Difference between revisions of "2013 USAJMO Problems/Problem 1"
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− | + | ==Problem== | |
+ | |||
+ | Are there integers <math> a </math> and <math> b </math> such that <math> a^5b+3 </math> and <math> ab^5+3 </math> are both perfect cubes of integers? | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | No, such integers do not exist. This shall be proven by contradiction, by showing that if <math>a^5b+3</math> is a perfect cube then <math>ab^5+3</math> cannot be. | ||
+ | |||
+ | Remark that perfect cubes are always congruent to <math>0</math>, <math>1</math>, or <math>-1</math> modulo <math>9</math>. Therefore, if <math>a^5b+3\equiv 0,1,\text{ or} -1\pmod{9}</math>, then <math>a^5b\equiv 5,6,\text{ or }7\pmod{9}</math>. | ||
+ | |||
+ | If <math>a^5b\equiv 6\pmod 9</math>, then note that <math>3|b</math>. (This is because if <math>3|a</math> then <math>a^5b\equiv 0\pmod 9</math>.) Therefore <math>ab^5\equiv 0\pmod 9</math> and <math>ab^5+3\equiv 3\pmod 9</math>, contradiction. | ||
+ | |||
+ | Otherwise, either <math>a^5b\equiv 5\pmod 9</math> or <math>a^5b\equiv 7\pmod 9</math>. Note that since <math>a^6b^6</math> is a perfect sixth power, and since neither <math>a</math> nor <math>b</math> contains a factor of <math>3</math>, <math>a^6b^6\equiv 1\pmod 9</math>. If <math>a^5b\equiv 5\pmod 9</math>, then <cmath>a^6b^6\equiv (a^5b)(ab^5)\equiv 5ab^5\equiv 1\pmod 9\implies ab^5\equiv 2\pmod 9.</cmath> Similarly, if <math>a^5b\equiv 7\pmod 9</math>, then <cmath>a^6b^6\equiv (a^5b)(ab^5)\equiv 7ab^5\equiv 1\pmod 9\implies ab^5\equiv 4\pmod 9.</cmath> Therefore <math>ab^5+3\equiv 5,7\pmod 9</math>, contradiction. | ||
+ | |||
+ | Therefore no such integers exist. | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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> 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. | ||
+ | |||
+ | ==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}} |
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.