Working out circulant graphs that have eigenvalue 0 of multiplicity 1 (the so-called "nut" graphs), I came to the polynomial
$$
x^6 - x^5 - 4x^4 + 3x^2 + 3x - 1,
$$
for which I would like to show that it does not have roots in the form $2\cos(a\pi/b)$, where $a/b$ is an arbitrary rational number. Any suggestion how to do this?



Polynomial roots in the form 2cos of a rational multiple of pi
If $2\cos \frac{a\pi}b$ is a root of your polynomial $f(x)$, then $e^{i\pi a/b}$ is a root of $g(x):=x^6f(x+1/x)$. But the minimal polynomial of $e^{2i\pi k/n}$, where $k$ and $n$ are coprime integers (and $n>0$), is $\Phi\_n(x)$, the cyclotomic polynomial. So we should check whether $g(x)$ has cyclotomic divisors. Since $\deg \Phi\_n=\varphi(n)$ and there are only finitely many $n$ with $\varphi(n)\leqslant 12=\deg g(x)$, you should check finitely many divisibilities of integer polynomials. This may be done by hands or by computer quite routinely.


