1. Solve the equation in integers

$$
2 x^{2}+2 x^{2} z^{2}+z^{2}+7 y^{2}-42 y+33=0
$$
Answer: $(\mathbf{1} ; \mathbf{5} ; \mathbf{0}),(-\mathbf{1} ; \mathbf{5} ; \mathbf{0}),(\mathbf{1} ; \mathbf{1} ; \mathbf{0}),(-\mathbf{1} ; \mathbf{1} ; \mathbf{0})$.

Solution: Transform the equation to the form:

$$
2 x^{2}+2 x^{2} z^{2}+z^{2}+7(y-3)^{2}=30
$$

There are 3 cases:

1) $|y-3|=0$, 2) $|y-3|=1$, 3) $|y-3|=2$

In the first case, we get the equation $2 x^{2}+2 x^{2} z^{2}+z^{2}=30$. Solving it as a quadratic equation in $x$, we get $\left(2 x^{2}+1\right)\left(1+z^{2}\right)=31$. By trying the positive divisors of 31, we conclude that there are no integer solutions.

In the second case, we get the equation $2 x^{2}+2 x^{2} z^{2}+z^{2}=23$. Solving it as a quadratic equation in $x$, we get $\left(2 x^{2}+1\right)\left(1+z^{2}\right)=24$. By trying the positive divisors of 24, we conclude that there are no integer solutions.

In the third case, we get the equation $2 x^{2}+2 x^{2} z^{2}+z^{2}=2$. Solving it as a quadratic equation in $x$, we get $\left(2 x^{2}+1\right)\left(1+z^{2}\right)=3$. By trying the positive divisors of 3, we arrive at the solutions:

1) $x=1, z=0$
2) $x=-1, z=0$.

Recall that in this case $|y-3|=2$, and therefore, $y=5, y=1$.

Thus, we obtain 4 solutions: $(1 ; 5 ; 0),(-1 ; 5 ; 0),(1 ; 1 ; 0),(-1 ; 1 ; 0)$.