Problem: (1999 Canada – #3) Determine all positive integers with the property that . Here denotes the number of positive divisors of .
Solution: So, testing some small numbers yields as a solution. We claim that this is the only such solution.
Clearly, since is a square, we can use a variant of the usual prime decomposition and say that .
Furthermore, again since is a square, we know
is odd, so must be odd as well, i.e. is not one of the . Then we use the equation given to us to get
Note, however, that by Bernoulli’s Inequality (overkill, I know) we have for
with equality iff and . So
Since we want equality, we must have and for all . But since the primes are supposed to be distinct we can have exactly one prime so that is the only solution. QED.
Comment: Another one of those problems that you kind of look at the result and you’re like huh, that’s interesting. But anyway, just throwing in some weak inequalities led to a pretty straightforward solution. As long as you know how to find the number of divisors of a positive integer it isn’t too much of a stretch to figure the rest out, though it make take some time to get in the right direction since the problem is quite open-ended.
Practice Problem: (1999 Canada – #4) Suppose are eight distinct integers from . Show that there is an integer such that the equation has at least three different solutions. Also, find a specific set of distinct integers from such that the equation does not have three distinct solutions for any .
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