5. How many solutions in natural numbers does the equation $(a+1)(b+1)(c+1)=2 a b c$ have?
Solution. Rewrite the equation as $(1+1 / a)(1+1 / b)(1+1 / c)=2$. Due to symmetry, it is sufficient to find all solutions with $a \leqslant b \leqslant c$. Then $(1+1 / a)^{3} \geqslant 2$, which means $a \leqslant(\sqrt[3]{2}-1)^{-1}<4$ and $a \in\{1,2,3\}$. In the case $a=1$, the inequality $2(1+1 / b)^{2} \geqslant 2$ holds, so there are no solutions. If $a=2$, then $\frac{3}{2}(1+1 / b)^{2} \geqslant 2$, which means $2 \leqslant b \leqslant\left(\frac{2 \sqrt{3}}{3}-1\right)^{-1}<7$. In this case, there are 3 solutions $(a, b, c)=(2,4,15),(2,5,9),(2,6,7)$ (for $b=2$ and $b=3$, the equation for $c$ has no solutions in natural numbers). Finally, if $a=3$, then $\frac{4}{3}(1+1 / b)^{2} \geqslant 2$, which means $3 \leqslant b \leqslant\left(\sqrt{\frac{3}{2}}-1\right)^{-1}<5$. This gives 2 more solutions $(a, b, c)=(3,3,8),(3,4,5)$. Taking into account permutations, there are a total of 27 solutions.

Criteria. If only part of the solutions is found, no more than 2 points are given.