2011 JBMO Problems/Problem 2
Find all primes such that there exist positive integers that satisfy .
Rearrange the original equation to get Since , , and are all integers, and must be a factor of . Now there are four cases to consider.
Case 1: and
Since and are positive integers, , so the first case can not happen.
Case 2: and
Rearranging the first equation results in . By substitution, we get Applying Simon's Favorite Factoring Trick results in . From this equation, and can equal 1, 2, 13, or 26. Since no value of or work, there are no solutions, so the second case can not happen.
Case 3: and
Rearranging the first equation results in . By substitution and Simon's Favorite Factoring Trick, we get From this equation, and can equal 1, 2, 3, or 6. The ordered pairs of can be , , , or . Since is prime, can equal in this case.
Case 4: and
Rearranging the first equation results in . Since are positive integers, we can simply take the few ordered pairs that satisfy the second equation and plug the values into the first equation to check for values of .
If is or , then , so . If is or , then , so . Both and are prime numbers, so both of the numbers work for .
In summary, the values of that satisfy the original conditions are .
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