Problem 11822
(American Mathematical Monthly, Vol.122, February 2015) Proposed by G. Stoica (Canada).
Call a polynomial real if all its coefficients are real. Let P and Q be polynomials with complex coefficients such that the composition P ◦ Q is real. Show that if the leading coefficient of Q and its constant term are both real, then P and Q are real.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Our proof is inspired by A. Horwitz’s paper Compositions of polynomials with coefficients in a given field, J. Math. Anal. Appl. 267 (2002), no. 2, 489-500. The same proof still holds if we replace R and C with two fields of characteristic 0, F1 and F2, such that F1⊂ F2.
We assume that P and Q are not constant, otherwise the statement is false:
P (x) = 0, Q(x) = x2+ ix, P (Q(x)) = 0 and P (x) = ix, Q(x) = 0, P (Q(x)) = 0.
Let P, Q ∈ C[x] such that P ◦ Q ∈ R[x] with P (x) =
n
X
k=0
akxk, Q(x) =
m
X
j=0
bjxj and P (Q(x)) =
mn
X
i=0
cixi. where n, m ≥ 1, bm, b0∈ R, and an6= 0, bm6= 0. Note that an= cmn/bm∈ R.
i) We show that Q ∈ R[x].
Assume by contradiction that Q is not real, and let d = min{j : bm−j6∈ R} ∈ [1, m − 1].
Since mn − d > m(n − 1), it follows that cmn−d= an[xmn−d]Q(x)n = an
X
k0+···km=n 0k0+···mkm=mn−d
n
k0, . . . km
(b0)k0· · · (bm)km.
Moreover
m−1
X
i=0
(m − i)ki= m
m
X
i=0
ki−
m
X
i=0
iki= mn − (mn − d) = d,
which implies that ki = 0 for i ∈ [0, m − d − 1)] (otherwise the above sum is greater than d).
Hence
dkm−d+
m−1
X
i=m−d+1
(m − i)ki= d
If km−d> 0 then km−d= 1 and ki= 0 for i ∈ [m − d + 1, m − 1]. Therefore cmn−d= annbm−dbn−1m + an
X n
km−d+1, . . . km
(bm−d+1)km−d+1· · · (bm)km.
Since an6= 0, bm6= 0, and cmn−d, an, bm−d+1, . . . , bm∈ R, it follows that also bm−d ∈ R which contradicts the definition of d.
ii) We show that P ∈ R[x].
Assume by contradiction that P is not real, and let d = min{j : an−j6∈ R} ∈ [1, n]. Then cmn−md= [xm(n−d)]
n
X
k=n−d+1
akQ(x)k+ an−dbn−dm .
Since an 6= 0, bm6= 0, Q ∈ R[x] and cmn−md, an−d+1, . . . , an∈ R, it follows that also an−d∈ R which contradicts the definition of d.
