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 $2x^2 - x +2$.
Asked
Active
Viewed 279 times
0
-
1Have you worked out the rational polynomial that this number satisfies? Your post is pretty terse. – hardmath Nov 18 '21 at 13:23
-
1@MartinR Sorry, the polynomial is $2x^2-x+2$ – IntegrableSystemsEnthusiast Nov 18 '21 at 13:35
-
2This may help https://math.stackexchange.com/a/68176/42969 – Martin R Nov 18 '21 at 13:44
-
2For 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 Nov 18 '21 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? – hardmath Nov 18 '21 at 18:19
1 Answers
3
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.
lulu
- 76,951