8.3. Find all non-negative integer solutions of the equation

$$
2 x^{2}+2 x y-x+y=2020
$$
Answer: $x=0, y=2020 ; x=1, y=673$.

Solution. Rewrite the equation as

$$
y(2 x+1)=2020-2 x^{2}+x
$$

For integer values of $x$, the expression $(2 x+1) \neq 0$, so by dividing it by $2 x+1$ and isolating the integer part, we get:

$$
y=\frac{2019}{2 x+1}-x+1
$$

For $y$ to be an integer, $2 x+1$ must be a divisor of 2019. By considering the positive divisors of 2019, we examine four cases: $2 x+1=1 ; 3 ; 673 ; 2019$. Only in two of these cases do we obtain two pairs of non-negative integer solutions: $(0,2020),(1 ; 673)$.

Comments. 2 points for the answer found by trial without justification.