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Suppose $P(x) \in \mathbb{Z}[x]$ is irreducible, and such that at least one of its roots has modulus $1.$ Is there anything we can say about the reduction of $P(x)$ modulo primes? Do these have some magical property?

Some words of explanation This question comes from the observation (not new) that a random palindromic polynomial has A LOT of roots on the unit circle (the fraction is $1/\sqrt{3},$ or around 58%)! Now, palindromic polynomials are characteristic polynomials in $Sp(2n, \mathbb{Z}),$ so this says that symplectic matrices have a lot of eigenvalues of modulus $1$ (who knew..) in some sense. Now, there must be some morally analogous statement for $Sp(2n, \mathbb{Z}/p \mathbb{Z}),$ but the question is: what is morally analogous?

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    $\begingroup$ Even in the quadratic case, $P$ can be anything of the form $Ax^2+Bx+A$, with $-2A<B<2A$, which doesn't translate into any sort of congruence condition mod primes (other than the trivial condition that two coefficients that are equal remain equal after reduction). So I'm unclear on what you're looking for. $\endgroup$
    – Steven Landsburg
    Commented Oct 6, 2014 at 3:43
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    $\begingroup$ Well there's always two roots that multiply to $1$, but beyond that, probably not. For example, in degree $2$ any polynomial $a x^2 + b x + a \bmod p$ lifts in many ways to a polynomial $\tilde a x^2 + \tilde b x + \tilde a \bmod p$ lifts in many ways to a polynomial satisfying your property (just make sure $|\tilde b| < 2 |\tilde a|$). $\endgroup$
    – Noam D. Elkies
    Commented Oct 6, 2014 at 3:44
  • $\begingroup$ (I see that Steven Landsburg posted the second half of my comment as I was editing it...) $\endgroup$
    – Noam D. Elkies
    Commented Oct 6, 2014 at 3:45
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    $\begingroup$ Of course irreducibility may not transfer. The following is true and some remains so following reduction. I don't know that it counts as magic: If there is a root on the unit circle then the degree $n=2m$ is even and $P(x)=x^nP(1/x)$ , equivalently $P(x)=x^mR(x+1/x)$ for a polynomial of degree $m.$ Equivalently, the coefficients are symmetric. $c_{n-i}=c_i$. The converse fails, $x^2+3x+1.$ If all the roots are on the unit circle then the polynomial is cyclotomic. Lehmers polynomial of degree $10$ has 2 real roots and $8$ on the unit circle. $\endgroup$
    – Aaron Meyerowitz
    Commented Oct 6, 2014 at 5:09

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