Integer roots

Since the positive integers $1,a,b,c$ follows an arithmetic progression , then $a-1 = b-a = c-b$ . Solving this equation in terms of $a$ gives $b = 2a-1, c = 3a-2$ .

We want to find the maximum number of integer roots (in $x$ ) of $0 =x^3 + ax^2 + (2a-1)x + (3a-2) = a(x^2 + 2x + 3) + (x^3 - x- 2) .$

Since $a$ is an integer, then $\dfrac{ x^3 - x-2}{x^2 + 2x+3} = \dfrac4{x^2 + 2x+3} + (x-2)$ is also integer.

Thus, $x^2 + 2x + 3 = (x+1)^2 + 2$ must divides 4. So, the possible values of $(x+1)^2 + 2$ are $\pm 1, \pm 2, \pm 4$ . Trial and error shows that $x = -1$ is the only solution (with $a=b=c=1$ ). Our answer is $\boxed1$ .

The common difference of the coefficients is $a-1$ , so we have $b=2a-1$ and $c=3a-2$ .

First we are going to show that not all of the roots are integers. Suppose on the contrary the opposite. Clearly the polynomial does not have nonnegative roots (since we would get $x^3+ax^2+bx+c \geq c > 0$ ), so the roots are $-x_1$ , $-x_2$ and $-x_3$ where $x_1$ , $x_2$ and $x_3$ are positive integers.

By Vieta's formulas we obtain the following equations:

$x_1 + x_2 + x_3 = a$

$x_1x_2 + x_1x_3 + x_2x_3 = 2a-1$

$x_1x_2x_3 = 3a-2$

From the last equation it can be seen that the product of the roots gives a remainder of 1 divided by 3, which is only possible if all of the roots are 1 mod 3, or two of them are 2 mod 3 and the third is 1 mod 3. This yields us two cases.

1. $x_1, x_2, x_3 \equiv 1 \pmod{3}$ In this case we would have $x_1 + x_2 + x_3 = a$ and $x_1x_2 + x_1x_3 + x_2x_3 = 2a-1$ to be divisible by 3 which is a contradiction.

2. $x_1 \equiv 1 \pmod{3}$ and $x_2, x_3 \equiv 2 \pmod{3}$ From this we would obtain $x_1x_2 + x_1x_3 + x_2x_3 = 2a-1$ to give a remainder of $2 + 2 + 4 \equiv 2 \pmod{3}$ which implies $3|a$ , but at the same time $a = x_1 + x_2 + x_3 \equiv 2 \pmod{3}$ , so this is also a contradiction. Thus the answer to the question can not be 3.

The assumption that exactly 2 of the roots are integers would lead us to the fact that $a = x_1 + x_2 + x_3$ is not an integer, yielding a contradiction. This means that the answer is at most 1.

One can easily see that 1 is possible, for example considering the equation below with root -1:

$x^3+x^2+x+1=0$

Thus the answer is 1.