3. Solve the equation in integers: $x^{2}+y^{2}=x+y+2$.

Answer: ( $-1 ; 1),(-1 ; 0),(0 ;-1),(0 ; 2),(1 ;-1),(1 ; 2),(2 ; 1),(2 ; 0$).
Solution. The equation is equivalent to the following: $x^{2}-x=-y^{2}+y+2$. The function $f(x) = x^{2}-x=(x-0.5)^{2}-0.25$ takes values in the interval $[-0.25 ;+\infty)$, $g(y)=-y^{2}+y+2=-(y-0.5)^{2}+2.25-$ on the interval $[2.25 ;-\infty)$, that is, they have only three common integer values: $0, 1, 2$. Since $x$ and $y$ are integers, $f(x)$ and $g(y)$ can only take even values, so the common value 1 is excluded. The equation $f(x)=0$ has roots 0 and 1, and the equation $g(y)=0$ has roots -1 and 2. This gives 4 pairs of solutions. The equation $f(x)=2$ has roots -1 and 2, and the equation $g(y)=2$ has roots 1 and 0. This gives another 4 pairs of solutions, all of which, obviously, are valid.

Criteria. If the solution is incorrect or only the answer without explanation - 0 points.

If the reasoning is correct and there is a computational error, or one answer is lost - 3 points.

If the solution is correct - 7 points.

Criteria. If the solution is incorrect - 0 points.

If the reasoning is correct and there is a computational error - 3 points. If the solution is correct - 7 points.