If $x$ is the root of $P$, then $x^2$ is also the root. Now spamming this we get that $x^{2^k}$ is eventually $1$ for fixed root $x$ (and some positive integer k, of course assuming that $x \ne 0$). But thats where i got stuck.
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There are multiple polynomials of every degree that satisfy this relation and they are difficult to list explicitly. Are the polynomials allowed to have complex coefficients? – DinosaurEgg Feb 23 '24 at 21:32
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I don't think $x^{2^k}$ must necessarily be 1; for example, $P(x) = x^2 + x + 1$ satisfies the condition and the sequence $x^{2^k}$ is periodic instead of eventually constant. (Also, there's the case of a root of 0 - $P(x) = x$ works, for example.) So, the fact $P$ has finitely many roots means $(x^{2^k})_{k=0}^\infty$ must eventually have repeated entries, so any root is either 0 or a root of unity. – Daniel Schepler Feb 23 '24 at 22:02
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On the other hand, I think any cyclotomic polynomial $\Phi_n(x)$ should work if $n$ is odd; and similarly any product of such cyclotomic polynomials. I don't know if that would cover all possible solutions. – Daniel Schepler Feb 23 '24 at 22:04
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2Oh, wait, if complex coefficients are allowed (rather than just rational coefficients) then I think $P(x) = (x - e^{2\pi i/7}) (x - e^{4\pi i/7}) (x - e^{8\pi i/7})$ would work, and it's not a product of cyclotomic polynomials. – Daniel Schepler Feb 23 '24 at 22:07