A number theory problem by Thành Đạt Lê

We are given that $2n+1|2n^2-n+2 \quad n\in\mathbb{Z}$

In other words:

$\dfrac{2n^2-n+2}{2n+1}\in\mathbb{Z}$

Let's simplify the fraction

$\begin{aligned} \dfrac{2n^2-n+2}{2n+1}&=\dfrac{\mathbin{\color{#D61F06}2n^2+n}-2n+2}{2n+1}\\ &=\mathbin{\color{#D61F06}n}+\dfrac{-2n+2}{2n+1}\\ &=n+\dfrac{\mathbin{\color{#3D99F6} -2n-1}+3}{2n+1}\\ &=n\mathbin{\color{#3D99F6}-1}+\dfrac{3}{2n+1} \end{aligned}$

So we need to have $2n+1|3$ . The smallest divisor of $3$ is $-3$

$\Longrightarrow 2n+1=-3 \iff n=\boxed{-2}$