# 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 .