<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] $1/a+1/b=1/p$, $bp+ap=ab$, $(a+b)p=ab$, so $p$ divides $a$ or $p$ divides $b$ (or both). Let's say $p$ divides $a$, so $a=cp$ for some $c$. Now $cp+b=bc$, $b=bc-pc=(b-p)c$, so $c$ divides $b$. Let $b=cd$. Then $d=cd-p$, so $p=cd-d=(c-1)d$. So either $c-1=p$ and $d=1$, or $c-1=1$ and $d=p$. Now work your way back up to $a$ and $b$. 
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