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f(P \cdot Q) = f(P) + f(Q) f(P(Q(x))) f(P \circ Q)

Source: Iran MO Third Round 2021 F3

September 25, 2021
functionalgebra

Problem Statement

Find all functions f:Q[x]→Rf: \mathbb{Q}[x] \to \mathbb{R} such that: (a) for all P,Q∈Q[x]P, Q \in \mathbb{Q}[x], f(P∘Q)=f(Q∘P);f(P \circ Q) = f(Q \circ P); (b) for all P,Q∈Q[x]P, Q \in \mathbb{Q}[x] with PQ≠0PQ \neq 0, f(P⋅Q)=f(P)+f(Q).f(P\cdot Q) = f(P) + f(Q).
(P∘QP \circ Q indicates P(Q(x))P(Q(x)).)
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