# What are the necessary conditions for a polynomial Q(X) such that the roots of Q(X) - X are equal to the real roots of a polynomial P?

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 $$Q(X)$$ ) ?

• Can you explain the question with an example, say, with the polynomial $Q(X)=X$? – Dietrich Burde Mar 18 at 16:33
• 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... – Lexi S. Mar 18 at 17:04

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 $$P(X) = \prod_{s \in S} (X - s)$$.