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

Correct Options:
1. c - b is an integer multiple of a
2. Both the roots of the equation $a{x}^2+bx+c=0$ are odd integers
3. If c = 15, then ab = 8

Step-by-step Explanation:

Let $a$, $b$, and $c$ be positive integers in arithmetic progression (A.P.).
Since $\mathrm{a,\; b,\; c}$ are in A.P., we have:

$2b=a+c$

Let the roots of the quadratic equation $a{x}^2+bx+c=0$ be integer roots $\alpha$ and $\beta$.
By Vieta's formulas for sum and product of roots:

$\alpha +\beta =-\frac{b}{a}$

$\alpha \beta =\frac{c}{a}$

Substituting $b=\frac{a+c}{2}$ into the sum of roots equation:

$\alpha +\beta =-\frac{a+c}{2a}=-\frac{1}{2}-\frac{c}{2a}=-\frac{1}{2}-\frac{1}{2}\alpha \beta$

Multiplying both sides by 2:

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

Rearranging the terms:

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

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 integer factor pairs of 3 are:
Case 1: $(\alpha +2)=1$ and $(\beta +2)=3$ ⇒ $\alpha =-1,\beta =1$
Case 2: $(\alpha +2)=-1$ and $(\beta +2)=-3$ ⇒ $\alpha =-3,\beta =-5$

Since $\mathrm{a,\; b,\; c}$ are positive integers, the product of roots $\alpha \beta =\frac{c}{a}$ must be positive.
For Case 1 ($\alpha =-1,\beta =1$), $\alpha \beta =-1<0$, which would make $c/a$ negative (impossible since $a>0,c>0$).
Therefore, Case 2 must be true:

$\alpha =-3$ and $\beta =-5$

Thus, both roots are odd integers ($-3$ and $-5$), which proves Statement 2 is TRUE.

Now, using these root values:

$\alpha +\beta =-8=-\frac{b}{a} \Rightarrow b=8a$

$\alpha \beta =15=\frac{c}{a} \Rightarrow c=15a$

Let us evaluate each statement:

Evaluating Statement 1:
$c-b=15a-8a=7a$
Since $7a$ is clearly an integer multiple of $a$, Statement 1 is TRUE.

Evaluating Statement 3:
If $c=15$, then since $c=15a$, we have $a=1$.
Then $b=8a=8$.
So $ab=1\times 8=8$.
Thus, Statement 3 is TRUE.

Evaluating Statement 4:
If $b=8$, then $a=1$ and the roots are $x=-3$ and $x=-5$ (not $x=3$).
Thus, Statement 4 is FALSE.