# Task 6.

$$
\begin{gathered}
4 x^{2}=y^{2}+2 y+1+3 \\
(2 x)^{2}-(y+1)^{2}=3 \\
(2 x-y-1) \cdot(2 x+y+1)=3
\end{gathered}
$$

Since $x, y$ are integers, we get:

1) $\left\{\begin{array}{l}2 x-y-1=1 \\ 2 x+y+1=3\end{array}\right.$
2) $\left\{\begin{array}{l}2 x-y-1=3 \\ 2 x+y+1=1\end{array}\right.$
3) $\left\{\begin{array}{l}2 x-y-1=-3 \\ 2 x+y+1=-1\end{array}\right.$
4) $\left\{\begin{array}{l}2 x-y-1=-1 \\ 2 x+y+1=-3\end{array}\right.$

From the first system: $x=1, y=0$

From the second system: $x=1, y=-2$

From the third system: $x=-1, y=0$

From the fourth system: $x=-1, y=-2$
Answer: $\{(1,0),(1,-2),(-1,0),(-1,-2)\}$.