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Another question I saw recently:

Find all triples of positive integers $(a,b,c)$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$.

Can someone help me with it?

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closed as off-topic by barak manos, alans, Brevan Ellefsen, quid, Namaste Aug 9 '16 at 20:50

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – alans, quid
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I'm voting to close this question as off-topic because just like every other question of yours, you show no effort whatsoever in attempting to answer it on your own. Asking 'Can anyone help me with it?' is simply not enough. $\endgroup$ – barak manos Aug 9 '16 at 9:59
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As $\frac{1}{n}>0$ for all natural $n$, all of $a,b,c$ have to be at least two. Let $a\leq b\leq c$. Unless $a=b=c=3$, $a=2$. Then we have $\frac{1}{b}+\frac{1}{c}=\frac{1}{2}$, so similarly either $b=c=4$ or $b=3$. In the latter case, we get $c=6$. So, $(a,b,c)=(3,3,3),(2,4,4),(2,3,6)$ are all the solutions.

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Hint

Without loss of generality, let $a<b<c$ and set $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.

  • If $a=1$ then $t>1$
  • If $a\ge 3$ then $t<1$

Thus $a=2$

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    $\begingroup$ Generality is lost in assuming that $a,b,c$ are all distinct, as the list of solutions in another answer indicates. $\endgroup$ – Karl Kronenfeld Jul 25 '16 at 2:52
  • $\begingroup$ @Karl Kronenfeldyes trivial solution is $(3,3,3)$ $\endgroup$ – Behrouz Maleki Jul 25 '16 at 10:40

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