# 2021 JMPSC Invitationals Problems/Problem 13

## Contents

## Problem

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 .

## Solution

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

## Solution 2

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

~Geometry285

## See also

- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.