Smallest $n$ with given number of solutions of $\frac{1}{u} + \frac{1}{v} = \frac{1}{n}$ With $n,u,v$ being positive integers, let $T(n)$ be the number of ordered pair of positive integers $(u,v)$ such that
$$ \frac{1}{u} + \frac{1}{v} = \frac{1}{n}$$
What is the smallest $N$ such that such that $N$ is a prime power and $T(N) = 13$?
What about a general $k$? How do we find the smallest $n$ such that $T(n) = k$?

This problem is posed by Sreejato B.
Details: A prime power is of the form $p^n$, where  is a prime number and  is a positive integer

 A: If I calculated correctly:


*

*T(1) = 1

*T(2) = 3

*T(3) = 3

*T(4) = 5

*T(5) = 3

*T(6) = 9

*T(7) = 3

*T(8) = 7

*T(9) = 5

*T(10) = 9

*T(11) = 3

*T(12) = 15


So, if you meant to ask for the smallest $n$ where $T(n) \ge 12$, that would be $n = 12$.  But $T(n)$ can never be exactly 12 because it's always odd.
tsillke gave one explanation.  Another way to see this is that, because of symmetry, solutions come in pairs: If $(u, v) = (a, b)$ is a solution, then so is $(u, v) = (b, a)$.  However, there's also always one solution with $u = v$, namely $(u, v) = (2n, 2n)$.  This is counted as a single solution instead of a pair, making the number of solutions odd.
A: This problem is essentially equivalent to solving 
$$\tau(n^2) = k$$
where $\tau(m)$ is the number of divisors of $m$
This we see by rewriting as
$$ (u-n)(v-n) = n^2$$
Notice that, this necessarily implies that $k$ must be odd (number of divisors of a perfect square is odd), and so $T(n) = 12$ has no solutions.
There might not be a simple solution for the general case.
A: I am writing it as an answer so that it is more visible.
This is a live Brilliant problem, and in fact proposed by me! I am requesting the other answerers to kindly delete their solutions till $5/8/2013$. For your information, there is a typo in the question. It should have been $f(n)= 13$ 
