let P(x) equals R(x) plus iS(x) then S(Q(x)) equals 0 for infinite number of x hence S(x) is equal to 0 .hence P(x) is real for all x . Now assume Q(x) equals M(x) plus ixN(x) (because the leading coeff and constant of Q(x) do not contribute any imaginary part ,hence the imaginary part of Q(x) has x as a factor). now using the fact that P(Q(x)) has real coeff and assuming some coeff for M(x) & N(x),equate every coeff of PoQ to real to prove by induction that all coeff of N(x) are 0. hence Q(x) is real. P.S:sry cannot use latex

Ankoganit wrote: Call a non-constant polynomial real if all its coecients are real. Let $P$ and $Q$ be polynomials with complex coefficients such that the composition $P \circ Q$ is real. Show that if the leading coefficient of $Q$ and its constant term are both real, then $P$ and $Q$ are real. Oh c'mon, this is again from the Monthly. Last time it was KöMaL.

YESMAths wrote: Oh c'mon, this is again from the Monthly. Last time it was KöMaL. I don't get your point?!