Difference between revisions of "2013 USAJMO Problems/Problem 1"
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==Solution== | ==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. | |
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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 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.