I have some doubts about this following problem, if you can please try to answer the congruence step:
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!!
$n(n+1)$ is always even and so the RHS can only be $\pm1$.
Moreover, $n(n+1)/2$ is odd iff $n \equiv 1 \text{ or } 2 \bmod 4$.