Integer Roots of $x(a-1)-x(a^2-1)=a^2+a$

Question

If $a$ is an integer, such that $a\neq0$, when does the following equation have integer roots? $$x(a-1)-x(a^2-1)=a^2+a$$

Answer 1

Note that $$x(a-1)-x(a^2-1)=a^2+a \implies x = \frac{1+a}{1-a} = -1+\frac{2}{1-a},$$ where $a \neq 0$, and $x$ is an integer. Since $x$ can be integer only if $(1-a)|2$, we have $1-a = \pm 1, \pm 2$. That is, $a=-1,2,3$.

Answer 2

$$x=\dfrac{1+a}{1-a}$$

For integers avoiding fractions we should have $a=(-1,0,1,2,3).$

Answer 3

Factor the LHS RHS as $x((a-1)-(a^2-1))=x(a-a^2)=(ax(1-a))$ and $a(1+a)$ respectively to get $$ax(1-a)=a(a+1) \longrightarrow x=\frac{a(1+a)}{a(1-a)}=x=\frac{1+a}{1-a}=\frac{2}{1-a}-1$$ Which is integral for $x=\{-1, 0, 2, 3\}$.