2022 USAMO Problems/Problem 4
Contents
Problem
Find all pairs of primes for which and are both perfect squares.
Solution 1
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 .
Solution 2
Let , , where are positive integers. . So,
For , . Then and . and . Thus, and we find . Hence .
For , ( integer), by , . Let's examine in , . But we know that . This is a contradiction and no solution for .
For , ( integer), by , . Let , where and are integers. Since , we see . Thus, by , . and are same parity and is even integer. So, and are both even integers. Therefore,
or Therefore, or . For each case, . But , this gives a contradiction. No solution for .
We conclude that the only solution is .
(Lokman GÖKÇE)
See also
2022 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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