Find all polynomials $P(x)$ such that $ x P(x-n)=(x-1) P(x) $ Question - 
Find all polynomials $P(x)$ with real coefficients such that
$$
x P(x-n)=(x-1) P(x)
$$
for some $n \in \mathbb{N}$ and for all $x \in \mathbb{R}$
My attempt - 
First by putting $x=0$, I get $p(0)=0$ ..
then the hint says that for $n>1$ show that $p(x)=0$ has infinitely many zeros...
I first let that another root $R$ is not equal to $0$..then by putting  $R$ in equation I get that $R-n$ is also root ...but I am unable to prove that there are infinitely roots...
Any hints ???
Thank You
 A: Case $n>1$
As you noticed $P(0)=0$. Using this fact and evaluate the equality in $x=n$ you have:
\begin{gather}
nP(n-n) = (n-1)P(n)\\
0 = P(n)
\end{gather}
This procedure suggests (in some sense) the following statement: 

If $k\in \mathbb N$ and $kn$ is a root of $P(x)$, then $(k+1)n$ is a root of $P$.

Infact, evaluating the eqaulity in $(k+1)n$ knowing that $P(kn)=0$ we have:
\begin{gather}
(k+1)n P((k+1)n-n)) = ((k+1)n-1)P((k+1)n)\\
0 = P((k+1)n)
\end{gather}
Thanks to this fact, you have that the set $\{0,n,2n,3n, 4n,...\} = \{kn\}_{k\in \mathbb N}$ is a set of roots of $P$. Since it is infinity, $P(x)=0$.

Case $n=1$
Again we have $P(0)=0$ so $P(x)=xQ(x)$ for a certain polynomial $Q(x)$. Sobstituting this equality in the equality of the text we have:
\begin{gather}
x(x-1)Q(x-1)=x(x-1)Q(x)\\
Q(x-1)=Q(x)
\end{gather}
And this implies that $Q(x)=c$ with $c\in \mathbb R$. Then the polynomial $P(x)$ is necessarly of the form $P(x)=cx$ for some $c\in \mathbb R$ and every polynomial of this form works.
Edit: In the case $1$ we have to take linear increment and not exponential.
A: For $n>1$,
Given : $$
x P(x-n)=(x-1) P(x)
$$
Firstly put $x=1$ in above equation to get $P(1-n)=0$. 
Then put $x=1-n$ again in that equation to conclude $P(1-2n)=0$ .Now put $x=1-2n$ and so on.
Can you continue this process indefinitely unless P is the zero polynomial?
For $n=1$, we have :
$xP(x-1)=(x-1)P(x)$
This implies $P(0)=0$. Now let $P(x)=xf(x)$, then we get :
$x(x-1)f(x-1)=(x-1)xf(x)$
This implies $f(x)=f(x-1)$ for all $x$ which is only possible when $f(x)$ is constant. Therefore $P(x)=cx$ for some constant $c$.
A: Let's do it for $n=2$. Suppose that $xP(x-2) = (x-1)P(x)$ for all $x$ real. By substitution of $x=0$, $0 = -P(0)$ therefore $P(0) = 0$.
Next, note that $2P(2-2) = (2-1)P(2)$, the LHS is $0$, so the RHS is $0$ i.e. $P(2) = 0$.
Next, $4P(4-2) = (4-1)P(4)$, the LHS is $0$, therefore so is the RHS i.e. $P(4) = 0$.
By induction, $P(2n) = 0$ for all $n$. This is impossible if $P$ is a polynomial unless $P \equiv 0$.
Can you do something similar for other $n$?

Suppose $n = 1$. Then, we want a polynomial such that $xP(x-1) = (x-1)P(x)$.
Note that $x-1$ is relatively prime to $x$ as a polynomial, therefore $x$ must be a divisor of $P(x)$. Let $Q(x) = \frac{P(x)}{x}$ (as a polynomial, so at $0$ it will be well defined), then from $\frac{P(x)}{x} = \frac{P(x-1)}{x-1}$ we get that $Q$ has infinitely many values all equal to each other. Hence, $Q$ is a constant polynomial.
Thus , $P(x) = Cx$ for some constant $C$. Clearly, any such real constant works.
