5. Solve the equation in integers: $5 x^{2}-2 x y+2 y^{2}-2 x-2 y=3$. In the answer, write the sum of all solutions $(x, y)$.

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5. Transforming the equation, we get: $(x+y)^{2}-2(x+y)+1+y^{2}-4 x y+4 x^{2}=4 \Leftrightarrow(x+y-1)^{2}+(y-2 x)^{2}=4$. Since $x$ and $y$ are integers, this equality is possible only in four cases:
$\left\{\begin{array}{c}x+y-1=0 \\ y-2 x=2\end{array}, \quad\left\{\begin{array}{c}x+y-1=2 \\ y-2 x=0\end{array}, \quad\left\{\begin{array}{c}x+y-1=0 \\ y-2 x=-2\end{array}, \quad\left\{\begin{array}{c}x+y-1=-2 \\ y-2 x=0\end{array}\right.\right.\right.\right.$

The first and fourth systems do not have integer solutions. The second system has the integer solution $x=1, y=2$, the third system has the solution $x=1, y=0$. $\underline{\text { Answer: }}\{4\}$