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.
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 $f(x)$.