[ equations in integers $]$

Solve the equation $9 x+2=(y+1) y$ in integers.

#
Let's rewrite the equation as $x=1 / 9(y-1)(y+2)$. It is necessary that $(y-1)(y+2)$ is divisible by 9. Since $(y+2)-(y-1)=3$, it is sufficient that $y-1$ is divisible by 3. Thus, $y=3 k+1$.

## Answer

$(k(k+1), 3 k+1), k \in \mathbf{Z}$.