Find polynomials such that $(x-16)p(2x)=16(x-1)p(x)$ Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$
I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.
Link: http://web.mit.edu/rwbarton/Public/func-eq.pdf

Putting $x=1$ and $x=16$ we see that $p(2)=p(16) = 0$.
Let $p(x) = (x-2)(x-16)g(x)$.
Thus we see that
$$(x-16)(2x-2)(2x-16)g(2x) = 16(x-1)(x-2)(x-16)g(x)$$
Thus $$ (x-8)g(2x) = 4(x-2)g(x)$$
We now see that $g(4) = g(8) = 0$
Thus $g(x) = (x-4)(x-8)h(x)$
Thus
$$ (x-8)(2x-4)(2x-8)h(2x) = 4(x-2)(x-4)(x-8)h(x)$$
i.e
$$ h(2x) = h(x) $$
This implies that $h(x)$ is a constant (as otherwise the non-identically zero polynomial $h(2x) - h(x)$ will have an infinite number of roots.)
Thus
$$p(x) = C(x-2)(x-4)(x-8)(x-16)$$
