Which polynomial $p(x)$ satisfies the equation

$$
(x-16) p(2 x)=16(x-1) p(x)
$$

? (H)
When $x=16$, the left side is 0, so the right side must also be 0. Since $16(16-1) \neq 0$, it follows that $p(16)=0$, meaning the polynomial $p(x)$ must be of the form $(x-16) q(x)$. Substituting this in:

$$
(x-16)(2 x-16) q(2 x)=16(x-1)(x-16) q(x),(2 x-16) q(2 x)=16(x-1) q(x),(x-8) q(2 x)=8(x-1) q(x)
$$

Similarly, we can factor out $(x-8)$, $(x-4)$, and finally $(x-2)$:

$$
q(x)=(x-8) r(x) ; \quad(x-4) r(2 x)=4(x-1) r(x), r(x)=(x-4) s(x) ; \quad(x-2) s(2 x)=2(x-1) s(x), s(x)=(x-2)
$$

Thus, for some polynomial $t(x)$, $t(2 x)=t(x)$. Clearly, $t(1)=t(2)=t(4)=t(8)=\cdots=c$ is a constant, so the polynomial $t(x)-c$ has infinitely many roots.

This is only possible if $t(x)-c \equiv 0$, that is, $t(x) \equiv c$. Therefore:

$$
s(x)=c(x-2), \quad r(x)=c(x-2)(x-4), \quad q(x)=c(x-2)(x-4)(x-8), p(x)=c(x-2)(x-4)(x-8)(x-16)
$$

Polynomials of this form indeed satisfy the equality required by the problem.