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
 A: 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 $z_i = 2^i$.
A: Since $\rm\;p(2x)\;$ has roots being half the roots $\rm\;r_i\;$ of $\rm\;p(x)\;$, and $\rm\;\frac{p(x)}{p(2x)} = \frac{x-16}{16(x-1)}\;$, for the roots to all cancel like that leaving only one term in the numerator and denominator, we must have simply that the map $\rm\;r\to r/2\;$ is a left shift map on the (intermediate) roots. The numerator shows the largest root is 16, and the denomimator shows half the smallest root is 1. The other intermediate roots must obey the left shift $\rm\;r_i/2 = r_{i-1},\;$ so the roots are $\rm\;2,4,8,16,\;$ i.e. $\rm\;p(x) = c(x-2)(x-4)(x-8)(x-16)\;$
EDIT$\:$ I purposely omitted a detail above since I thought it might be a homework problem. Based on the comments, this caused some confusion, so I will elaborate. Namely, notice that each linear linear factor of $\rm\;p(x)\;$ introduces a constant factor of $2\:$ in the denominator of $\rm\: p(x)/p(2x)$. But said denominator has constant factor $16$, so the $4$ factors found above are complete. In particular, $0$ cannot be a root of $\rm\:p(x).$
A: I love Moron's approach.  Because it is usually interesting to look at other solutions, consider what happens if you just write $p(x) = p_n x^n + p_{n-1}x^{n-1} + \cdots + p_0$ and write out the equation.  Comparing the coefficients of $x^k$ ($k \gt 0$) yields the recursion
$p_k= \frac{2^{k-5}-1}{2^k - 1} p_{k-1}$,
whence $p_5 = 0$, implying $p_6 = p_7 = \cdots = 0$, and the rest of the recursion yields $p_4, p_3, \ldots, p_0$ as multiples of $p_4$.  This approach, although pedestrian, also demonstrates that we need only assume $p$ is analytic near 0.
A: 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
$$p(x) = C(x-2)(x-4)(x-8)(x-16)$$
A: Hint $\:$ Note the equation is $\displaystyle \rm\; \frac{\sigma\:f}f \:=\; \frac{\sigma^4 g}g\ $ for $\rm\ g = x-16 \:,\:\;$ shift  $\rm\;\;\sigma\:f(x) = f(\sigma\:x) = f(2\:x)$
Or, written additively $\rm\  (\sigma-1) \; f \;=\: (\sigma^4 - 1) \; g \quad\;$ [see Note below on this additive notation] 
Thus  $\rm\;\; f\, =\,\smash{\dfrac{\sigma^4-1}{\sigma-1}}\,g\, =\,  (1+\sigma+\sigma^2\!+ \sigma^3) \: g \;=\: (x-16)\:(2x-16)\:(4x-16)\:(8x-16) \;$
unique up to a factor of $\rm\;h \in ker(\sigma-1) = \{h: \sigma\:h = h\:\} = \;$  constants, i.e. $\rm deg\;h = 0 \ \ $ QED
Remark $\;$ This reformulation of my prior answer is intended to dramatically illustrate the innate symmetry. Its striking simplicity arises precisely from the simple structure of the orbits of the shift automorphism $\rm\: \sigma.\;$  Namely, by orbit decomposition the problem reduces to one on the single orbit of an irreducible polynomial. By exploiting polynomial structure there, we reduce the problem to a trivial polynomial division by $\rm\:\sigma - 1.\:$ This nicely illustrates the essence of the ideas employed to solve general difference equations (recurrences) over rational function fields - ideas at the foundation of algorithms employed in computer algebra systems.
Hopefully the symmetry is clearer in the above reformulation. Based on prior comments, and someone reposting my prior solution stripped of the symmetry, I fear my prior answer did not succeed in explicitly emphasizing the beautiful innate symmetry lying at the heart of this problem  - a germ of the Galois theory of difference fields. Probably the problem was devised to help spur one to discover this beautiful structure.
Note $\;\;$ The point of the additive notation is to exploit the natural polynomial structure of the action of $\rm\:\sigma\:$ on the multiplicative group generated by the elements $\rm\: \sigma^n \,f\;$ in the orbit of $\rm\!\: f \!\:$ under $\rm\:\sigma.\:$ This action is best comprehended by examining it on a specific example.$ $ Recalling  $\rm\,\sigma\,f(x) = f(2\:\!x)$
$\quad\quad\begin{align}{}
\rm \sigma\:(\:f(2^{-2}\: x)^a \; f(2^3 x)^b \;\: f(2^5 x)^c)\; =& \rm\;\;\: f(\:2^{-1} x)^a \;\: f(2^4 x)^b \;\: f(2^6 x)^c \\\\
\rm \iff \quad\;\;\: \sigma\;\:(\:(\:\sigma^{-2}\: f\;)^a \;\; (\:\sigma^3\: f\:)^b \;\; (\:\sigma^5\: f\:)^c)\; =& \rm\;\; \;\; (\:\sigma^{-1} \: f\:)^a \;\; (\:\sigma^4 \: f\:)^b \;\; (\:\sigma^6 \: f\;)^c \\\\
\rm \iff \quad\;\; \sigma\; (\:a\:\sigma^{-2} \;+\:\;\; b\:\sigma^3 \;+\;\; c\:\sigma^5)\:\; f\quad\; =& \rm\; (\:a\:\sigma^{-1} \: + \:\;\: b\: \sigma^4 \; +\;\;  c\: \sigma^6)\; f \\\\
\end{align}$
In the Galois theory literature the above action is frequently written in a highly suggestive exponential form. To illustrate this, below is the key identity in the problem at hand, expressed in this exponential notation.
$\quad\quad\begin{align}{}
\rm g^{\:\sigma^4-\,1} \;=\;& \rm g^{\:(1\:+\;\sigma\:+\;\sigma^2\:+\:\,\sigma^3)\:(\sigma\,-\,1)} \\\\
\iff\quad\quad\rm  \frac{\sigma^4 g}g \;=\;& \rm (g \;\: \sigma\:g \;\:\sigma^2 g \;\:\sigma^3 g)^{\sigma - 1} \;=\; \frac{\phantom{g\;\;\:} \sigma\:g \;\;\: \sigma^2 g \;\;\:\sigma^3 g \;\;\:\sigma^4 g}{g \;\;\:\sigma\:g \;\;\:\sigma^2 g \;\;\:\sigma^3 g\phantom{\;\;\:\sigma^4 g}} \\\\
\end{align}$
