Find all polynomials $P \in \mathbb{R}[x]$ Find all polynomials $P \in \mathbb{R}[x]$ such that $$P(x^2-2x)=P(x-2)^2.$$
I think we should replace $x$ with $x+1$, I really don't know, any help?
 A: Yes you're right.
Replacing $x$ by $x+1$, this becomes $P(x^2-1)=P(x-1)^2$
Writing $P(x)=Q(x+1)$, this becomes $Q(x^2)=Q(x)^2$ which is a very classical problem whose solutions are :
$Q(x)=0$ and so $\boxed{\text{S1 : }P(x)=0\text{  }\forall x}$ which indeed is a solution
$Q(x)=x^n$ and so $\boxed{\text{S2 : }P(x)=(x+1)^n\text{  }\forall x}$ which indeed is a solution whatever is the non-negative integer $n$.
Proof for the "classical problem"

Assume $x$ is a complex root of $P$. Then clearly also $x^2$ is a root of $P$ and similarly $x^4, x^8, x^{16},\dotsc$ are such roots.
  Assume $|x| \ne 0;1$. Then these roots have either strictly increasing or strictly decreasing absolute values and are hence distinct i.e. $P$ has infinitely many roots i.e. $P(x)$ is identically zero which is indeed a solution.
  So from now on assume that all roots of $P$ have absolute value 1 or are 0 and that $P$ is not identically zero.
  If $x=0$ is a root of $P$ then we can write $P(x)=x^kQ(x)$ with $Q(0) \ne 0$. Then the original condition implies that $x^{2k}Q(x^2)=x^{2k}Q^2(x)$ and hence $Q(x^2)=Q^2(x)$. So $Q$ satisfies the same functional equation as $P$ and has hence only roots with absolute value 1.
  So take any root (if it exists) $x_0=e^{iz}$ of $Q$ with $0 \le z <2\pi$. Then by plugging in $x=e^{\frac{iz}{2}}$ we obtain that $e^{\frac{iz}{2}}$ is also a root of $P$. Similarly, $e^{\frac{iz}{2^k}}$ is a root for all positive integers $k$. If $z \ne 0$ then all these roots have distinct arguments and hence are distinct. Therefore $P$ had infinitely many roots and must be constantly zero (a case we already considered). So for any other solution we must have $z=0$ and hence $x_0=1$ is the only root of $P$. But plugging in $x=-1$ we obtain that $P(-1)=0$ is also a root. Absurd! Hence $Q$ can't have any roots and hence $P(x)=cx^k$ for some constant $c$. But plugging this into the original equation we directly see that $c=0$ or $c=1$ i.e. $P(x)=0$ and $P(x)=x^k$ are the only solutions.

