I was given the following set $A = \{ n \in \mathbb{Z} \mid \frac{n^2-n+2}{2n+1} \in \mathbb{Z}\}$
I have to list all the elements of $A$.
I started using the Euclidean division:
$$n^2-n+2=(2n+1)(\frac{1}{2}n-\frac{3}{4}) + \frac{11}{4}$$
In order to eliminate the fractions, I multiplied the whole expression with 4:
$$4(n^2-n+2)=(2n+1)(2n-3) + 11$$
I´m stuck here, maybe I´m on the wrong way. Could anyone give me some hints how can I approach this problem. Thank you in advance.
As have you have noticed, $$ 4\frac{n^2-n+2}{2n+1}= 2n-3+ \frac{11}{2n+1} $$ Therefore, if $\dfrac{n^2-n+2}{2n+1}$ is an integer, then so is $\dfrac{11}{2n+1}$.
This leaves very few candidates for $n$.
You need to check each candidate to see whether $\dfrac{n^2-n+2}{2n+1}$ is an integer.
The answer based on the hint of @lhf
$$4\frac{n^2-n+2}{2n+1}= 2n-3+ \frac{11}{2n+1}$$ $$4\frac{n^2-n+2}{2n+1} \in \mathbb{Z} \Rightarrow \frac{11}{2n+1} \in \mathbb{Z}$$ $$\Leftrightarrow 2n+1 \in D_{11} \Leftrightarrow 2n+1 \in \{-11, -1, 1 , 11\}$$ $$\Leftrightarrow 2n \in \{-12, -2, 0 , 10\}$$ $$\Leftrightarrow n \in \{-6, -1, 0 , 5\}$$
Another approach:
$$\frac{n^2-n+2}{2n+1} \in \mathbb{Z}$$ $$\Leftrightarrow n^2-n+2 \equiv 0[2n+1]$$ $$\Rightarrow 4(n^2-n+2) \equiv 0[2n+1]$$ $$\Leftrightarrow (2n+1)(2n-3) + 11 \equiv 0[2n+1]$$ $$\Rightarrow 11 \equiv 0[2n+1]$$ $$\Rightarrow (2n+1) \mid 11$$ $$\Leftrightarrow (2n+1) \in D_{11}$$
Another approach:
$$4(n^2-n+2)=(2n+1)(2n-3) + 11$$ According to the successive division: $$4(n^2-n+2)\wedge (2n+1)=(2n+1) \wedge 11$$ We have $$(\forall n \in \mathbb{Z}) 4 \wedge (2n+1)= 1$$ Since $(\forall n \in \mathbb{Z}) n^2 - n + 2 \in \mathbb{Z^*}$, then $$4 \wedge (2n+1)= 1 \Rightarrow 4(n^2-n+2)\wedge (2n+1) = (n^2-n+2)\wedge (2n+1)$$ $$\Leftrightarrow (n^2-n+2)\wedge (2n+1) = (2n+1) \wedge 11$$ We have $ \frac{n^2-n+2}{2n+1} \in \mathbb{Z}$, then $(2n+1) \mid (n^2-n+2)$ $$\Leftrightarrow (n^2-n+2)\wedge (2n+1) = \mid2n+1\mid$$ $$\Leftrightarrow (2n+1) \wedge 11 = \mid2n+1\mid$$ $$\Leftrightarrow (2n+1) \in D_{11} $$