Triples of Numbers I have a question:
How many triples $(a,b,c)$ are there such that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$$ and $a <b<c$? They have to be positive integers. Also find those triples.
I know that all of them have to be $\geq 2$. So do I just fix a number and count the other pairs?
If I choose $a = 3$ then I count the other pairs $(b,c)$? If I choose a very large $a$ then it seems that no triples will satisfy the condition since the sum will be too small.

Well, $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} < 1$, so we must have $a=2$. So we really just need $\frac{1}{b} + \frac{1}{c} = \frac{1}{2}$.
Since $\frac{1}{4} + \frac{1}{5} < \frac{1}{2}$, $b = 3$. That leaves $c = 6$.
I think a nice way to think about it is to view the number $1$ as $\frac{1}{1}$. We can decompose $\frac{1}{n}$ into $\frac{1}{n+1} + \frac{1}{n^2+n}$, then decompose one of those to get an expression for $\frac{1}{n}$ as the sum of three harmonic numbers. In this case, we see that $\frac{1}{1} = \frac{1}{2} + \frac{1}{2} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$.
See the Leibniz Harmonic Triangle.
