Let $a$, $b$, $c$ be positive integers in arithmetic progression such that the equation $ax^{2} + bx + c = 0$ has only integer solutions. Then which of the following statements is (are) TRUE ?

$a,\ b,\ c\ \text{positive integers in AP} \implies 2b = a + c$ $ax^{2}+bx+c=0\ \text{has only integer roots}$ $(1)\ c-b\ \text{multiple of } a,\qquad (2)\ \text{both roots odd integers}$ $(3)\ c=15 \implies ab=8,\qquad (4)\ b=8 \implies x=3\ \text{root}$ $1.\ \text{roots} = -m, -n \quad (m, n \in \mathbb{Z}^{+})$ $2.\ ax^2+bx+c = a(x+m)(x+n) \Rightarrow b,\ c$ $3.\ \text{apply AP, then check all options}$ $x \ge 0:\ ax^2+bx+c > 0 \Rightarrow \text{no non-negative root}$ $x_1 x_2 = \dfrac{c}{a} > 0, \quad x_1 + x_2 = -\dfrac{b}{a} < 0$ $\therefore x_1, x_2 \text{ both negative}$ $\text{Let roots be } -m,\ -n, \quad m, n \in \mathbb{Z}^{+}$ $ax^2+bx+c = a(x+m)(x+n)$ $= ax^2 + a(m+n)x + amn$ $\Rightarrow b = a(m+n), \quad c = amn$ $c - b = b - a$ $amn - a(m+n) = a(m+n) - a$ $mn - (m+n) = (m+n) - 1$ $\Rightarrow mn - 2m - 2n + 1 = 0$ $mn - 2m - 2n + 4 = 3$ $(m-2)(n-2) = 3$ $(m-2, n-2) = (1, 3) \ \text{or}\ (3, 1)$ $\Rightarrow \{m, n\} = \{3, 5\}$ $\text{roots} = -3, -5$ $b = a(3+5) = 8a, \quad c = a\cdot 3\cdot 5 = 15a$ $(a, b, c) = (a, 8a, 15a)$ $(1)\ c-b = 15a - 8a = 7a \Rightarrow \text{multiple of } a \ \checkmark$ $(2)\ \text{roots } -3, -5 \text{ are both odd integers} \ \checkmark$ $(3)\ c = 15 \Rightarrow 15a = 15 \Rightarrow a = 1$ $b = 8, \quad ab = 8 \ \checkmark$ $(4)\ b = 8 \Rightarrow 8a = 8 \Rightarrow a = 1, \quad c = 15$ $x^2+8x+15 = (x+3)(x+5); \quad x = 3:\ 48 \neq 0 \ \times$ $\boxed{(1),\ (2),\ (3) \text{ are TRUE}}$