844. Solve the equation in natural numbers

$$
x^{2}+x y+y^{2}=x+y+9
$$
$\triangle$ Let's gather all terms on the left side and multiply the resulting equation by 2, highlighting three squares of sums in the left part:

$$
\begin{gathered}
2 x^{2}+2 x y+2 y^{2}-2 x-2 y-18=0 \\
\left(x^{2}+2 x y+y^{2}\right)+\left(x^{2}-2 x+1\right)+\left(y^{2}-2 y+1\right)-2-18=0
\end{gathered}
$$

$$
(x+y)^{2}+(x-1)^{2}+(y-1)^{2}=20
$$

The sum of three perfect squares equals 20 in only one case: $16+4+0=20$. Several possibilities arise depending on which square is 16, which is 4, and which is 0. Try them out yourself.

In $\S 18$ (section 18.3), when solving similar equations with two unknowns in integers, we used other methods, for example, using the discriminant of a quadratic equation.

Answer: $(3 ; 1),(1 ; 3)$