Question

2. Find all integers x #= 3 such that x-3Ix³-3.

Answer 1

$\dfrac {x^3-3}{x-3}=~ \dfrac {x^3-27+24}{x-3}= ~\dfrac {(x-3)(x²+3x+9)+24}{x-3}=~\underbrace{x²+3x+9}_{\text{integer}}~+~\dfrac {24}{x-3}$

So, we are looking for values of x, such that:

$x-3 ~|~24 \implies \\ \implies ~x-~3\in~\begin{Bmatrix} ±1,±2,±3,±4,±6,±8,±12,±24 \end{Bmatrix} \implies \\ \implies ~x \in ~ \begin{Bmatrix} −21,−9,−5,-3,−1,0,1,2,4,5,6,7,9,11,15,27 \end{Bmatrix}$

Corresponding x values are

$−21,−9,−5,−3,−1,0,1,2,4,5,6,7,9,11,15 ~and~ 27$