<s>[INST] How many ordered pairs of positive integers 
For a prime integer p, how many ordered pairs of positive integers
  (a, b) are there that satisfy $$\frac{1}{a} + \frac{1}{b} =\frac{1}{p}$$
For example, for p = 5, $$\frac{1}{6} + \frac{1}{30}$$ and
  $$\frac{1}{30} + \frac{1}{6}$$ are two different ways of getting
  $\frac{1}{5}$.

Ok, so I tried this for prime numbers from 2 to 19. It seems like that for any prime number there are only 3 such ordered pairs. 
(Question: $\frac{1}{10} + \frac{1}{10}$ is just one ordered pair for p=5, right?)
But I don't know how to go about proving this. I can see that:
After $\frac{1}{p}$, take the next smallest fraction of the form $\frac{1}{a}$. Now we always get a fraction $\frac{1}{a} + \frac{1}{b}$. Also, now b is the upper bound for the numbers we need to check. But there is always just one other integer between a and b which satisfies for this property, this integer is is 2p
The question does not specifically ask for a prove, but they always expect a prove for everything. 
 [/INST] If we multiply through by $pab$, we find that for $a, b \ne 0$, our equation is equivalent to $pb+pa=ab$, which we can rewrite as
$$(a-p)(b-p)=p^2.$$
We are looking for non-zero solutions of this equation. Conveniently, $p^2$ does not have many factorizations! 
We can have $a-p=-1$, $b-p=-p^2$, which yields a negative $b$, or $a-p=-p^2$, $b-p=-1$, which yields a negative $a$.  
We could have $a-p=-p$, $b-p=-p$, but that yields the impossible $a=b=0$.
We can have $a-p=1$, $b-p=p^2$, or $a-p=p^2$, $b-p=1$, which respectively yield the solutions $a=p+1$, $b=p^2+p$, and $a=p^2+p$, $b=p+1$.
Finally, we can have $a-p=p$, $b-p=p$, which yields $a=b=2p$.  
Comment: The same idea can be used to find all integer solutions of the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{n}$, where $n$ is a given positive integer.  The factorizations of $n^2$, with the exception of $n^2=(-n)(-n)$, supply the integer solutions.  
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