What are the roots of this equation? I have a quadratic equation $ ax^2 +bx+c =0 $, where $ a,b,c $ all are positives and are in Arithmetic Progression. 
Also, the roots $\alpha$ and $\beta$ are integers.
I need to find out $ \alpha + \beta + \alpha\beta $.
I have tried taking $ a = a' - d , b = a' , c = a' + d $ because I have supposed $ a' $ is the first term and $ d $ is common difference, I used sum of roots and product of roots rule, but nothing helped..
 A: We see that the equation $$ax^2+(a+d)x+a+2d=0$$
has integer roots, which says $1+\frac{d}{a}\in\mathbb Z$.
Let $\frac{d}{a}=k$.
Hence, we have $$x^2+(1+k)x+1+2k=0$$
has integer roots, which gives
$$(1+k)^2-4(1+2k)=n^2,$$ where $n$ is non-negative integer number,
which gives
$$k^2-6k-3=n^2$$ or
$$(k-3)^2=n^2+12$$ or
$$(k-3-n)(k-3+n)=12$$ and the rest is smooth:
Since $n$ is non-negative integer, we obtain two cases only.


*

*$n-k+3=6$ and $n+k-3=-2$, which gives $k=-1$;

*$n-k+3=-2$ and $n+k-3=6$, which gives $k=7$.
A: If the roots are integers, we can write $x^2 + bx + c = (x-\alpha)(x-\beta)$. Note that it's okay to take $a=1$, since $b-a = c-b \implies \frac{b-a}{a} = \frac{c-b}{a}$, i.e. the arithmetic progression is preserved.
Now write $c = b + (b-1) = 2b-1$.
Use Vieta's: $\alpha + \beta = -b$ and $\alpha\beta  = 2b-1$.
$\alpha = \frac{-b\pm \sqrt{b^2 - 8b +4}}{2}$. But $\alpha$ is an integer, so $(b-4)^2 -12$ is a perfect square means that we must have $b = 8$ (can be checked manually by trying squares up to 6, since difference between consecutive squares is $2n+1$). Then $\alpha + \beta + \alpha\beta = -b + (2b-1) = b - 1 = 7$.
