Advanced book on Mathematics Olympiad

2.2 Polynomials 53

P 1 (xj)P 2 (xj)···Pk(xj)= 1 , forj= 1 , 2 ,...,n.

HencePi(xj) =±1, which then implies Pi(x^1 j) = Pi(xj), i = 1 , 2 ,...,k,j =
1 , 2 ,...,n.
Now let us see how derivatives come into play. The key observation is that the
zerosxjof(P (x))^2 appear with multiplicity greater than 1, and so they are zeros of the
derivative. Differentiating with the product rule, we obtain

∑k

i= 1

P 1 (xj)···Pi′(xj)···Pk(xj)= 0 , forj= 1 , 2 ,...,n.

This sum can be simplified by taking into account thatP 1 (xj)P 2 (xj)···Pk(xj)=1 and
1
Pi(xj)=Pi(xj)as

∑k

i= 1

Pi′(xj)Pi(xj)= 0 , forj= 1 , 2 ,...,n.

It follows thatxjis a zero of the polynomial

∑k

i= 1

2 Pi′(x)Pi(x)=

( k

∑

i= 1

Pi^2 (x)

)′

.

Let us remember thatPi(xj) =±1, which then implies

∑k
i= 1 P
2
i(xj)−n =0 for
j= 1 , 2 ,...,n. The numbersxj,j= 1 , 2 ,...,n, are zeros of both

∑k

i= 1 P

2
i(x)−n
and its derivative, so they are zeros of order at least 2 of this polynomial. Therefore,

∑k

i= 1

Pi^2 (x)=(x−x 1 )^2 (x−x 2 )^2 ···(x−xn)^2 Q(x)+n,

for some polynomialQ(x)with integer coefficients. We deduce that there exists an index
i 0 such that the degree ofPi 0 (x)is greater than or equal ton. Forneven,n= 2 n+ 21 , and
we are done. Fornodd, since(P (x))^2 +1 does not have real zeros, neither doesPi 0 (x),
so this polynomial has even degree. Thus the degree ofPi 0 (x)is at leastn+ 1 = 2 n+ 21 .
This completes the solution. 

166.Find all polynomials P(x)with integer coefficients satisfying P(P′(x)) =
P′(P (x))for allx∈R.

167.Determine all polynomialsP(x)with real coefficients satisfying(P (x))n=P(xn)
for allx∈R, wheren>1 is a fixed integer.