Polynomials$ P(x^2)=(x+1)P(x)$ Find all polynomials $P(x) \in\mathbb{R}[x]$ satisfying $$ P(x^2)=(x+1)P(x)$$
Please check my work :
Let $r$ be root of $P(x)$ so $P(r)=0$
Substitute $x=r$, $ P(r^2)=(r+1)P(r)$ so $P(r^2)=0$, i.e., if $r$ is root then $r^2$ is also root.
Substitute $x=-1$, so $P(1)=0$, i.e., if $r$ is root then $r^2$ is also root.
If $ \mid\; r\mid >1$ then  $ \mid\; r^4\mid > \mid\; r^2\mid > \mid\; r\mid  $ so $P(x)$ has infinitely many roots.
If $ \mid\; r\mid <1$ then  $ \mid\; r^4\mid < \mid\; r^2\mid < \mid\; r\mid  $ so $P(x)$ has infinitely many roots.
Hence $ \mid\; r\mid =1$ so $P(x)=c, \;c(x-1), \;c(x+1), \;c(x-1)(x+1)$
By checking, my answer is $P(x)=c(x-1)$
Edited work :
If deg$(P(x))=0$, then $P(x)=0$.
Assume deg$(P(x))=k \not= 0$ so deg$(P(x^2))=2k$ and deg$((x+1)P(x))=k+1$
so $2k=k+1$ we have $k=1$ so $P(x)=ax+b$
so $P(x^2)=ax^2+b=(x+1)(ax+b)$, we obtain $a=-b$
$P(x)=ax+b=ax-a=a(x-1)$
Answer : $P(x)=a(x-1)$, $P(x)=0$.

Suppose $P(x)$ has a root other than $1$ and $-1$ (say $r$). Then $r^2$ is also a root of the polynomial, i.e. $r^4$ is also a root of the polynomial which leads $P$ has infinitely many roots, which is not possible.
Then only roots $P$ can have are $1$ and $-1$. It is easy to see that $1$ is a root of $P$.
$c\neq 0$
Let $P(x)=c(x-1)(x+1)\Rightarrow P(x^2)=\color{red}{c(x^2-1)(x^2+1)}=(x+1)P(x)=\color{green}{c(x-1)(x+1)^2}$, which is not possible, since degree of the red coloured polynomial is $\color{red}{4}$ and degree of the green coloured polynomial is $\color{green}{3}$.
Let $P(x)=c(x-1)\Rightarrow P(x^2)=c(x^2-1)=(x+1)P(x)=(x+1)c(x-1)=c(x^2-1)$
Hence $P(x)$ is constant or $c(x-1)\space\space\space\blacksquare$
