Determine the Polynomial $P$ for which $16P(x^2) = P(2x)^2$ Determined the Polynomial $P$ for which $16P(x^2) = P(2x)^2$ 
Okay, this problem was given as an example on an Olympiad training material I'm Self-studying, however I do not understand most of the arguments in the solution. Here's the solution.
Plugging $x = 0$ in   $16P(x^2) = P(2x)^2$ , we obtain $P(0) = 0$ or $16$
$(i)$, suppose $P(0) = 0$. Then $P(x) = xQ(x)$, for some Polynomial $Q$, hence
$16x^2Q(x^2) = 4x^2Q(2x)^2$, which reduces to $4Q(x^2) = Q(2x)^2$, setting $4Q(x) = R(x)$ for some Polynomial $R$, gives $16R(x^2) = R(2x)^2$ hence $P(x) = \frac {1}{4}xR(x)$, with R satisfying the same relation as $P$
I understand the above argument very well, but the argument presented below is what I do not understand at all.
$(ii)$, Suppose that $P(0) = 16$ putting $P(x) = xQ(x) + 16$ in the given relation, we obtain $4xQ(x^2) = xQ(2x)^2 + 16Q(2x)$. Hence $Q(0) = 0$, i.e. $Q(x) = xQ_1(x)$ for some Polynomial $Q_1$
Furthermore; $x^2Q(2x)^2 = x^2Q_1(2x)^2 + 8Q_1(2x)$, implying that $Q_1(0) = 0$ so $Q_1$ too is divisible by $x$, thus $Q(x) = x^2Q_1(x)$, now, suppose $x^n$ is the highest degree of $x$ dividing $Q_1$, and $Q(x) = x^nR(x)$, where $R(0) \ne 0$, then $R$ satisfies 
$4x^{n-1}R(x^2) = 2^{2n}x^{n+1}R(2x)^2 + 2^{n+4}R(2x) . . (*)$ which implies that $R(0) = 0$, a contradiction, it follows that $Q(0) = 0$ and $R(x) = 16.
Please I need someone to break down $(ii)$ for me, especially the (*) part. Thanks in advance.

I assume you understood how the relation $4xQ(x^2)=xQ(2x)^2+16Q(2x)$ was obtained. Now you just have to substitute $Q(x)=x^nR(x)$.
In particular, $4xQ(x^2)=4x^{2n+1}R(x^2)$, $xQ(2x)^2=2^{2n}x^{2n+1}R(2x)^2$ and $16Q(2x)=2^{n+4}x^nR(2x)$.
After simplifying $x^n$ on both sides, you find $4x^{n+1}R(x^2)=2^{2n}x^{n+1}R(2x)^2+2^{n+4}R(2x)$, which is (the correct form of) (*).
