Divisibility in polynomial equations Let $p$ be a prime number. Let $a, b$, and $c$ be integers that are divisible by p such that the equation $x^3 + ax^2 + bx + c = 0$ has at least two diﬀerent integer roots. Prove that c is divisible by $p^3$.
I have the solution, but I don't get parts of it
It said that: Let r and s be two diﬀerent integral roots of the equation
$r^3 + ar^2 +br +c=0$ 
and
$s^3 + as^2 + bs +c =0$ because of the remainder theorem
It then said that since a, b, and c are divisible by p, then $r^3$ and $s^3$ are divisible by p, I don't quite the reasoning behind this, can someone provide an explanation.
 A: Here is the reasoning. Consider $r^3 + ar^2 +br +c=0$ . The RHS, $0$ is divisible by $p$ so the LHS must also be divisible by $p$. Now in the LHS, the terms $ar^2$, $br$ and $c$ are all divisible by $p$ since $a$, $b$ and $c$ are all divisible by $p$. Hence for the LHS to be divisible by $p$ the remaining term $r^3$ must also be divisible by $p$. Similarly you can show that $s^3$ is divisible by $p$.
A: Alternative solution. 
First we see that if $x_1,x_2\in\mathbb{Z}$ then by Vieta formula we have $$x_3 = -a-x_1-b_2\implies x_3\in \mathbb{Z}$$
Now, if $x_i$ is a solution to a given equation we have $$x_i^3 = -x_i^2a-x_ib-c = p\cdot\underbrace{(...)}_{\in \mathbb{Z}} \implies p\mid x_i$$
So, again by third Vieta formula we have $$c =-x_1x_2x_3 = p^3\cdot (...)$$
and we are done.
A: $$r^3+a'pr^2+b'pr+c'p=0\iff r^3=-p(a'r^2+b'r+c')$$
and $p$ divides $r^3$.
A: If the polynomial has at least two solutions, then it can be factored as follows, where $\alpha$, $\beta$, $\gamma$ are numbers (that are not necessarily unique). Also, let's write $a = pa'$, $b = pb'$, and $c = cp'$ where $a',b',c' \in \mathbb Z$.
$$x^3 + pa'x^2 + pb'x + pc' = (x + \alpha)(x + \beta)(x + \gamma)$$
$$x^3 + pa'x^2 + pb'x + pc' = x^3 + (\alpha + \beta + \gamma)x^2 + (\alpha \beta + \alpha \gamma + \beta \gamma) x + \alpha \beta \gamma$$
This gives a system of equations.
$$a' = \frac{\alpha + \beta + \gamma}{p}$$
$$b' = \frac{\alpha \beta + \alpha \gamma + \beta \gamma}{p}$$
$$c' = \frac{\alpha \beta \gamma}{p} \tag{1}$$
Equation (1) implies that one of the $\alpha, \beta, \gamma$ must be divisible by $p$. Without loss of generality, say $\alpha = p\alpha'$ where $\alpha' \in \mathbb Z$. The three equations now read as follows.
$$a' = a' + \frac{\beta + \gamma}{p}$$
$$b' = \alpha' \beta + \alpha' \gamma + \frac{\beta \gamma}{p} \tag 2$$
$$c' = \alpha' \beta \gamma$$
Equation (2) implies that one of the $\beta$, $\gamma$ must be divisible by $p$. Again, without loss of generality, write $\beta = p\beta'$ where $\beta' \in \mathbb Z$.
$$a' = a' + \beta' + \frac{\gamma}{p} \tag 3$$
$$b' = \alpha' \beta + \alpha' \gamma + \beta' \gamma$$
$$c' = \alpha' \beta \gamma$$
Equation (3) implies $p|\gamma$. Thus, we can write $\gamma = p\gamma'$ for some $\gamma' \in \mathbb Z$. Therefore,
$$c = pc' = \alpha \beta \gamma = p^3 \alpha' \beta' \gamma',$$
which implies $p^3 | c$.
