<s>[INST] Find all polynomials such that $p(x^2-2x)=p(x-2)^2$ 
Find all polynomials $p\in \mathbb{C}[x]$ such that $$p(x^2-2x)=p(x-2)^2$$

We can not say anything specific about the degree since both sides are of the degree $2n$.
Also by copering the coeficients we see that the leading coefficent must be $1$.
Setting $$x^2-2x = x-2\implies x\in\{1,2\}$$
If $x= 2$ we get $p(0)=p(0)^2$ so $p(0)\in\{0,1\}$.
If $x= 1$ we get $p(-1)=p(-1)^2$ so $p(-1)\in\{0,1\}$.
Now it sems there is a lot of options. Also if $p$ is linear we have $$a(x^2-2x)+b= (ax-2a+b)^2 =a^2x^2+2a(b-2a)x+(b-2a)^2$$
so $a=a^2$ and $b=(b-2a)^2$ and $-2a=2a(b-2a)$ so $a=1$ and $b=1$ thus $p(x)=x+1$. But how to finish in general?
 [/INST] Substitute $y = x-1$, we see that $p(y^2-1) = (p(y-1))^2$. Take $q(x) = p(x-1)$ we get $q(x^2) = (q(x))^2$. Take $r(x) = q(e^x))$. we get $r(2x) = r(x)^2$. Take $s(x) = \log(r(x))$ we get $s(2x) = 2s(x)$. As $s(x)$ is defined at $(0,\infty)$ and continuous we know that the only possibility is $s(x) = cx$ for some constant $c$. Plugging all the way back you see $p(x) = (x + 1)^n$ for some $n$.
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