# 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.

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If $a = 3$, then we must have $b \geq \ldots$ and $c \geq \dots$. Is this possible? –  Srivatsan Sep 2 '11 at 3:15
You can just fix a number and count. There's a clear upper bound for $a$, and for each $a$, see my comment to mixedmath's answer. –  Soarer Sep 2 '11 at 3:24
@James Your question is well-written. +1 for telling us your thoughts. But I feel the tags should be changed. [combinatorics] doesn't seem to be relevant, and [number-theory] should be replaced by [elementary-number-theory]. –  Srivatsan Sep 2 '11 at 3:29

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.

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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 $\lt$.

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+1 I wish this were the only hint given for a while. A good one. –  Ross Millikan Sep 2 '11 at 4:13

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.

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For two variable case, one should handle it this way: $Kbc = b+c \Rightarrow K^2bc - K(b+c) + 1 = 1 \Rightarrow (Kb - 1)(Kc -1) = 1$. –  Soarer Sep 2 '11 at 3:23
Clever. Now I leave my answer so that your comment is also left. –  mixedmath Sep 2 '11 at 3:37
@Soarer One can also observe that $Kbc = b+c$ does not have any solution unless $K\leq 2$, since $b+c \leq b \cdot c+ b \cdot c < 3bc$. –  Srivatsan Sep 2 '11 at 3:43
@Srivatsan Nice. –  Soarer Sep 2 '11 at 3:45