Find all polynomials $P(x)$ such that $2xP(x)=(x+1)P(x-1)+(x-1)P(x+1)$. Find all polynomials $P(x)$ such that $2xP(x)=(x+1)P(x-1)+(x-1)P(x+1)$.

Well, if $\deg P\le 3$ this is easy since we can deduce $P(0)=P(1)=P(-1)$ by letting $x=0,1,-1$
 A: First some simple observations:


*

*setting $x:=0$ and $x:=1$ one finds $P[-1]=P[0]=P[1]$ in all cases,

*constant polynomials are solutions,

*the equation is linear in $P$,

*so we may add constants freely, after which we may assume $P[-1]=P[0]=P[1]=0$,

*and $P=(X-1)X(X+1)$ is a solution by easy explicit computation.


So the solutions up to degree$~3$ are $P=\lambda(X-1)X(X+1)+\mu$ where $\lambda,\mu$ are arbitrary constants (I suppose you found these solutions).
Now in the general case let $P =a_dX^d+a_{d-1}X^{d-1}+a_{d-2}X^{d-2}+(\text{lower order terms})$, where $d=\deg P$ (so $a_d\neq0$). A simple (but somewhat tedious) computation gives
$$
  (X+1)P[X-1]+(X-1)P[X+1]=2a_dX^{d+1}+2a_{d-1}X^d+\left((2\tbinom d2-2d)a_d+2a_{d-2}\right)X^{d-1}+(\text{lower order terms}).
$$
Comparing with $2XP$, we see that the third term only matches when $2\tbinom d2-2d=0$, in other words when $d(d-1)-2d=d^2-3d=0$, or $d\in\{0,3\}$. There are no other solutions than the ones above.
A: We can just use the fact that the only polynomial which has infinitely many roots is the zero polynomial $p(x)=0$. (without calculating the coefficient)
Since $P(0)=P(-1)=P(1)$(as you have already done), there is some constant $c$ and $d$ such that $$P(x)=cx(x+1)(x-1)+d$$ holds for $x=-1,0,1$ and $2$. (Set $d=P(0)$ and $c=(P(2)-d)/6$)
Let $Q(x) = cx(x+1)(x-1)+d$ and $R(x) = P(x) - Q(x)$. Then obviously, $R(x)=0$ for $x=-1,0,1$ and $2$. It could be easily checked that $R(x)$ also satisfies the equation $$2xR(x)=(x+1)R(x-1)+(x-1)R(x+1). $$
If $R(n-1) = R(n) = 0$ for some positive integer greater than $1$, put $x=n$ and we obtain $R(n+1) = 0$. Since we have the initial conditions $R(1)=R(2)=0$, we can prove inductively that for all positive integers $n$, $R(n) = 0$. Therefore $R(x)=0$ have infinitely many roots and hence, $R(x)$ equals zero. 
Thus $P(x)$ equals $Q(x) = cx(x+1)(x-1)+d$ and these are the only solutions. 
A: (Marc van Leeuwen was 8 minutes faster.)
Assume $n\geq3$ and write $p$ in the form
$$p(x)=x^n+a x^{n-1}+b x^{n-2}+ r_{n-3}\ ,$$
where here and in the following $r_k$ denotes an unspecified polynomial of degree $k\geq0$. In particular 
$$(x\pm 1)^m=x^m\pm m x^{m-1}+{m(m-1)\over2} x^{m-2}+r_{m-3}\ .$$
Computation then shows that
$$(x+1)p(x-1)+(x-1)p(x+1)-2x p(x)=n(n-3) x^{n-1}+ r_{n-2}\ .$$
It follows that there is no polynomial of degree $>3$ having the required property. 
