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Does there exist a monic self-reciprocal integer polynomial $p\in\mathbb{Z}[x]$ of degree 10 with roots $r\textrm{e}^{\pm\textrm{i}\theta_1}, r^{-1}\textrm{e}^{\pm\textrm{i}\theta_1}, r\textrm{e}^{\pm\textrm{i}\theta_2}, r^{-1}\textrm{e}^{\pm\textrm{i}\theta_2}, r,$ and $r^{-1}$ with the following properties:

  1. modulus $r>1$, and
  2. no quotient of two distinct roots of $p$ is a root of unity
Zephos
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  • By self-reciprocal, do you intend that $\forall z \in \Bbb{C},~ f[f(z)] = z ~$? If so, then you have $(10)$ different values $r_1, r_2, \cdots r_{10}$ such that $f(r_i) = 0 \implies f[f(r_i)] = f(0) = r_i$. However, $f(0)$ can not equal more than 1 value. Therefore, since the function has more than 1 root, isn't the constraint contradictory? Have I misinterpreted the phrase self-reciprocal? – user2661923 Jul 05 '21 at 07:56
  • Here by a self-reciprocal polynomial I mean that $\forall z\in\mathbb{C},\ z^d f(z^{-1}) = f(z)$ where $d$ is the degree of the $f$. – Zephos Jul 05 '21 at 07:57
  • Very interesting. Just to be clear, then if $s$ is a root of $f(z) - 1 = 0$, then $$f(s) = 1 \implies s^{(10)} f\left[s^{(-1)}\right] = 1 \implies f\left[s^{(-1)}\right] = \frac{1}{s^{(10)}}.$$ Have I understood this correctly? – user2661923 Jul 05 '21 at 08:07
  • For a self-reciprocal polynomial $f$, $\alpha$ is a root of $f$ if and only if $\alpha^{-1}$ is a root of $f$. The Wikipedia page for such polynomials gives the common characterisations: https://en.wikipedia.org/wiki/Reciprocal_polynomial – Zephos Jul 05 '21 at 08:30

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