KVPY 2016 SA Question 18

If the positive integer coefficients $a, b, c$ are in arithmetic progression, then we have $b = a + \delta, c = a + 2\delta$ . Taking $P(x) = 0$ with integer roots $\alpha, \beta$ produces:

$P(x) = x^2 + \frac{b}{a}x + \frac{c}{a} = (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha\beta = 0;$

or $\alpha + \beta = -\frac{b}{a} = -\frac{a + \delta}{a}$ and $\alpha\beta = \frac{c}{a} = \frac{a + 2\delta}{a}.$

The expression $\alpha + \beta + \alpha\beta$ computes to: $-\frac{a + \delta}{a} + \frac{a + 2\delta}{a} = \frac{\delta}{a}.$ Taking each answer choice and substituting it back into $P(x) = x^2 + \frac{a+\delta}{a}x + \frac{a+2\delta}{a} = x^2 + (1 + \frac{\delta}{a})x + (1 + \frac{2\delta}{a}) = 0$ gives:

$\frac{\delta}{a} = 3$ , or $x^2 + 4x + 7 = 0 \Rightarrow x = -2 \pm \sqrt{3}i;$

$\frac{\delta}{a} = 5$ , or $x^2 + 6x + 11 = 0 \Rightarrow x = -3 \pm \sqrt{2}i;$

$\frac{\delta}{a} = 7$ , or $x^2 + 8x + 15 = 0 \Rightarrow x = -3, -5;$

$\frac{\delta}{a} = 14$ , or $x^2 + 15x + 29 = 0 \Rightarrow x = \frac{-15 \pm \sqrt{109}}{2}$

Hence, choice C is correct.

Since $a$ , $b$ , $c$ are in arithmetic progression $\implies 2b=a+c$

$\begin{aligned} 2b&=a+c \\ 2 \dfrac{b}{a}&=1+\dfrac{c}{a} & \small \color{#3D99F6}{ \text{ divided by } a} \\ 2(- (\alpha + \beta )) &= 1+ \alpha \beta &\small \color{#3D99F6}{ \text{ by Vieta} } \\ -2 \alpha -2 \beta &= 1 + \alpha \beta \\ \alpha \beta +2 \alpha +2 \beta + 1&=0 \\ ( \alpha +2) ( \beta+2) &= 3 \implies ( \alpha, \beta)=(1,-1),(-1,1),(-3,-5),(-5,-3) \end{aligned}$

$(1,-1),(-1,1)$ is rejected because $c= \alpha \beta$ must be a positive integer.

Putting $( \alpha, \beta)=(-3,-5),(-5,-3)$ to $\alpha +\beta +\alpha \beta$ equals to $\boxed{7}$