2020 CIME II Problems/Problem 10
First suppose that . Then from whence we have .
Now suppose that . Since is a positive integer, from the equation we have . Hence , and since we have . Since the original equation is symmetric in it follows that as well. Adding the inequalities gives . From the original equation we know that ; hence is a multiple of which is no more than . It follows that , for if we have ; a contradiction since .
We now check each of these 5 cases using the original equation, keeping in mind the two solutions already found.
Case I) .
Case II) .
Case III) .
Case IV) for which there are no solutions.
Case V) for which there are 2 solutions (corresponding to the factors 9 and 14) however they have ; already covered.
Computing for each of the 9 solutions and adding the results we have .