<s>[INST] All solutions of $a+b+c=abc$ in natural numbers I was observing some nice examples of equalities containing the numbers $1,2,3$ like $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ and $\log 1+\log 2+ \log 3=\log (1+2+3)$. I found out this only happens because $1+2+3=1*2*3=6$. I wanted to find other examples in small numbers, but I failed. How can we find all of the solutions of $a+b+c=abc$ in natural numbers?The question seemed easy, but it seems difficult to find. I would prefer an elementary way to find them! What I did: We know if $a+b+c=abc$, $a|a+b+c$ so $a|b+c$. Similarly, $b|a+c$ and $c|a+b$. Other than that, if we multiply both sides by $b$, we get $b^2+1=(bc-1)(ab-1)$. If we also divide both sides by $abc$, we get $\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1$. I don't know how to go further using any of these, but I think they are a good start. I would appreciate any help.
 [/INST] leaving aside the solution in which $a=b=c=0$ order the numbers so $a \le b \le c$
as OP shows, $c|(a+b)$ so we have:
$$
\frac{a+b}{c} \le 2
$$
thus only the values $1$ and $2$ are possible for $\frac{a+b}c$. these give $a+b=c$ and $a+b=2c$ respectively. however the latter is only possible if all three numbers are equal, as $c$ cannot be the arithmetic mean of two smaller numbers. 
if all three numbers are equal they must each be $\sqrt{3}$, not a natural number solution.
plugging $a+b=c$ back into the equation $a+b+c=abc$ gives $ab=2$ hence $a=1, b=2, c=3$
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