2013 USAJMO Problems/Problem 1

Revision as of 23:38, 8 April 2015 by Bobthesmartypants (talk | contribs) (Solution 2)


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


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.

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