Difference between revisions of "2013 USAJMO Problems/Problem 1"
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Revision as of 17:01, 3 July 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. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.