Let a, b, c be positive integers in arithmetic progression such...

Given $a, b, c$ are in arithmetic progression, we have $2b = a + c$.

Let the integer roots of the equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$.

Sum of the roots is $\alpha + \beta = -\dfrac{b}{a}$ and product of the roots is $\alpha \beta = \dfrac{c}{a}$.

Dividing the arithmetic progression condition by $a$, we get:

$\dfrac{2b}{a} = 1 + \dfrac{c}{a}$

Substituting the sum and product of the roots:

$-2(\alpha + \beta) = 1 + \alpha \beta$

$\alpha \beta + 2\alpha + 2\beta + 1 = 0$

Adding $3$ to both sides to factorize:

$\alpha \beta + 2\alpha + 2\beta + 4 = 3$

$(\alpha + 2)(\beta + 2) = 3$

Since $\alpha$ and $\beta$ are integers, $(\alpha + 2)$ and $(\beta + 2)$ must be integer factors of $3$. The possible pairs are $(1, 3)$, $(3, 1)$, $(-1, -3)$, and $(-3, -1)$.

If $(\alpha + 2, \beta + 2) = (1, 3)$ or $(3, 1)$, the roots are $-1$ and $1$. The sum of the roots is $0$, which implies $b = 0$. This is rejected because $b$ is a positive integer.

If $(\alpha + 2, \beta + 2) = (-1, -3)$ or $(-3, -1)$, the roots are $-3$ and $-5$. The sum of the roots is $-8$, which implies $-\dfrac{b}{a} = -8 \Rightarrow b = 8a$.

The product of the roots is $15$, which implies $\dfrac{c}{a} = 15 \Rightarrow c = 15a$.

Since $a$ is a positive integer, $b = 8a$ and $c = 15a$ are also positive integers. The roots of the equation are always $-3$ and $-5$.

Evaluating the given statements:

$c - b = 15a - 8a = 7a$, which is an integer multiple of $a$.

The roots are $-3$ and $-5$, both of which are odd integers.

If $c = 15$, then $15a = 15 \Rightarrow a = 1$. Consequently, $b = 8(1) = 8$. Thus, $ab = 1 \times 8 = 8$.

If $b = 8$, then $8a = 8 \Rightarrow a = 1$. The roots are $-3$ and $-5$, so $x = 3$ is not a root.

Answer: $c - b$ is an integer multiple of $a$; Both the roots of the equation $ax^2 + bx + c = 0$ are odd integers; If $c = 15$, then $ab = 8$