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

Revision as of 13:14, 15 June 2020

Problem

Are there integers $a$ and $b$ such that $a^5b+3$ and $ab^5+3$ are both perfect cubes of integers?

Solution

No, such integers do not exist. This shall be proven by contradiction, by showing that if $a^5b+3$ is a perfect cube then $ab^5+3$ cannot be.

Remark that perfect cubes are always congruent to $0$ , $1$ , or $-1$ modulo $9$ . Therefore, if $a^5b+3\equiv 0,1,\text{ or} -1\pmod{9}$ , then $a^5b\equiv 5,6,\text{ or }7\pmod{9}$ .

If $a^5b\equiv 6\pmod 9$ , then note that $3|b$ . (This is because if $3|a$ then $a^5b\equiv 0\pmod 9$ .) Therefore $ab^5\equiv 0\pmod 9$ and $ab^5+3\equiv 3\pmod 9$ , contradiction.

Otherwise, either $a^5b\equiv 5\pmod 9$ or $a^5b\equiv 7\pmod 9$ . Note that since $a^6b^6$ is a perfect sixth power, and since neither $a$ nor $b$ contains a factor of $3$ , $a^6b^6\equiv 1\pmod 9$ . If $a^5b\equiv 5\pmod 9$ , then $a^6b^6\equiv (a^5b)(ab^5)\equiv 5ab^5\equiv 1\pmod 9\implies ab^5\equiv 2\pmod 9.$ Similarly, if $a^5b\equiv 7\pmod 9$ , then $a^6b^6\equiv (a^5b)(ab^5)\equiv 7ab^5\equiv 1\pmod 9\implies ab^5\equiv 4\pmod 9.$ Therefore $ab^5+3\equiv 5,7\pmod 9$ , contradiction.

Therefore no such integers exist.

Solution 2

We shall prove that such integers do not exist via contradiction. Suppose that $a^5b + 3 = x^3$ and $ab^5 + 3 = y^3$ for integers x and y. Rearranging terms gives $a^5b = x^3 - 3$ and $ab^5 = y^3 - 3$ . Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = $(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}$ and b = $(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}$ . Consider a prime p in the prime factorization of $x^3 - 3$ and $y^3 - 3$ . If it has power $r_1$ in $x^3 - 3$ and power $r_2$ in $y^3 - 3$ , then $5r_1$ - $r_2$ is a multiple of 24 and $5r_2$ - $r_1$ also is a multiple of 24.

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

Therefore no such integers exist.

Solution 3

Let $a^5b+3=x^3$ and $ab^5+3=y^3$ . Then, $a^5b=x^3-3$ , $ab^5=y^3-3$ , and $(ab)^6=(x^3-3)(y^3-3)$ Now take $\text{mod }9$ (recall that perfect cubes $\equiv -1,0,1\pmod{9}$ and perfect sixth powers $\equiv 0,1\pmod{9}$ ) on both sides. There are $3\times 3=9$ cases to consider on what values $\text{mod }9$ that $x^3$ and $y^3$ take. Checking these $9$ cases, we see that only $x^3\equiv y^3\equiv 0\pmod{9}$ or $x\equiv y\equiv 0\pmod{3}$ yield a valid residue $\text{mod }9$ (specifically, $(x^3-3)(y^3-3)\equiv 0\pmod{9}$ ). But this means that $3\mid ab$ , so $729\mid (ab)^6$ so $729\mid (x^3-3)(y^3-3)\iff 729\mid (27x'^3-3)(27y'^3-3)\iff 81\mid (9x'^3-1)(9y'^3-1)$ contradiction.

Solution 4

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

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

Solution 5

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

Solution 6

I claim there are no such a or b such that both expressions are cubes.

Assume to the contrary $a^5b +3$ and $ab^5 + 3$ are cubes.

Lemma 1: If $a^5b +3$ and $ab^5 + 3$ are cubes $ab^5,a^5b \equiv 5,7 \pmod 9$

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

Lemma 2: If k is a perfect 6th power, $k \equiv 0,1 \pmod 9$

Proof: since cubes are congruent to $0, 1, -1 \pmod 9$ , 6th powers are congruent to $0, 1 \pmod 9$ .

Since $ab^5 \cdot a^5b = a^6 b^6 = (ab)^6$ , which is a perfect 6th power, by lemma 2, $ab^5 \cdot a^5b \equiv 0,1 \pmod 9$ .

But, by lemma 1, $ab^5 \cdot a^5b \equiv 5 \cdot 5, 5 \cdot 7, 7 \cdot 7 \equiv 7, 8, 4 \pmod 9$ .

So, $ab^5 \cdot a^5b$ , 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. $\blacksquare$

-AlexLikeMath

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

@@ Line 41: / Line 41: @@
 Assume to the contrary <math>a^5b +3</math> and <math>ab^5 + 3</math> are cubes.
-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.
+'''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>
-A similar argument can be made for <math>ab^5 \neq 6 \pmod 9</math>. Thus, we get that:
+'''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>.
-<cmath>ab^5,a^5b \equiv 5,7 \pmod 9</cmath>
-A critical idea is that since cubes are congruent to <math>0, 1,  -1 \pmod 9</math>, 6th powers are congruent to <math>0, 1 \pmod 9</math>. And, <math>ab^5 \cdot a^5b = a^6 b^6 = (ab)^6</math>, making the the product of <math>ab^5, a^5b</math> a perfect 6th power.
+'''Lemma 2''': If k is a perfect 6th power, <math>k \equiv 0,1 \pmod 9</math>
-But, we can substitiute our possible values of <math>ab^5, a^5b</math>. <math>5 \cdot 5 \equiv 7 \pmod 9</math>, <math>5 \cdot 7 \equiv 8 \pmod 9</math>, and <math>7 \cdot 7 \equiv 4 \pmod 9</math>. None of these are congruent to <math>0, 1 \pmod 9</math>, but perfect 6th powers have to be congruent to <math>0, 1 \pmod 9</math>. We have a contradiction. <math>\blacksquare</math>
+'''Proof''': since cubes are congruent to <math>0, 1,  -1 \pmod 9</math>, 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
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