Let $f(x)\in \mathbb Z[x]$ be a monic polynomial such that $f(a)=f(3a)=0$ for some $a\in \mathbb C$. Then, is it true that $3|f(0)$ ?
If $g(x)\in \mathbb Q[x]$ is the minimal polynomial of $a$ , then $3^n g(x/3)$ is the minimal polynomial of $3a$, where $n=\deg g(x)$. So $f(x)=3^n g(x/3) h(x)$ for some $h(x)\in \mathbb Q[x]$. Now if we knew $h(0)\in \mathbb Z$ we would be done, but I'm not sure if that is true. Please help.
Suppose $f$ is a monic polynomial in $\mathbb{Z}[X]$ and suppose $K$ is its splitting field. Then its roots are algebraic integers in the ring $\mathcal{O}_K$, so $a$ and $3a$ are in $\mathcal{O}_K$.
Now by Vieta's formulas, the constant term of $f$ is $\pm$ the product of all roots of $f$, including $a$ and $3a$. Therefore the constant term $f(0)$ is an integer (in $\mathbb{Z}$) divisible by $3$ (in $\mathcal{O}_K$).
We write this as $f(0) = 3\alpha$ where $\alpha \in \mathcal{O}_K$.
Taking field norms we find $N_{K/\mathbb{Q}}(f(0)) = N_{K/\mathbb{Q}}(3)N_{K/\mathbb{Q}}(\alpha)$, so $f(0)^n = 3^n N_{K/\mathbb{Q}}(\alpha)$ (where $n$ is the degree of $K/\mathbb{Q}$).
Since $N_{K/\mathbb{Q}}(\alpha)$ is an integer, this shows that $3$ divides $f(0)$ in $\mathbb{Z}$.
Alternative proof:
Starting from $f(0) = 3\alpha$ where $\alpha \in \mathcal{O}_K$, we conclude that $\frac{f(0)}{3} \in \mathcal{O}_K \cap \mathbb{Q}$. However, $\mathbb{Z}$ is integrally closed so $\mathcal{O}_K \cap \mathbb{Q} = \mathbb{Z}$ and we are done.