4 Find all polynomials $f$ that satisfy the equation
$$
\frac{f(3 x)}{f(x)}=\frac{729(x-3)}{x-243}
$$
for infinitely many real values of $x$.
4 Solution: The above equation holds for infinitely many $x$ if and only if
$$
(x-243) f(3 x)=729(x-3) f(x)
$$
for all $x \in \mathbb{C}$, because $(x-243) f(3 x)-729(x-3) f(x)$ is a polynomial, which has infinitely many zeroes if and only if it is identically 0 .

We now plug in different values to find various zeroes of $f$ :
$$
\begin{array}{l}
x=3 \quad \Longrightarrow \quad f(9)=0 \\
x=9 \quad \Longrightarrow \quad f(27)=0 \\
x=27 \quad \Longrightarrow \quad f(81)=0 \\
x=81 \quad \Longrightarrow \quad f(243)=0 \\
\end{array}
$$

We may write $f(x)=(x-243)(x-81)(x-27)(x-9) p(x)$ for some polynomial $p(x)$. We want to solve
$$
\begin{array}{l}
(x-243)(3 x-243)(3 x-81)(3 x-27)(3 x-9) p(3 x) \\
=729(x-3)(x-243)(x-81)(x-27)(x-9) p(x) .
\end{array}
$$

Dividing out common factors from both sides, we get $p(3 x)=9 p(x)$, so the polynomial $p$ is homogeneous of degree 2. Therefore $p(x)=a x^{2}$ and thus $f(x)=a(x-243)(x-81)(x-27)(x-9) x^{2}$, where $a \in \mathbb{C}$ is arbitrary.

Alternative solution. As above, we wish to find polynomials $f$ such that
$$
(x-243) f(3 x)=729(x-3) f(x) \text {. }
$$

Suppose $f$ is such a polynomial and let $Z$ be its multiset of zeroes.

Both $(x-243) f(3 x)$ and $729(x-3) f(x)$ have the same multiset of zeroes, i.e. $\{243\} \cup \frac{1}{3} Z=\{3\} \cup Z$ or $\{729\} \cup Z=\{9\} \cup 3 Z$. Thus
$$
\begin{aligned}
9 \in Z & \Rightarrow 27 \in 3 Z \\
& \Rightarrow 27 \in Z \\
& \Rightarrow 81 \in 3 Z \\
& \Rightarrow 81 \in Z \\
& \Rightarrow 243 \in 3 Z \\
& \Rightarrow 243 \in Z .
\end{aligned}
$$

Let $Y$ be the unique multiset with $Z=\{9,27,81,243\} \cup Y$. This gives
$$
\{9,27,81,243,729\} \cup Y=\{9,27,81,243,729\} \cup 3 Y,
$$
so $Y \subset \mathbb{C}$ is a finite multiset, invariant under multiplication by 3 . It follows that $Y=\{0,0, \ldots, 0\}$ with some multiplicity $k$. Hence
$$
f(x)=a x^{k}(x-9)(x-27)(x-81)(x-243)
$$
for some constant $a \in \mathbb{C}$. Plugging this back into the original equation, we see that $k=2$ and that $a \in \mathbb{C}$ can be arbitrary.