Integer Solutions of Quadratic Equation with Coefficients in Arithmetic Progression

Given that $a, b, c$ are positive integers in an arithmetic progression (AP), we have: $2b = a + c \implies c = 2b - a$

Let $\alpha$ and $\beta$ be the integer roots of the quadratic equation $ax^2 + bx + c = 0$. By Vieta's formulas: $\alpha + \beta = -\frac{b}{a} \implies b = -a(\alpha + \beta)$ $\alpha \beta = \frac{c}{a} \implies c = a\alpha\beta$

Substituting the expressions for $b$ and $c$ into the AP relationship $c = 2b - a$: $a\alpha\beta = 2(-a(\alpha + \beta)) - a$

Since $a$ is a positive integer ($a \neq 0$), we can divide both sides by $a$: $\alpha\beta = -2(\alpha + \beta) - 1$ $\alpha\beta + 2\alpha + 2\beta + 1 = 0$

Adding $3$ to both sides allows us to factor the expression: $(\alpha + 2)(\beta + 2) = 3$

Since $\alpha$ and $\beta$ are integers, $(\alpha + 2)$ and $(\beta + 2)$ must be integer factors of $3$. The possible pairs for $(\alpha + 2, \beta + 2)$ are:

1. $(1, 3) \implies \alpha = -1, \beta = 1$
2. $(3, 1) \implies \alpha = 1, \beta = -1$
3. $(-1, -3) \implies \alpha = -3, \beta = -5$
4. $(-3, -1) \implies \alpha = -5, \beta = -3$

Now, let's analyze these cases:

• Case 1: $\{\alpha, \beta\} = \{-1, 1\}$ $\alpha + \beta = 0 \implies b = -a(0) = 0$ This contradicts the condition that $b$ is a positive integer ($b > 0$). Thus, this case is invalid.

• Case 2: $\{\alpha, \beta\} = \{-3, -5\}$ $\alpha + \beta = -8 \implies b = 8a$ $\alpha\beta = 15 \implies c = 15a$ For any positive integer $a$, $b = 8a$ and $c = 15a$ are positive integers, and they satisfy $2b = 16a = a + 15a = a + c$. Hence, the roots of the equation must be $\alpha = -3$ and $\beta = -5$.

Now we evaluate the given options:

• Option (A): $c - b = 15a - 8a = 7a$ Since $a$ is an integer, $c - b$ is an integer multiple of $a$. Therefore, Option (A) is TRUE.

• Option (B): The roots are $-3$ and $-5$, both of which are odd integers. Therefore, Option (B) is TRUE.

• Option (C): If $c = 15$, then $15a = 15 \implies a = 1$. Then $b = 8a = 8(1) = 8$. Thus, $ab = 1 \cdot 8 = 8$. Therefore, Option (C) is TRUE.

• Option (D): If $b = 8$, then $8a = 8 \implies a = 1$, which gives $c = 15$. The roots are $x = -3$ and $x = -5$. Thus, $x = 3$ is not a root of the equation. Therefore, Option (D) is FALSE.

Hence, the correct statements are (A), (B), and (C).