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

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

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+ | ==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. | ||

+ | |||

+ | ==Solution 4== | ||

+ | If <math>a^5b+3</math> is a perfect cube, then <math>a^5b</math> can be one of <math>5,6,7 \pmod 9</math>, so <math>(a^5b)^5 = a^{25} b^5</math> can be one of <math>5^5 \equiv 2</math>, <math>6^5 \equiv 0</math>, or <math>7^5 \equiv 4 \pmod 9</math>. If <math>a</math> were divisible by <math>3</math>, we'd have <math>a^5 b \equiv 0 \pmod 9</math>, which we've ruled out. So <math>\gcd(a,9) = 1</math>, which means <math>a^6 \equiv 1 \pmod 9</math>, and therefore <math>a^{25} b^5 \equiv ab^5 \pmod 9</math>. | ||

+ | |||

+ | We've shown that <math>a b^5</math> can be one of <math>0, 2, 4 \pmod 9</math>, so <math>ab^5 + 3</math> can be one of <math>3, 5, 7 \pmod 9</math>. None of these are possibilities for a perfect cube, so if <math>a^5b+3</math> is a perfect cube, <math>ab^5+3</math> cannot be. | ||

+ | |||

+ | ==Solution 5== | ||

+ | |||

+ | As in previous solutions, notice <math>ab^5,a^5b \equiv 5,6,7 \pmod 9</math>. Now multiplying gives <math>a^6b^6</math>, which is only <math>0,1 \pmod 9</math>, so after testing all cases we find that <math>ab^5\equiv a^5b \equiv 6 \mod 9</math>. Then since <math>\phi (9) = 6</math>, <math>ab^5\equiv \frac{a}{b}\pmod 9</math> and <math>a^5b \equiv \frac{b}{a}\pmod 9</math> (Note that <math>a,b</math> cannot be <math>0\pmod 9</math>). Thus we find that the inverse of <math>6</math> is itself under modulo <math>9</math>, a contradiction. | ||

+ | |||

+ | ==Solution 6== | ||

+ | I claim there are no such a or b such that both expressions 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, 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>. | ||

+ | |||

+ | |||

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

+ | |||

+ | '''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>. | ||

+ | |||

+ | |||

+ | Since <math>ab^5 \cdot a^5b = a^6 b^6 = (ab)^6</math>, which is a perfect 6th power, by lemma 2, <math>ab^5 \cdot a^5b \equiv 0,1 \pmod 9</math>. | ||

+ | |||

+ | But, by lemma 1, <math>ab^5 \cdot a^5b \equiv 5 \cdot 5, 5 \cdot 7, 7 \cdot 7 \equiv 7, 8, 4 \pmod 9</math>. | ||

+ | |||

+ | So, <math>ab^5 \cdot a^5b</math>, 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. <math>\blacksquare</math> | ||

+ | |||

+ | -AlexLikeMath | ||

+ | |||

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

## Latest revision as of 15:10, 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

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