# 2002 IMO Shortlist Problems/A3

## Problem

Let be a cubic polynomial given by , where are integers and . Suppose that for infinitely many pairs , of integers with . Prove that the equation has an integer root.

## Solutions

We note that the condition is equivalent to

.

Since , we may remove the factor to obtain

,

or

.

We can denote and as and , respectively, rewriting this as

.

### Solution 1

We claim that can only assume finitely many values. We note that , so , which brings us to

,

which is clearly true for at most finitely many integer values of .

We denote by . The condition is then equivalent to , or . But can only assume finitely many values, so for some value of , are quartic polynomials equivalent at infinitely many points and are therefore equivalent.

Now, if , then we have , so is an even function. Since is divisible by , it must therefore also be divisible by , implying that is divisible by , which means that 0 is a root of , as desired.

If , then we have , so , implying that is equal to 0 since . Thus is the desired root of .

### Solution 2

Consider the function . If has roots , then is bounded on the interval , and injective on each of the intervals . Because each value of yields at most finitely many solutions , there are at most finitely many solutions such that and have the same sign.

Since , if , then each value of yields at most one solution. But for arbitrarily large values of , and have the same sign, making impossible if and have different signs. Thus there must be some integer such that . (Indeed, if , then any pair of integers with sum is a solution.)

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*