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 : $P(x)=a(x-1)$, $P(x)=0$.
 A: Assuming the degree of a solution $P(x)$ is $d\geq 1$, the degree of $P(x^2)$ is $2d$ while the degree of $(x+1)P(x)$ is $d+1$. It follows that the solutions can only be constant or linear polynomials, and there is not much work left.
A: Assume that $P(x)\neq 0$.  It is easy to see that $P(x)$ is linear (using the degree-balancing argument).  Thus, if $P(x)=ax+b$, then the functional equation leads to ....
A: 4th line - I think you mean 'if $-1$ is a root, then $1$ is also a root'. I don't quite see why this line is necessary though.
In addition, I think you mean to write $|r|<|r^2|<|r^4|<\cdots<|r^{2^k}|<\cdots$ for the inequality (and similar with $>$).
Everything else seems fine, but you might want to explicitly show the calculations when you say 'By checking'.
Also it is best to start your proof by saying 'assume there is a polynomial which satisfies the equation' or similar.
A: 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 $c(x-1)\space\space\space\blacksquare$
