TITLE: Find all polynomials : $ P(x^2-x)=xP(x-1)$
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: The answer is fine but it can be made more precise in the same direction.
$$Q(x^2-x)=Q(x-1),x\neq0,1$$
Write this by substitution $y=x-1$ as
$$Q((y+1)y)=Q(y),y\neq0,-1$$
Now let us define the recursion
$$y_n=(y_{n-1}+1)y_{n-1},\ y_1=1$$
Obviously 
$$y_n \gt y_{n-1}$$
$$Q(y_{n})=Q((y_{n-1}+1)y_{n-1})=Q(y_{n-1})=Q(1)$$ 
for any $n$, leading to your conclusion.