11.5. Solve the equation in integers: $\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3$.
Answer. $(\mathrm{x}, \mathrm{y}, z)=(2,2,3),(1,3,8),(1,4,5),(-4,1,1)$ and all permutations of these triples. In total, there are 18 solutions.

Solution. Without loss of generality, we can assume that $a \leq b \leq c$.

Notice that if $x<0$, then $1+\frac{1}{x}<1$, and if $x \geq 1$, then $1+\frac{1}{x} \leq 2$. Therefore, among the numbers $a \leq b \leq c$, there is at most one negative number.

1) Suppose $a<0<b \leq c$. If $1<b \leq c$, then $\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)<\left(1+\frac{1}{2}\right)^{2}=\frac{9}{4}<3$. If $1=b<c$, then $\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)<2\left(1+\frac{1}{2}\right)=3$. Therefore, in this case, we have the only possibility $a=-4, b=c=1$.
2) Suppose $1 \leq a \leq b \leq c$. If $3 \leq a \leq b \leq c$, then $\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)<\left(1+\frac{1}{3}\right)^{3}=\frac{64}{27}<3$, hence $a \leq 2$.

A) Suppose $a=1$, then $\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{3}{2}$. If $5 \leq b \leq c$, then

Checking the cases $b=1,2,3,4$ gives two solutions: $a=1, b=3, c=8$ and $a=1, b=4, c=5$.

B) Suppose $a=2$, then $\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=2$. If $4 \leq b \leq c$, then $\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) \leq \frac{25}{16}<2$.

Checking the cases $b=2,3$ gives one solution: $a=2, b=2, c=3$.