# 1993 IMO Problems/Problem 1

Let , where is an integer. Prove that cannot be expressed as the product of two non-constant polynomials with integer coefficients.

## Solution

For the sake of contradiction, assume that for polynomials and in . Furthermore, let with if and with if . This gives that .

We have that , or . WLOG, let (and thus ). Since and divides but not , we need that . We can keep on going up the chain until we get that . Then, by equating coefficients once more, we get that . Taking the equation gives that . This implies that . Thus, the degree of is at least . However, because is a non-constant factor of , we have that the degree of is at most . Thus, the degree of is , implying that the degree of is .

From this fact, we have that there must exist a rational root of . The only candidates are , , , and , though. , , , and , so none of these work. Thus, there are no linear factors of .

In other words, cannot be expressed as for polynomials and in . This means that cannot be expressed as the product of two non-constant polynomials with integer coefficients.

Q.E.D.

## Alternate Solution

Trivial by Perron's Criterion lol

Note: Quoting Perron's Criterion on the actual IMO will very likely result in a score in the set , since it was not a well-known result back then.