Prove that if $a$ and $b$ are integers with $a\not= 0$ and $x$ is a positive integer such that $ax^2 + bx + b − a = 0$, then $a|b$. Prove that if $a$ and $b$ are integers with $a\not= 0$ and $x$ is a positive integer such that $ax^2 + bx + b − a = 0$, then $a|b$.
I will use the backwards and forwards method to prove the proposition. A represents the hypothesis, and B represents the conclusion.
 A: $a, b$ are integers with $a \not= 0$ and $x$ is a positive integer such that $ax^2 + bx + b − a = 0$.
B: $a|b$ ($b$ is divisible by $a$)
B1: 
How can I show that one integer (namely, $b$) is divisible by another (namely, $a$)?
Show that there is an integer $k$ such that $\dfrac{b}{a} = k$.
A1: $ax^2 + bx + b − a = 0$ (A)
$\therefore x = \dfrac{ -b \pm \sqrt{b^2 - 4a(b-a)} }{ 2a }$
$ \implies x = \dfrac{ -b \pm \sqrt{b^2 - 4ab + 4a^2} }{ 2a }$
$b^2 - 4ab + 4a^2 = (2a - b)^2$
$ \therefore x = \dfrac{ -b \pm \sqrt{(2a - b)^2} }{ 2a }$
$ \implies x = \dfrac{ -b \pm (2a - b) }{ 2a }$
$ \implies x = \dfrac{ 2a - 2b }{ 2a }$
$ = \dfrac{ a - b }{ a }$
$ = 1 - \dfrac{ b }{ a }$
$\implies \dfrac{b}{a} = 1 - x$ where $x$ is a positive integer (A)
$\therefore \dfrac{b}{a} = k$ where $k$ is an integer (B1).
$Q.E.D.$
I would greatly appreciate it if the members of MSE could please take the time to review my proof.
A: Simpler: $(x+1) b = (1-x^2) a = (x+1)(1-x) a$ so (since $x \ne -1$) $b = (1-x) a$ is divisible by $a$.
A: Reorder 
$$ax^2-a=-bx-b $$
$$a (x-1)(x+1)=-b (x+1) $$
All are integers 
$$ a (x-1)= -b $$
Hence $a|b $
A: Alternatively, one can approach the problem as follows:
Rewrite the equation: $ax^2-a = -b(x+1)\implies a(x+1)(x-1) = -b(x+1)\implies b = -a(x-1)\implies a \mid b$
