Exercise 1231Algebra › PolynomialsOlympiadAdded 4 Nov 2026

Divisibility of values of an integer polynomial

#integer polynomials#a−b divides P(a)−P(b)#integer roots

Statement

Let $P$ be a polynomial with integer coefficients.

Show that for all integers $a\ne b$ , $a-b$ divides $P(a)-P(b)$ . Deduce that no such $P$ satisfies $P(7)=11$ and $P(11)=13$ .
Suppose that $P(a)=P(b)=P(c)=-1$ for three distinct integers $a,b,c$ . Show that $P$ has no integer root.

Hint

For part 1, use $x^k-y^k=(x-y)(x^{k-1}+\dots+y^{k-1})$ . For part 2, write $P(x)+1=(x-a)(x-b)(x-c)Q(x)$ and look at an integer root $n$ .

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Solution

1. If $P(x)=\sum c_kx^k$ , then $P(a)-P(b)=\sum c_k(a^k-b^k)$ and each $a^k-b^k$ is a multiple of $a-b$ . So $(a-b)\mid P(a)-P(b)$ .

If $P(7)=11$ and $P(11)=13$ , then $7-11=-4$ must divide $11-13=-2$ : false. So no such polynomial exists.

2. The polynomial $P(x)+1$ has the integer roots $a,b,c$ , so $P(x)+1=(x-a)(x-b)(x-c)Q(x)$ where $Q$ has integer coefficients. If $n$ were an integer root of $P$ , then
$(n-a)(n-b)(n-c)Q(n)=P(n)+1=1.$
So $n-a,n-b,n-c$ are three integers whose product divides $1$ : each of them is $\pm1$ . But they are pairwise distinct, and there are only two values $\pm1$ . Contradiction.

Lesson. For integer polynomials, always think of the divisibility $a-b\mid P(a)-P(b)$ and of factoring $P(x)-k$ over the integers.

Divisibility of values of an integer polynomial

Statement

Solution

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