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?


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

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  • 2
    $\begingroup$ 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

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.



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 $a=2$

  • 1
    $\begingroup$ 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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