How to prove that a given complex number of modulus 1 is not a root of unity?

How to prove that the complex number $$\frac{1+\sqrt{15}i}{4}$$ of absolute value $$1$$ is not a root of unity of any order? Just in case, this number is a root of the polynomial $$2x^2 - x +2$$.

• Have you worked out the rational polynomial that this number satisfies? Your post is pretty terse. Commented Nov 18, 2021 at 13:23
• @MartinR Sorry, the polynomial is $2x^2-x+2$ Commented Nov 18, 2021 at 13:35
• This may help math.stackexchange.com/a/68176/42969 Commented Nov 18, 2021 at 13:44
• For that to happen $\,p(x)=x^2-\frac{1}{2}x+1\,$ would have to be a factor of $x^n-1$ for some $n$. But Monic Factors in $\mathbb Q[x]$ of a Monic $f \in \mathbb Z[x]$ are also in $\mathbb Z[x]$ while $p(x)$ is obviously not.
– dxiv
Commented Nov 18, 2021 at 18:11
• It's an interesting kind of problem, but I imagine there is some context for it that could be shared to improve your Question. E.g. where did this problem arise? Commented Nov 18, 2021 at 18:19

For simplicity, consider $$\alpha=1+\sqrt{-15}$$. We will argue that the imaginary part of $$\alpha^n$$ is never $$0$$ (for $$n≥1$$). Clearly, that will suffice. We remark that the minimal polynomial of $$\alpha$$ is $$x^2-2x+16$$
If we define $$\alpha^n=a_n+b_n\sqrt {-15}$$
with $$a_n, b_n\in \mathbb Z$$, we must have $$b_n=2b_{n-1}-16b_{n-2}$$
with $$b_1=1, b_2=2$$.
A routine induction tells us that the order of $$2$$ in $$b_n$$ increases by exactly $$1$$ as $$n$$ increases by $$1$$, hence $$b_n$$ never vanishes, and we are done.