2. Find all integer solutions of the equation

$$
x+\frac{1}{y+\frac{1}{z}}=\frac{7}{3}
$$
Answer: $(2 ; 2 ; 1) ;(2 ; 4 ;-1) ;(3 ;-1 ;-2) ;(3 ;-2 ; 2) ;(1 ; 1 ;-4)$.

Solution. First, consider the case when $\left|y+\frac{1}{z}\right| \geqslant 1 \Leftrightarrow \frac{1}{\left|y+\frac{1}{z}\right|} \leqslant 1$.

Possible scenarios: 1) $x=2 \Rightarrow \frac{1}{y+\frac{1}{z}}=\frac{1}{3} \Rightarrow y+\frac{1}{z}=3 \Rightarrow\left\{\begin{array}{l}y=2 \\ z=1\end{array}\right.$ or $\left\{\begin{array}{l}y=4 \\ z=-1 .\end{array}\right.$
2) $x=3 \Rightarrow \frac{1}{y+\frac{1}{z}}=-\frac{2}{3} \Rightarrow y+\frac{1}{z}=-\frac{3}{2} \Rightarrow\left\{\begin{array}{l}y=-1 \\ z=-2\end{array}\right.$ or $\left\{\begin{array}{l}y=-2 \\ z=2 .\end{array}\right.$

It remains to consider the case $\left|y+\frac{1}{z}\right|<1$, which is possible only when $\left[\begin{array}{l}y=0 \\ y=1 \\ y=-1 .\end{array}\right.$
3) $y=0-$ does not work, since $x+\frac{1}{1}=x+z$ is an integer, while $\frac{7}{3}$ is not.

$$
y+\frac{1}{z}
$$

4) $y=1$. Then $x+\frac{1}{y+\frac{1}{z}}=x+\frac{z}{z+1}=x+1-\frac{1}{z+1}=\frac{7}{3} \Rightarrow z=-4, x=1$.
5) $y=-1$. Then $x+\frac{1}{y+\frac{1}{z}}=x+\frac{z}{-z+1}=x-1+\frac{1}{1-z}=\frac{7}{3} \Rightarrow z=-2$. But then $y+\frac{1}{z}=-\frac{3}{2}$, which does not fall under the considered case.