2. Find all integer solutions of the inequality

$$
x^{2} y^{2}+y^{2} z^{2}+x^{2}+z^{2}-38(x y+z)-40(y z+x)+4 x y z+761 \leq 0
$$
# Solution:

$\left(x^{2} y^{2}+2 x y z+z^{2}\right)+\left(y^{2} z^{2}+2 x y z+x^{2}\right)-38(x y+z)-40(y z+x)+761 \leq 0$.

$(x y+z)^{2}+(y z+x)^{2}-38(x y+z)-40(y z+x)+761 \leq 0$,

$(x y+z)^{2}-2 \cdot 19(x y+z)+361-361+(y z+x)^{2}-2 \cdot 20(y z+x)+400-400+761 \leq 0$,

$(x y+z-19)^{2}+(y z+x-20)^{2} \leq 0,\left\{\begin{array}{l}x y+z=19, \\ y z+x=20,\end{array} \Leftrightarrow\left\{\begin{array}{c}x y+z=19, \\ y(z-x)-(z-x)=20,\end{array} \Leftrightarrow\left\{\begin{array}{c}x y+z=19, \\ (z-x)(y-1)=1 .\end{array}\right.\right.\right.$

Since we need to find integer solutions to the inequality, there are only two cases:

1) $\left\{\begin{array}{c}x y+z=19, \\ z-x=1, \\ y=2,\end{array} \Leftrightarrow\left\{\begin{array}{c}2 x+x+1=19, \\ z=x+1, \\ y=2,\end{array} \Leftrightarrow\left\{\begin{array}{l}x=6, \\ y=2, \\ z=7 .\end{array}\right.\right.\right.$
2) $\left\{\begin{array}{c}x y+z=19, \\ z-x=-1, \\ y=0,\end{array} \Leftrightarrow\left\{\begin{array}{c}x-1=19, \\ z=x-1, \\ y=0,\end{array} \Leftrightarrow\left\{\begin{array}{c}x=20, \\ y=0, \\ z=19 .\end{array}\right.\right.\right.$

Answer: $(6 ; 2 ; 7),(20 ; 0 ; 19)$.