The question most likely has a very simple solution
Unfortunately I am not familiar with number theory as I have just decided to start studying it so I do not see the solution.
If someone wouldn't mind giving a little more of an explanation it would me most helpful. Thanks!
Well consider the set of all pairs where is a divisor of . What you want to show is that every divisor occurs not only as a first element of some pair (which it does by definition) but also as the second element of some pair. Hint : show that when is a divisor of , is also a divisor of . Which pair has as a first element?