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It is known that for every palindromic polynomial $f(x)$ of even degree $2d$ there is a polynomial $g$ of degree $d$ such that

$f(x) = x^d g\left(x + \frac{1}{x}\right)$.

For $n>2$ cyclotomic polynomials are of even degree and are palindromic.

So, what can one say on the polynomial $g$ for a cyclotomic polynomial?

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    $\begingroup$ @FrançoisBrunault I think that would be correct if $f(x)$ were $x^n-1$, but in this case $f(x)$ is the cyclotomic polynomial $\prod_{d\mid n}(x^d-1)^{\mu(n/d)}$. So as you say, the answer is closely related to the Chebyshev polynomials (maybe a product of them), but they're not equal to the Chebyshev polynomials. $\endgroup$
    – Joe Silverman
    Commented Sep 15, 2020 at 16:54
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    $\begingroup$ @JoeSilverman Right, let me try again: starting from the polynomial $x^{2n}+1$, you get essentially the Chebyshev polynomials. Now you can write $x^{2n}+1$ as a product of cyclotomic polynomials, and try to apply Möbius inversion. $\endgroup$
    – François Brunault
    Commented Sep 15, 2020 at 16:55
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    $\begingroup$ $g$ is just the minimal polynomial for $2\cos(2\pi/n)$, no? $\endgroup$
    – Gerry Myerson
    Commented Sep 16, 2020 at 0:01
  • $\begingroup$ Thanks for setting me on the right track! I found chapter 3 of Antonio Cafure, & Eda Cesaratto. (2017). Irreducibility Criteria for Reciprocal Polynomials and Applications. The American Mathematical Monthly, 124(1), 37-53 quite helpful. The functions $f_k$ there are more or less Chebyshev polynomials. $\endgroup$
    – borntomath
    Commented Oct 20, 2020 at 7:17

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