Advanced book on Mathematics Olympiad

2.2 Polynomials 57

Proof.We argue by contradiction. Suppose thatP(x)=Q(x)R(x), withQ(x)and
R(x)not identically equal to±1. Let

Q(x)=bkxk+bk− 1 xk− 1 +···+b 0 ,

R(x)=cn−kxn−k+cn−k− 1 xn−k−^1 +···+c 0.

Let us look closely at the equalities

∑i

j= 0

bjci−j=ai,i= 0 , 1 ,...,n,

obtained by identifying the coefficients. From the first of them,b 0 c 0 =a 0 , and because
a 0 is divisible bypbut not byp^2 it follows that exactly one ofb 0 andc 0 is divisible by
p. Assume thatb 0 is divisible bypand take the next equalityb 0 c 1 +b 1 c 0 =a 1. The
right-hand side is divisible byp, and the first term on the left is also divisible byp. Hence
b 1 c 0 is divisible byp, and sincec 0 is not,b 1 must be divisible byp.
This reasoning can be repeated to prove that all thebi’s are divisible byp.Itis
important that bothQ(x)andR(x)have degrees greater than or equal to 1, for the fact
thatbkis divisible bypfollows from

bkc 0 +bk− 1 c 1 +··· =ak,

whereakis divisible bypfork<n. The contradiction arises in the equalityan=bkcn−k,
since the right-hand side is divisible byp, while the left-hand side is not. This proves
the theorem. 

The first three problems listed below use this result, while the others apply simi-
lar ideas.

182.Prove that the polynomial

P(x)=x^101 + 101 x^100 + 102

is irreducible overZ[x].

183.Prove that for every prime numberp, the polynomial

P(x)=xp−^1 +xp−^2 +···+x+ 1

is irreducible overZ[x].

184.Prove that for every positive integern, the polynomialP(x)=x^2
n
+1 is irreducible
overZ[x].