# 1998 APMO Problems/Problem 2

## Problem

Show that for any positive integers and , cannot be a power of .

## Solution 1

First, assume that is a power of . Let and . Then

Consider . Factoring out gives

Because contains odd factors and divides , must also divide , so .

Testing values shows that divides 18. It can be easily shown that , so the least possible value of is 18. But since , we reach a contradiction.

## Solution 2

Assume that is a power of . Then must also be a solution to for some positive integer . WLOG, assume and let be minimal. Then the least possible value of is . For all positive integers , . So both must divide 4 as well. Then must also be a solution. But is minimal and we find a smaller integer solution (because divides 4), so we reach a contradiction.