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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 ? 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?2011-09-02
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    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.2011-09-02
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    @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].2011-09-02

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