1. Solve the equation in natural numbers

$$
2 y^{2}-x y-x^{2}+2 y+7 x-84=0
$$
Answer: $(1 ; 6),(14 ; 13)$.

Solution: The original equation can be represented as

$$
2 y^{2}-x y-x^{2}+2 y+7 x-12=72
$$

Next, we factorize the left side of the equation using the discriminant:

$$
(x+2 y-4)(y-x+3)=72
$$

Since $72=1 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3$, we can reduce the number of cases to check.

Let $x+2 y-4=a, y-x+3=b$. Then $a \cdot b=72$ and $y=\frac{a+b+1}{3}$ and $y \in \mathbb{N}$.

Therefore, $a$ and $b$ cannot be negative, and neither of them should be divisible by 3, but one of them must be divisible by 9. Thus, we reduce the number of cases to:

1) $\left\{\begin{array}{l}a=9, \\ b=8 .\end{array}\right.$ 2) $\left\{\begin{array}{l}a=8, \\ b=9 .\end{array}\right.$, $)\left(\begin{array}{l}a=18, \\ b=4 .\end{array}\right.$ 4) $\left\{\begin{array}{l}a=4, \\ b=18 .\end{array}\right.$ ) $)\left\{\begin{array}{l}a=36, \\ b=2 .\end{array}\right.$, $\left\{\begin{array}{l}a=2, \\ b=36 .\end{array}\right.$ 7) $\left\{\begin{array}{c}a=72, \\ b=1 .\end{array},\left\{\begin{array}{l}a=1, \\ b=72 .\end{array}\right.\right.$ Considering $y=\frac{a+b+1}{3}, y \in \mathbb{N}$, we find that only two cases fit:

$$
\text { 1) }\left\{\begin{array} { l } 
{ a = 9 , } \\
{ b = 8 }
\end{array} \text { 2) } \left\{\begin{array}{c}
a=36 \\
b=2
\end{array}\right.\right.
$$

From this, we obtain two solutions: $(\mathbf{1} ; \mathbf{6}),(\mathbf{1 4} ; \mathbf{1 3})$.