Find all polynomials $P(x)$ with real coefficients such that for all real $x$ we have:

$$
(x-16) P(2 x)=16(x-1) P(x)
$$
Let's start by looking for obvious roots. Considering $x=1$ we have

$$
-15 \cdot P(2)=16 \cdot 0 \cdot P(1)=0
$$

so $P(2)=0$. By doing the same with $x=2, 4$, and 8:

$$
-14 \cdot P(4)=16 \cdot 1 \cdot P(2)=0 \text { so } P(4)=0
$$

$$
\begin{aligned}
& -12 \cdot P(8)=16 \cdot 3 \cdot P(4)=0 \text { so } P(8)=0 \\
& -8 \cdot P(16)=16 \cdot 7 \cdot P(8)=0 \text { so } P(16)=0
\end{aligned}
$$

We now have 4 roots, so $P(x)$ is divisible by $(x-2)(x-4)(x-8)(x-16)$.

Next, consider the leading coefficient $a_{n} x^{n}$ of $P(x)$. By looking at the leading coefficients on both sides of the equation, we get $a_{n} 2^{n} x^{n+1}=16 a_{n} x^{n+1}$. There are two possibilities: either $a_{n}=0$ and $P$ is the zero polynomial, or $a_{n} \neq 0$ and then $n=4$. Since we already have 4 roots for $P$, we have all the solutions:

$$
P(x)=\lambda(x-2)(x-4)(x-8)(x-16) \text { with } \lambda \text { a real number }
$$

To complete the proof, we need to show that all these solutions satisfy the equation, which is not very difficult.