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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?

VereX
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    If $x$ and $y$ are integers, then $xy$ is an integer. So $\frac{1}{a}$ must be an integer (note that your case of $a = 0$ results in a linear function, which we have to consider separately). This gives us only 2 cases to work with in the quadratic case; you can verify both by hand. – platty Aug 07 '17 at 21:21

2 Answers2

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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, $$a\in\{0,1\} $$

hamam_Abdallah
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a = 1, 0 , - 1/7

if c and b be the roots, then c + b = - 1 - 1/a and cb = 1 - 1/a

so, c + b + 1 = cb - 1 or, ( c - 1 ) ( b - 1 ) = 3 which is 1 x 3 or - 1 x - 3 (commuting will yield the same pair).

mathemanics
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