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

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

Line 17: | Line 17: | ||

==Solution 2== | ==Solution 2== | ||

We shall prove that such integers do not exist via contradiction. | We shall prove that such integers do not exist via contradiction. | ||

− | Suppose that <math>a^5b + 3 = x^3</math> and <math>ab^5 + 3 = y^3</math> for integers x and y. Rearranging terms gives <math>a^5b = x^3 - 3</math> and <math>ab^5 = y^3 - 3</math>. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = <math>(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}</math> and b = <math>(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}</math>. Consider a prime p in the prime factorization of <math>x^3 - 3</math> and <math>y^3 - 3</math>. If it has power <math>r_1</math> in <math>x^3 - 3</math> and power <math>r_2</math> in <math>y^3 - 3</math>, then <math>5r_1</math> - <math>r_2</math> | + | Suppose that <math>a^5b + 3 = x^3</math> and <math>ab^5 + 3 = y^3</math> for integers x and y. Rearranging terms gives <math>a^5b = x^3 - 3</math> and <math>ab^5 = y^3 - 3</math>. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = <math>(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}</math> and b = <math>(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}</math>. Consider a prime p in the prime factorization of <math>x^3 - 3</math> and <math>y^3 - 3</math>. If it has power <math>r_1</math> in <math>x^3 - 3</math> and power <math>r_2</math> in <math>y^3 - 3</math>, then <math>5r_1</math> - <math>r_2</math> is a multiple of 24 and <math>5r_2</math> - <math>r_1</math> also is a multiple of 24. |

+ | |||

+ | Adding and subtracting the divisions gives that <math>r_1</math> - <math>r_2</math> divides 12. (actually, <math>r_1 - r_2</math> is a multiple of 4, as you can verify if <math>\{^{5r_1 - r_2 = 24}_{5r_2 - r_1 = 48}</math>. So the rest of the proof is invalid.) Because <math>5r_1</math> - <math>r_2</math> also divides 12, <math>4r_1</math> divides 12 and thus <math>r_1</math> divides 3. Repeating this trick for all primes in <math>x^3 - 3</math>, we see that <math>x^3 - 3</math> is a perfect cube, say <math>q^3</math>. Then <math>x^3 - q^3 = 3,</math> and <math>(x-q)(x^2 + xq + q^2) = 3</math>, so that <math>x - q = 1</math> and <math>x^2 + xq + q^2 = 3</math>. Clearly, this system of equations has no integer solutions for <math>x</math> or <math>q</math>, a contradiction, hence completing the proof. | ||

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

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

## Revision as of 21:41, 26 July 2014

## 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.