2021 JMPSC Invitationals Problems/Problem 13
Let be a prime and be an odd integer (not necessarily positive) such that is an integer. Find the sum of all distinct possible values of .
Assume temporarily that . Then, and are both odd which implies that the numerator is odd and the denominator is even. Since an even number cannot divide an odd number, we have a contradiction. Since our claim is incorrect, and we now wish to make an integer. We see that the denominator is always odd, while the numerator is always even. Thus, the only possible values for are when the denominator is , which implies . These correspond with , so for an answer of . ~samrocksnature
Suppose is odd. We have that our expression can never be an integer, so . Now, is an integer, implying is a perfect power of . Now, we have is even if , and are the only values that work when testing for odd . The answer is
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