Find all real values of 'a' for which all the function roots are integers. $$ f(x) = ax^{2} + (a+1)x + a-1$$
I was thinking about Vieta's formula so: $$ \\xy = 1 - \frac{1}{a} \\x + y = -1 - \frac{1}{a} \\xy ∈ \Bbb Z \\x+y ∈ \Bbb Z. \\a ∈ \Bbb R $$
After all of that my answer is $$ a = 0, a = 1 $$
But I'm not sure, how should I prove that those are the only options?
Assume $a\ne 0$.
Sum $S $ of roots is then integer.
$$S=-\frac{a+1}{a}=-1-\frac {1}{a} \in \Bbb Z$$
$\implies a=1$ or $a=-1$.
for $a=1$, roots are $0,-2$
for $a=-1$ , no root.
If $a=0$, the root is $x=1$.
Finally, $$a\in\{0,1\} $$