers of the 18-th root of unity are never equal to $i,-i$ because 4 does not divide 18. – SquirtleSquad Jul 4 '16 at 1:29
• **Case**$\#1:$ If $(n,36)=1$ $2k\equiv9n+9\pmod{36}\implies9|k,2|(n+1)$ If $k=9a,n=2b-1;18a\equiv18b\pmod{36}\iff a\equiv b\pmod2$ **Case**$\#2:$ If $(n,36)=2,n=2m$(say) where $(m,36)=1$ $4km\equiv18m(2m+1)\pmod{36}\iff2(k-9m-4)\equiv1\pmod{18}$ which is impossible. **Case**$\#3:$ If $(n,36)=3,n=3m$(say) where $(m,36)=1$ and so on – lab bhattacharjee Jul 4 '16 at 16:02

$n(n+1)$ is always even and so the RHS can only be $\pm1$.
Moreover, $n(n+1)/2$ is odd iff $n \equiv 1 \text{ or } 2 \bmod 4$.