# 2013 USAJMO Problems/Problem 1

## Problem

Are there integers and such that and are both perfect cubes of integers?

## Solution

No, such integers do not exist. This shall be proven by contradiction, by showing that if is a perfect cube then cannot be.

Remark that perfect cubes are always congruent to , , or modulo . Therefore, if , then .

If , then note that . (This is because if then .) Therefore and , contradiction.

Otherwise, either or . Note that since is a perfect sixth power, and since neither nor contains a factor of , . If , then Similarly, if , then Therefore , contradiction.

Therefore no such integers exist.

## Solution 2

We shall prove that such integers do not exist via contradiction. Suppose that and for integers x and y. Rearranging terms gives and . Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = and b = . Consider a prime p in the prime factorization of and . If it has power in and power in , then - is a multiple of 24 and - also is a multiple of 24.

Adding and subtracting the divisions gives that - divides 12. (actually, is a multiple of 4, as you can verify if . So the rest of the proof is invalid.) Because - also divides 12, divides 12 and thus divides 3. Repeating this trick for all primes in , we see that is a perfect cube, say . Then and , so that and . Clearly, this system of equations has no integer solutions for or , a contradiction, hence completing the proof.

Therefore no such integers exist. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.