Difference between revisions of "2022 USAMO Problems/Problem 4"
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Revision as of 18:29, 31 March 2022
Problem
Solution
Since is a perfect square and is prime, we should have for some positive integer . Let . Therefore, , and substituting that into the and solving for gives Notice that we also have and so . We run through the cases
- : Then so , which works.
- : This means , so , a contradiction.
- : This means that . Since can be split up into two factors such that and , we get
and each factor is greater than , contradicting the primality of .
Thus, the only solution is .
See also
2022 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.