2. Find all pairs of natural numbers $x, y$ that satisfy the equation

$$
6 x^{2} y+2 x^{2}+3 x y+x-9 y=2016
$$
# Problem 2.

Answer: $x=4, y=20$.

Solution. Factoring the left side of the equation, we get

$$
(3 y+1)(2 x+3)(x-1)=2013
$$

Since $2013=33 \cdot 11 \cdot 61$ and the number $3 y+1$ has a remainder of 1 when divided by 3, then $3 y+1=61 \Longleftrightarrow y=20$. Since $2 x+3>x-1$, then $x-1=3 ; 2 x+3=11$, i.e., $x=4$.