10.5. How many solutions in integers \(x, y\) does the equation \(6 x^{2}+2 x y+y+x=2019\) have?
Answer. 4 solutions. Solution. Express $y$ from the given equation: $y=\frac{2019-6 x^{2}-x}{2 x+1}$. Dividing $6 x^{2}+x$ by $2 x+1$ with a remainder, we get $6 x^{2}+x=(3 x-1)(2 x+1)+1$. Thus, the expression for $y$ will take the form: $y=\frac{2019-(3 x-1)(2 x+1)-1}{2 x+1}=\frac{2018}{2 x+1}-3 x+1$. Therefore, for $y$ to be an integer, 2018 must be divisible by $2 x+1$. Since $2 x+1$ is an odd number, and 2018:2=1009 is a prime number, $2 x+1$ can take 4 values: $1, -1, 1009, -1009$. Then the corresponding values of $x$ and $y$ are: (0;2019), (-1;-2014), (504;-1509), (-505;1514).