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.