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If $P(X), Q(X) ∈ ℝ[X]$ , and $P(X) | P( Q(X) ) $ , what could be the necessary conditions for $Q(X)$ such that the set of the real roots of $P(X) $ to be equal to the set of the real roots of $Q(X) - X $( i.e. the set of fixed points of the polynomial function of $Q(X)$ ) ?

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    $\begingroup$ Can you explain the question with an example, say, with the polynomial $Q(X)=X$? $\endgroup$ – Dietrich Burde Mar 18 at 16:33
  • $\begingroup$ The example I used when I asked myself this question was actually when Q(X)=X^2. More specific, if we had P(X) to be a real monic polynomial, with simple roots, such that P(X^2) = ± P(X) *P(-X), I found out that the only such possibilities are P(X)=X, P(X)=X-1, or P(X)=X(X-1). I was wondering if I could somehow find a generalisation to this problem, I don't know if that makes any sense... $\endgroup$ – Lexi S. Mar 18 at 17:04
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One necessary condition is that $Q$ does not induce a permutation without fixed points on any finite subset of $\mathbb R$, i.e. there does not exist a finite set $S \subset \mathbb R$ such that $Q(S) = S$ but $Q(s) \ne s$ for all $s \in S$. Namely, if such $S$ existed we could take $P(X) = \prod_{s \in S} (X - s)$.

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  • $\begingroup$ I was wondering, could this condition actually be equivalent to the polynomial function Q being strictly monotone on the interval Im(Q) ? The left to right implication is certainly true, however I’m not quite sure about the right to left one... $\endgroup$ – Lexi S. Mar 20 at 5:48

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