<s>[INST] Number Theory : Find the group of A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ Find the group A such that  $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ ?
Polynomial Long Division we get
$\frac{x^{2}}{2}-\frac{x}{4}-\frac{11}{8}+\frac{27}{8\left(2x+1\right)}$
but how i can from here find all x such that $x \in \mathbb{Z}$ ??
i did use the hint for $y = 2x+1$ so that  $x =\frac{y-1}{2}$
made the equation to be $\frac{(y-1)^3}{8}-\frac{3(y-1)}{2}+2=\frac{y^3-3y^2-9y+27}{8}$
$\frac{\frac{y^3-3y^2-9y+27}{8}}{y}=\frac{y^3-3y^2-9y+27}{8y}$
I can not understand with the help of the clues, can some 1 some formal proof ?
 [/INST] Hint: Since $2x+1$ is odd for all integers $x$, we have that $\frac{x^3-3x+2}{2x+1}\in\mathbb Z$ if and only if $\frac{2(x^3-3x+2)}{2x+1}\in\mathbb Z$.
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