find all polynomials that satisfy $(x-16)P(2x)=16(x-1)P(x)$ find all polynomials that satisfy this functional equation
$$\forall x \in \mathbb{R} \\ (x-16)P(2x)=16(x-1)P(x)$$
I write the polynomial
$$P(x)=\sum_{k=0}^n a_k x^k$$
but I found by comparing the LHS ans RHS (after rearrangement) that all coefficients must be zero, except the fourth coefficient; $P(x)=a x^4$.
But when I re-substitute it into the equation I found that it's inconsistent but it make sense only if $a=0$.

By plugging in $x=16$, we find $P(16)=0$. So $P$ has some positive roots. If $P$ is not $\equiv 0$, it has a minimal positive root $x_0$.
Plug in $\frac 12x_0$ to find $0=16(\frac12x_0-1)P(\frac12x_0)$ and so $x_0=2$ as $P(\frac12x_0)\ne 0$. By plugging in $x=2$, we then find $P(4)=0$, next $P(8)=0$ and $P(16)=0$. Similar to above, we find that the maximal positive root is $16$.
So
$$P(x)=(x-2)(x-4)(x-8)(x-16)Q(x) $$
and therefore
$$ (x-16)P(2x)=(x-16)(2x-2)(2x-4)(2x-8)(2x-16)Q(2x)=16(x-1)(x-2)(x-4)(x-8)(x-16)Q(2x)$$
whereas
$$ 16(x-1)P(x)=16(x-1)(x-2)(x-4)(x-8)(x-16)Q(x).$$
Therefore $Q(2x)=Q(x)$ for all $x$ (except possibly for $x=1,2,4,8,16$). Hence $Q$ is bounded, hence constant.
