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

Moron's solution and Dubuque's solution are essentilly the same, but I think their presentations can be made neater.
Clearly, $p$ is not a constant polynomial (unless it is zero).  So we may write $p(x) = C(x-z_1)(x-z_2)\ldots(x-z_n)$.  Then the given equality implies that
$(2x-z_1)(2x-z_2)\ldots(2x-z_n)(x-16) = 16(x-1)(x-z_1)(x-z_2)\ldots(x-z_n)$.
By comparing the leading coefficient on both sides, one sees $n=4$.  Thus,
$(x-\frac{z_1}{2})(x-\frac{z_2}{2})(x-\frac{z_3}{2})(x-\frac{z_4}{2})(x-16)=(x-1)(x-z_1)(x-z_2)(x-z_3)(x-z_n)$.
Hence $\lbrace \frac{z_1}{2}, \frac{z_2}{2}, \frac{z_3}{2}, \frac{z_4}{2}, 16\rbrace = \lbrace 1, z_1, z_2, z_3, z_4\rbrace$.  Now one easily sees that, up to a relabelling of indices, we must have $z_i = 2^i$.
