Find all polynomials such that $p(x^2-2x)=p(x-2)^2$

Question

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?

Answer 1

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 $n$.

Answer 2

EDIT: (I made a change of variable $x-1\mapsto x$ as suggested by @Hw Chu.) We need to solve $\displaystyle p((x-1)^2-1)=p(x-1-1)^2$ or $p(x^2-1)=p(x-1)^2$ equivalently. Let $$\displaystyle q(x) =p(x-1)= cx^k\prod_{1\le i\le N\\a_i\ne 0} (x-a_i)^{n_i},$$ where $c\ne 0$ and $a_j\ne a_k$ for $j\ne k$. Plugging this into the equation $q(x^2)=q(x)^2$, we obtain $$ cx^{2k}\prod_{i=1}^N(x^2-a_i)^{n_i} =c^2x^{2k}\prod_{i=1}^N (x-a_i)^{2n_i},\\ \prod_{i=1}^N(x+\sqrt{a_i})^{n_i}(x-\sqrt{a_i})^{n_i}=c\prod_{i=1}^N (x-a_i)^{2n_i}. $$ Now, we find that the LHS has $2N$ linear factors because $$ x+\sqrt{a_j}\ne x-\sqrt{a_j},\\ x\pm\sqrt{a_j} \ne x\pm\sqrt{a_k} $$ for all $j\ne k$. Since the RHS has only $N$ linear factors, it follows $N=0$ and $c=1$. Thus, non-trivial solutions are $p(x)=q(x+1)=(x+1)^k$ for some $k\ge 0$.