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Difference between revisions of "2013 USAJMO Problems/Problem 1"
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==Solution==
 
==Solution==
  
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Consider the equation modulo <math>9</math>.
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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.
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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>.
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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.
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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.
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Therefore no such integers exist.

Revision as of 20:29, 11 May 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.

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