Solve the following equation over the set of integer pairs:

$$
x^{2}-3 y^{2}+2 x y-2 x-10 y+20=0
$$
I. solution. Let's try to write the left side as a sum or difference of complete squares. Since $2 x y$ and $-2 x$ appear, let's see what remains if we write the expression $(x+y-1)^{2}$:

$$
\begin{aligned}
x^{2}-3 y^{2}+2 x y-2 x-10 y+20 & =(x+y-1)^{2}-4 y^{2}-8 y+19= \\
& =(x+y-1)^{2}-(2 y+2)^{2}+23
\end{aligned}
$$

Thus, the equation becomes:

$$
\begin{aligned}
(x+y-1)^{2}-(2 y+2)^{2} & =-23 \\
(x+y-1+2 y+2)(x+y-1-2 y-2) & =-23 \\
(x+3 y+1)(x-y-3) & =-23
\end{aligned}
$$

The number -23 can be expressed as the product of two integers in two ways: $-23=(-1) \cdot 23=1 \cdot(-23)$.

Case I: $x+3 y+1=-1, x-y-3=23$. Subtracting the second from the first: $4 y+4=-24$, from which $y=-7$, and then $x=19$.

Case II: $x+3 y+1=1, x-y-3=-23$. From this, $4 y+4=24$, from which $y=5$ and $x=-15$ follow.

Case III: $x+3 y+1=23, x-y-3=-1$. From here, $4 y+4=24$, from which $y=5$ and $x=7$ follow.

Case IV: $x+3 y+1=-23, x-y-3=1$, from which $4 y+4=-24$, and thus $y=-7$ and $x=-3$ follow.

We performed equivalent transformations, so the four solutions are: $\left(x_{1} ; y_{1}\right)=(19 ;-7),\left(x_{2} ; y_{2}\right)=(-15 ; 5),\left(x_{3} ; y_{3}\right)=(7 ; 5)$, $\left(x_{4} ; y_{4}\right)=(-3 ;-7)$

II. solution. Write the quadratic formula for $x$:

$$
\begin{aligned}
0 & =x^{2}-3 y^{2}+2 x y-2 x-10 y+20=x^{2}+2(y-1) x-3 y^{2}-10 y+20 \\
x_{1 ; 2} & =\frac{-2(y-1) \pm \sqrt{4(y-1)^{2}-4\left(-3 y^{2}-10 y+20\right)}}{2}= \\
& =-y+1 \pm \sqrt{y^{2}-2 y+1+3 y^{2}+10 y-20}=-y+1 \pm \sqrt{4 y^{2}+8 y-19}
\end{aligned}
$$

Since $y$ is an integer, $x$ will be an integer if the expression under the square root is a perfect square: $4 y^{2}+8 y-19=n^{2}$, where $n \in \mathbb{Z}^{+}$. Further transforming this: $(2 y+2)^{2}-23=n^{2}$, or $(2 y+2)^{2}-n^{2}=23$. Since $2 y+2$ and $n$ are both integers, we need to find two perfect squares whose difference is 23. The sequence of perfect squares up to the first 13 elements is:

$0 ; 1 ; 4 ; 9 ; 16 ; 25 ; 36 ; 49 ; 64 ; 81 ; 100 ; 121 ; 144$

The difference increases after this, so the only solution (even considering non-adjacent elements) is 144 - 121 = 23. Thus, $(2 y+2)^{2}=121+23=144$, from which $2 y+2= \pm 12$.

Case I: $2 y+2=12$, from which $y=5$ and $x_{1 ; 2}=-5+1 \pm \sqrt{4 \cdot 25+8 \cdot 5-19}=4 \pm 11, x_{1}=7$ and $x_{2}=-15$.

Case II: $2 y+2=-12$, from which $y=-7$ and $x_{3 ; 4}=7+1 \pm \sqrt{4 \cdot 49-8 \cdot 7-19}=8 \pm 11, x_{3}=19$ and $x_{4}=-3$.

The solutions to the equation are: $(7 ; 5),(-15 ; 5),(19 ;-7),(-3 ;-7)$.