# Difference between revisions of "2013 USAJMO Problems/Problem 1"

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Assume to the contrary <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes. | Assume to the contrary <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes. | ||

− | '''Lemma 1''': If <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes <math>ab^5,a^5b \equiv 5,7 \pmod 9</math> | + | '''Lemma 1''': If <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes, then <math>ab^5, a^5b \equiv 5,7 \pmod 9</math> |

'''Proof''' Since cubes are congruent to any of <math>0, 1, -1 \pmod 9</math>, <math>ab^5,a^5b \equiv 5,6,7 \pmod 9</math>. But if <math>ab^5 \equiv 6 \pmod 9</math>, <math>3|a</math>, so <math>a^5b \equiv 0 \pmod 9</math>, contradiction. A similar argument can be made for <math>ab^5 \neq 6 \pmod 9</math>. | '''Proof''' Since cubes are congruent to any of <math>0, 1, -1 \pmod 9</math>, <math>ab^5,a^5b \equiv 5,6,7 \pmod 9</math>. But if <math>ab^5 \equiv 6 \pmod 9</math>, <math>3|a</math>, so <math>a^5b \equiv 0 \pmod 9</math>, contradiction. A similar argument can be made for <math>ab^5 \neq 6 \pmod 9</math>. | ||

− | '''Lemma 2''': If k is a perfect 6th power, <math>k \equiv 0,1 \pmod 9</math> | + | '''Lemma 2''': If k is a perfect 6th power, then <math>k \equiv 0,1 \pmod 9</math> |

− | '''Proof''': | + | '''Proof''': Since cubes are congruent to <math>0, 1, -1 \pmod 9</math>, we can square, and get 6th powers are congruent to <math>0, 1 \pmod 9</math>. |

## Latest revision as of 14:10, 15 June 2020

## 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.

## 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

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