A good number theory problem.

Let's assume, without loss of generality, that $a\leq b\leq c$ . Therefore, if $a\geq 3$ we have

$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) \leq \left(1+\frac{1}{a}\right)^3 \leq \left(1+\frac{1}{3}\right)^3 = \frac{64}{27} < 3$ ,

which shows that there are no positive integer solutions to the given equation such that $3\leq a\leq b\leq c$ . Hence, we have just two cases to consider:

(1) $a=1$ :

$\left(1+\frac{1}{1}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) = 3 \Rightarrow 2(b+1)(c+1) = 3bc \Rightarrow bc-2b-2c-2 = 0 \Rightarrow (b-2)(c-2) = 6$ .

There are only 2 positive integer solutions to the above equation satisfying the inequality $b\leq c$ : $(b,c) = (3,8)$ and $(b,c) = (4,5)$ .

(2) $a=2$ :

$\left(1+\frac{1}{2}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) = 3 \Rightarrow (b+1)(c+1) = 2bc \Rightarrow bc-b-c-1 = 0 \Rightarrow (b-1)(c-1) = 2$ .

There is only 1 positive integer solution to the above equation satisfying the inequality $b\leq c$ : $(b,c) = (2,3)$ .

Combining the results of cases (1) and (2), we conclude that there are only $\boxed{3}$ unordered triples $(a,b,c)$ of positive integers satisfying the given equation, namely $(1,3,8),\,(1,4,5)$ , and $(2,2,3)$ .

123 213 321

Does anyone have an elegant solution for this one?