<s>[INST] Find all polynomials : $ P(x^2-x)=xP(x-1)$ Find all polynomials $P(x) \in\mathbb{R}[x]$ satisfying $$ P(x^2-x)=xP(x-1)$$
Please check my answer :
$P(0) =0$, so $0$ is root of $P(x)$, there exists $Q(x)$ such that $P(x)=xQ(x)$
then $(x^2-x)Q(x^2-x)=x(x-1)Q(x-1)$, so $Q(x^2-x)=Q(x-1)$ where $x \not= 0, 1$
Substitute $x=2$, we have $Q(2) = Q(1)$
Substitute $x=3$, we have $Q(6) = Q(2)$
Substitute $x=7$, we have $Q(42) = Q(6)$
Substitute $x=43$, we have $Q(43^2-43) = Q(42)$
Since there are infinitely many $x$ such that $Q(x) = Q(1)$ so $Q(x)$ is constant polynomial.
Therefore, $P(x)=cx$ where $c$ is constant.
 [/INST] You can go also with canonical form of polynomial: $p(x) =a_nx^n+...$, where $a_n\ne 0$. So we have $$a_n(x^2-x)^n+...= xa_n(x-1)^n+...$$ so $a_nx^{2n}=a_nx^{n+1}$. Since $a_n\ne 0$ we have $n=1$ or $n=- \infty $. So $p(x)=ax+b$ for some $a,b$. Pluging in to starting equation we have:
$$a(x^2-x)+b= x(a(x-1)+b)\Longrightarrow ax^2-ax+b = ax^2+(-a+b)x$$
So $p(x)=ax$ is solution for any $a$.
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