Problem:
Are there integers and such that and are both perfect cubes of integers?
Solution:
The answer is negative. Modulo , a cube is or . Assuming that one of and is , it follows that at least one of the numbers and , say , is divisible by , hence is , not a perfect cube. If and are both perfect cubes of the form , then and are both or , and so their product, , is , or . But is the square of a perfect cube not divisible by , so is precisely , a contradiction.
This problem and solution were suggested by Titu Andreescu.
The problems on this page are the property of the MAA's American Mathematics Competitions