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

Mathcool2009 (talk | contribs) (→Solution 2) |
(→Solution 2) |
||

Line 22: | Line 22: | ||

Therefore no such integers exist. | Therefore no such integers exist. | ||

+ | |||

+ | ==Solution 3== | ||

+ | Let <math>a^5b+3=x^3</math> and <math>ab^5+3=y^3</math>. Then, <math>a^5b=x^3-3</math>, <math>ab^5=y^3=3</math>, and <cmath>(ab)^6=(x^3-3)(y^3-3)</cmath> | ||

+ | Now take <math>\text{mod }9</math> (recall that perfect cubes <math>\equiv -1,0,1\pmod{9}</math> and perfect sixth powers <math>\equiv 0,1\pmod{9}</math>) on both sides. There are <math>3\times 3=9</math> cases to consider on what values <math>\text{mod }9</math> that <math>x^3</math> and <math>y^3</math> take. Checking these <math>9</math> cases, we see that only <math>x^3\equiv y^3\equiv 0\pmod{9}</math> or <math>x\equiv y\equiv 0\pmod{3}</math> yield a valid residue <math>\text{mod }9</math> (specifically, <math>(x^3-3)(y^3-3)\equiv 0\pmod{9}</math>). But this means that <math>3\mid ab</math>, so <math>729\mid (ab)^6</math> so <cmath>729\mid (x^3-3)(y^3-3)\iff 729\mid (27x'^3-3)(27y'^3-3)\iff 81\mid (9x'^3-1)(9y'^3-1)</cmath> contradiction. | ||

+ | |||

{{MAA Notice}} | {{MAA Notice}} |

## Revision as of 23:38, 8 April 2015

## 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.

## 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.

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