Advanced book on Mathematics Olympiad

406 Algebra

a contradiction. We conclude thatP(x)is irreducible.

188.Assume the contrary, and let

(x^2 + 1 )n+p=Q(x)R(x),

withQ(x)andR(x)of degree at least 1. Denote byQ(x)ˆ ,R(x)ˆ the reduction of these

polynomials modulop, viewed as polynomials inZp[x]. ThenQ(x)ˆ R(x)ˆ =(x^2 + 1 )n.

The polynomialx^2 +1 is irreducible inZp[x], since−1 is not a quadratic residue in

Zp. This impliesQ(x)ˆ =(x^2 + 1 )kandR(x)ˆ =(x^2 + 1 )n−k, with 1≤k ≤n− 1

(the polynomials are monic and their degree is at least 1). It follows that there exist

polynomialsQ 1 (x)andR 1 (x)with integer coefficients such that

Q(x)=(x^2 + 1 )k+pQ 1 (x) and R(x)=(x^2 + 1 )n−k+pR 1 (x).

Multiplying the two, we obtain

(x^2 + 1 )n+p=(x^2 + 1 )n+p((x^2 + 1 )n−kQ 1 (x)+(x^2 + 1 )kR 1 (x))+p^2 Q 1 (x)R 1 (x).

Therefore,

(x^2 + 1 )n−kQ 1 (x)+(x^2 + 1 )kR 1 (x)+pQ 1 (x)R 1 (x)= 1.

Reducing modulopwe see thatx^2 +1 divides 1 inZp[x], which is absurd. The contra-

diction proves that the polynomial from the statement is irreducible.

189.We will show that all the zeros ofP(x)have absolute value greater than 1. Letybe

a complex zero ofP(x). Then

0 =(y− 1 )P (y)=yp+yp−^1 +yp−^2 +···+y−p.

Assuming|y|≤1, we obtain

p=|yp+yp−^1 +yp−^2 +···+y|≤

∑p

i= 1

|y|i≤

∑p

i= 1

1 =p.

This can happen only if the two inequalities are, in fact, equalities, in which casey=1.

ButP( 1 )>0, a contradiction that proves our claim.

Next, let us assume thatP(x)=Q(x)R(x)withQ(x)andR(x)polynomials with

integer coefficients of degree at least 1. Thenp=P( 0 )=Q( 0 )R( 0 ). Since bothQ( 0 )

andR( 0 )are integers, eitherQ( 0 )=±1orR( 0 )=±1. Without loss of generality, we

may assumeQ( 0 )=±1. This, however, is impossible, since all zeros ofQ(x), which

are also zeros ofP(x), have absolute value greater than 1. We conclude thatP(x)is

irreducible.

(proposed by M. Manea forMathematics Magazine)