# How to prove the following polynomial does not have root of a special form?

I'm working on a special kind of graphs. To prove some uniqueness, I need to prove that the polynomial \begin{equation} x^{8}-7x^{6}+14x^{4}-8x^{2}+1 \end{equation} does not have any root of the form \begin{equation} 2\cos\frac{(2k+1)\pi}{2n} \quad k\in \lbrace 0,1,\cdots , n-1 \rbrace , n \in \mathbb{E}. \end{equation} Can anybody help me?

Bests.

• Note that all roots of the above polynomial are real. – A. Mpi Dec 6 '17 at 12:53
• This polynomial is solvable, because it is equal to $g(x^2)$ where $g(x) = x^4 - 7x^3 + 14x^2 - 8x + 1$; and quartic polynomials are always solvable. Thus you can find the roots explicitly and therefore determine whether they are or are not of the form that you ask. – Stanley Yao Xiao Dec 6 '17 at 13:00
• In fact the polynomial $g(x) = x^4 - 7x^3 + 14x^2 - 8x + 1$ has Galois group isomorphic to $C_4$, and its roots can be found explicitly using the the identity $g(x) = x^2(x-2)^2 - 3x(x-1)(x-2) + (x-1)^2$. Since this polynomial is so special, I am very curious as to how it arose. What graph does this originate from, say? – Stanley Yao Xiao Dec 6 '17 at 13:40

The claim is simply not true. A simple check (e.g. following the suggestion of Stanley Yao Xiao, or by substituting $x = cos \alpha$ and using trigonometric identities) shows that this polynomial has 8 real roots and they are precisely of this form
$$2 cos\frac{\pi}{30}, 2 cos\frac{7\pi}{30}, 2 cos\frac{11\pi}{30},, 2 cos\frac{13 \pi}{30}, 2 cos\frac{17 \pi}{30}, 2 cos\frac{19\pi}{30}, 2 cos\frac{23 \pi}{30}, 2 cos\frac{29\pi}{30}$$
• Sorry, i forgot to say $n$ is even. – A. Mpi Dec 6 '17 at 13:59
• If $n$ is even, i checked by computer and it doesn't have root of the given form. – A. Mpi Dec 6 '17 at 14:04
• Indeed, if you take $x = 2 \cos(\alpha)$, $z = e^{i\alpha}$ is a root of $z^{16}+z^{14} - z^{10} - z^8 - z^6 + z^2 + 1$, which is $C_{30}(z^2)$ where $C_{30}$ is the 30'th cyclotomic polynomial. – Robert Israel Dec 6 '17 at 23:47