Similar to Find polynomials : $ xP(x-1)=(x-11)P(x)$ but my solution is different.
Bob asks us to find all real polynomials $P$ such that $$xP(x-1)=(x-2)P(x)$$
Here is my solution, is it right?
Replace $x$ with $0$ and $1$ and clearly, they equal $0$.
Suppose there is another positive root $r\neq 1$.
Then replace $x$ with $r+1$, which means $(r+1)P(r)=(r-1)P(r+1)\implies P(r+1)=0$ since $r-1$ and $r+1$ are not $0$.
Thus if there exists another positive root $r\neq 1$, then we have an infinite number of roots which means $P(x)=0$.
Now suppose there is another negative root $c$.
Replace $x$ with $c$, sos $cP(c-1)=(c-2)P(c)\implies P(c-1)=0$ since $c$ and $c-2$ are not $0$.
Thus if $c$ exists, then we have an infinite number of roots which again means $P(x)=0$.
So the only two polynomials that work are $P(x)=x(x-1)$ and $P(x)=0$.
If you put $x=2$ you get $P(1)=0$ and if you put $x=0$ you get $P(0)=0$ so 1 and 0 are zeroes so we can write: $$P(x)= x(x-1)Q(x)$$ for some polynomial $Q$. If you put this in original equation we get $$Q(x)=Q(x-1)$$ which is valid for all $x\ne 0,1,2$ and there for it is valid for all $x$, so $Q$ must be constant (since it is periodic).
So general soution is $$P(x) = cx(x-1)$$ for arbitrary constant $c$.