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.
 A: 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.
A: HINT
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \lt \frac{1}{a} + \frac{1}{a} + \frac{1}{a},
$$
which shows that $a \leq \ldots$.1 
(After fixing $a$, you can use the same idea again to complete the proof.)

1EDIT: Corrected the first inequality sign from $\gt$ to $\lt$.
A: Suppose we fix a, small enough that solutions exist. Then we have some sort of equation of the form $\frac{1}{b} + \frac{1}{c} = K$, or that $1 + \frac{b}{c} = bK \implies \frac{b}{c} - bK = b(\frac{1}{c} - K) = -1 \implies b = \dfrac{-1}{\frac{1}{c} - K}$
That is to say, that there are still infinitely many solutions for just 2 variables (under the assumption that $c \not = 0$ and $\frac{1}{c} - K \not = 0$. I note that there are infinitely because these solutions fall in a range, and if the ordering is backwards then we simply switch the roles of the variables. So that does not play a big role.
