How to find all polynomials $P(x)$ with real coefficents satisfying $P^2(x)-1=4P(x^2-4x+1)$  
Find all polynomials $P(x)$ with real coefficents satisfying
  $P^2(x)-1=4P(x^2-4x+1)$.

My solution:
Let the first term of $P(x)$ be $ax^n$. We see first term of left side is easily $a^2x^{2n}$ and right side's is
$4ax^{2n}$ . As we see there's absolutely no solution. If $n=0$ then easily, $a^2-4a-1=0$ and then we can find only 2 values for $P(x)=a$. So the solution is $2$.
 A: Let $Q(x+3)=P(x)$. Hence $$Q^2(x+3)-1=4Q(x^2-4x+4)=4Q((x-2)^2)$$
Let $x\rightarrow x+2$. $$Q^2(x+5)=4Q(x^2)+1$$
So $Q(5+x)=\pm Q(5-x)$
Let $R(x)=Q(5+x)$. $R$ is either odd or even and $$R^2(x)=4R(x^2-5)+1$$
If $R$ is odd then $R(0)=0$ and $R(5)=\frac{1}{4}$ and $R(-20)=\frac{15}{64}$ and more generally :
$S_0=0$, $S_{n+1}=5-S_n^2$, $T_0=0$ and $T_{n+1}=\frac{1-T_n^2}{4}$
It's easy to see that $P(S_n)=T_n$, but $\lim(T_n)$ is finite and $\lim(|S_n|)=\infty$. So $R$ is not a polynomial function.
If $R$ is even, $R(0)$ is an optimum (either maximum or minimum, $R'(0)=0$). Hence $R(\sqrt{5})$ is an optimum too, and $S_0=0$, $S_{n+1}=\sqrt{5+S_n}$ , then $S_n$ are all optimum... But all $S_n$ are different, so $R'$ has an infinite number of roots. This is not a polynomial function, except for a degree 0 polynomial.
So the only solutions are constant.
I hope there is no mistake :)
A: Suppose that P is not constant. Fixing $degP=n$ and comparing coefficients of both sides we deduce that the coefficients of polynomial $P$ must be rational. On the other hand, setting $x=a$ with $a=a^2−4a+1$, that is, $a=\frac{5±\sqrt{21}}{2}$, we obtain $P(a)=b$, where $b^2−4b−1=0$, i.e. $b=2±\sqrt{5}$. However, this is impossible because P(a) must be of the form $p+q\sqrt{21}$ for some rational p,q for the coefficients of P are rational. It follows that $P(x)$ is constant.
That is solution from http://www.imomath.com/index.php?options=346&lmm=0
But I don't know why the coefficients of polynomial $P$ must be rational
