Constant Term of a Monic Polynomial in $\mathbb{Z}[x]$ Is Divisible by 3 
Given a monic polynomial $f(x)$ in $\mathbb{Z}[x]$ such that $\alpha$ and $3 \alpha$ are complex roots of $f(x),$ prove that the constant term of $f(x)$ is divisible by 3.

I have attempted this problem several times and have not found a satisfactory proof. We will denote the minimal polynomial of $\alpha$ over $\mathbb{Q}$ by $p(x)$ and the minimal polynomial of $3 \alpha$ over $\mathbb{Q}$ by $q(x).$ Clearly, we have that $\mathbb{Q}(\alpha) = \mathbb{Q}(3 \alpha),$ from which it follows that $\deg(p) = \deg(q).$ Consider the polynomial $r(x) = p(\frac{x}{3})$ in $\mathbb{Q}[x].$ We note that $r(3 \alpha) = p(\alpha) = 0$ and $\deg(r) = \deg(p) = \deg(q),$ from which I would like to conclude that $r(x) = 3^{-k} q(x)$ for $k = \deg(p).$ But I don't believe this does anything to solve the problem, and this is where I am now stuck. Thanks for your time.
 A: Consider the splitting field $F$ of the polynomial, and the ring of integers $\mathcal{O}_F$ of this field (i.e. the integral closure of $\mathbb{Z}$ in $F$).  Then the constant coefficient $f_0$ of $f$ is equal to the product of the roots of $f$ in $F$, which are all integral over $\mathbb{Z}$; therefore, if $\beta$ is the product of the roots other than $\alpha$ and $3\alpha$, then $f_0 = 3 \alpha^2 \beta$ and $\beta \in \mathcal{O}_F$.  It follows that $\frac{1}{3} f_0 \in \mathcal{O}_F \cap \mathbb{Q} = \mathbb{Z}$.
A: You can replace $3$ by any integer $m\neq 0$, and $\alpha$ is not necessarily a non-real complex number.  I do not assume either that $f(x)\in\mathbb{Z}[x]$ is monic except that the leading coefficient of $f(x)$ is coprime to $m$.  
Let $f(x)\in\mathbb{Z}[x]$ is a polynomial such that, for some algebraic number $\alpha$, both $\alpha$ and $m\alpha$ are roots of $f(x)$.  Suppose further that the leading coefficient of $f(x)$ is relatively prime to $m$.  Let $p(x)\in\mathbb{Z}[x]$ be a primitive minimal polynomial of an arbitrary algebraic number $\alpha$.  Let $n$ be the degree of $p(x)$.  
Show that $q(x):=m^n\,p\left(\frac{x}{m}\right)\in\mathbb{Z}[x]$ is a primitive minimal polynomial of $m\alpha$.  Verify that, if $\alpha\neq 0$ and $|m|>1$, then $p(x)$ and $q(x)$ are relatively prime over $\mathbb{Q}$.  Prove also that both $p(x)$ and $q(x)$ divide $f(x)$.  In the case that $\alpha$ is not a rational number, we know that $n\geq 2$.  Since $m^n$ divides the constant term of $f(x)$, we conclude that at least $m^2$ divides the constant term of $f(x)$.
