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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.

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  • $\begingroup$ Note that all roots of the above polynomial are real. $\endgroup$
    – A. Mpi
    Commented Dec 6, 2017 at 12:53
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    $\begingroup$ 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. $\endgroup$
    – Stanley Yao Xiao
    Commented Dec 6, 2017 at 13:00
  • $\begingroup$ 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? $\endgroup$
    – Stanley Yao Xiao
    Commented Dec 6, 2017 at 13:40

2 Answers 2

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The degree of the minimal polynomial of your cos is bounded below by phi(n)/2 since it generates a real field inside the cyclotomic field. If phi(n)/2 is bigger than 8 you are done. For small n you can check one by one with computer (or write the minimal polynomial and see it doesn’t divide that polynomial or compute discriminant, which is an easy thing for cosine as the real part of a root of unity)

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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

$$ 2 cos\frac{\pi}{30}, 2 cos\frac{7\pi}{30}, 2 cos\frac{11\pi}{30},, 2 cos\frac{13 \pi}{30}, 2 cos\frac{17 \pi}{30}, 2 cos\frac{19\pi}{30}, 2 cos\frac{23 \pi}{30}, 2 cos\frac{29\pi}{30} $$

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  • $\begingroup$ Sorry, i forgot to say $n$ is even. $\endgroup$
    – A. Mpi
    Commented Dec 6, 2017 at 13:59
  • $\begingroup$ If $n$ is even, i checked by computer and it doesn't have root of the given form. $\endgroup$
    – A. Mpi
    Commented Dec 6, 2017 at 14:04
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    $\begingroup$ 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. $\endgroup$
    – Robert Israel
    Commented Dec 6, 2017 at 23:47

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