For what integral values of $a$ does $x^2-x(1-a) - (a+2)=0$ have integral roots?

Question

For what integral values of $a$ does the equation $$x^2-x(1-a) - (a+2)=0$$ have integral roots?

I have tried making the discriminant to be a perfect square; but when it becomes perfect square, $a$ itself is not an integer.

Answer 1

The discriminat of this quadratic equation $$(1-a)^2+4(a+2) = a^2+2a+9$$ must be a perfect square.

So $$(a+1)^2+8 = b^2\implies \boxed{(b-a-1)(b+a+1)=8}$$

Answer 2

Write $$ a={x^2-x-2\over 1-x}$$

Let $y=1-x$, so $x=1-y$ and now we have $$a={1-2y+y^2-1+y-2\over y} = y-1-{2\over y}$$

So $y\mid 2\implies y\in \{-2,-1,1,2\}$ so $a\in \{...\}$.