# 2013 USAJMO Problems/Problem 1

## Contents

## 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.

Amkan2022

## 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.

## Solution 3

Let and . Then, , , and Now take (recall that perfect cubes and perfect sixth powers ) on both sides. There are cases to consider on what values that and take. Checking these cases, we see that only or yield a valid residue (specifically, ). But this means that , so so contradiction.

## Solution 4

If is a perfect cube, then can be one of , so can be one of , , or . If were divisible by , we'd have , which we've ruled out. So , which means , and therefore .

We've shown that can be one of , so can be one of . None of these are possibilities for a perfect cube, so if is a perfect cube, cannot be.

## Solution 5

As in previous solutions, notice . Now multiplying gives , which is only , so after testing all cases we find that . Then since , and (Note that cannot be ). Thus we find that the inverse of is itself under modulo , a contradiction.

## Solution 6

I claim there are no such a or b such that both expressions are cubes.

Assume to the contrary and are cubes.

**Lemma 1**: If and are cubes, then

**Proof** Since cubes are congruent to any of , . But if , , so , contradiction. A similar argument can be made for .

**Lemma 2**: If k is a perfect 6th power, then

**Proof**: Since cubes are congruent to , we can square, and get 6th powers are congruent to .

Since , which is a perfect 6th power, by lemma 2, .

But, by lemma 1, .

So, , which is an integer, can't go into any of the possible residue classes modulo 9, without breaking one of these lemmas. This is a contradiction, and the proof is complete.

-AlexLikeMath

## Solution 7

Note that . So .

.

Multiplying and we get . By substituting values for and we can see that but because cubes are not , and squares are not so .

- Case I: .This is only true when . If then , which is not a perfect cube. The same can be seen with if then and if then . So 6 mod 9 case is eliminated from the above relations of congruences
- Case II: for this we'll take a look at the above information regarding and . If then,

which is not possible. Similarly, if then,

which is not possible. So, no such integers.

-Lakshya Pamecha

## Note

After you get that , you can also bash by listing all possible residues for and their corresponding values for . At that point, you can easily see that all the solutions for and do not work in the other equation, meaning that there are no solutions that satisfy both equations.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.