All values $n$ s.t. ${1 \over x} + {1 \over y} = {1 \over n}$ has exactly 5 solutions Given
${1 \over x} + {1 \over y} = {1 \over n}$
where $x,y,n$ are positive integers.
Find all possible values of $n$ such that there are only $5$ ordered pairs of $(x,y)$ which satisfies the equation.
I tried taking LCM and trying to factorize it, but I am unable to do so. Please help. 
PS- Do not use calculus in your reasoning or solution as I am preparing for an Olympiad which does not allow calculus. You can still upload an answer involving calculus for other's to understand. 
 A: If 
$${1 \over x} + {1 \over y} = {1 \over n}$$
Then 
$$(x-n)(y-n)=n^2\, \qquad \forall xy \neq \, 0$$
A: Let $d=(x,y)$ and take $a,b$ such that $x=ad$ and $y=bd$ (so that $(a,b)=1$).  Then $$\frac{1}{da}+\frac{1}{db}=\frac{1}{n}$$ or $$n(a+b)=dab$$ $(a+b,a)=(a+b,b)=1$, so the solutions to this problem in terms of $n$ are generated from precisely the ways to choose $a$ and $b$ such that (a,b)=1 and $ab\mid n$. 
If $n$ is prime, there are three solutions:  (a,b) = (n, 1), (1,n), or (1,1).  If n=$p^2$ for some prime p, there are five solutions: ($p^2$,1), (1,$p^2$), (p,1), (1,p),(1,1).  The interested reader could go on to calculate that $n=p^k$ has $2k+1$ solutions and $n$ having two distinct prime factors has at least nine solutions.  
Therefore, the problem has five solutions if and only if $n$ is the square of a prime.
A: Some months ago I was thinking about the same problem and I have written a correct function, $\lambda(\xi)$ that takes in input an intger $\xi$ and calculate for how many values of $x\in N$, with $0\leq x < \xi$, the ratio: $$\psi = \frac{\xi x}{\xi -x}$$ is an integer $(\psi \in N)$.
Here there is a graph of $\lambda(\xi)$ for $2\leq \xi \leq113$, created in Excel:

This function can be rewritten (because $\psi = \frac{\xi^2}{\xi-x}-\xi\in\mathbb{N}\iff \frac{\xi^2}{\xi-x}\in\mathbb{N}$ and the latter is true only when $(\xi-x)|\xi^2$ i.e. once for each divisor of $ξ^2$) as $\lambda(\xi)=\frac{d(\xi^2)+1}2$, where $d(n)$ is the number of divisors of $n$.
