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

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Therefore no such integers exist.
 
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Revision as of 18:01, 3 July 2013

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

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