TITLE: Polynomial : $ P(x+1)-2P(x)+P(x-1)=6x $
QUESTION [5 upvotes]: Find all polynomials $P(x) \in \mathbb{R}[x]$ satisfying 
$$ P(x+1)-2P(x)+P(x-1)=6x $$
My attempt :
Since $ P(x+1)+P(x-1)-2P(x)=6x $, so $P(x)$ is not constant polynomial.
Let $P(x+1)-P(x)= Q(x)$
so $Q(x)-Q(x-1)=6x, \;\; \forall x \in \mathbb{R}$
by induction,  $Q(x+n)-Q(x)= 6((x+n)+(x+n-1)+...+(x+1))$
Continue from dxiv's answer,
$Q(n)-Q(0) = 6(n+(n-1)+...+1)$
$Q(n)=Q(0) + 6(n+(n-1)+...+1)=Q(0) + 3n(n+1) $
so $P(n) = Q(n-1) + Q(n-2)+ ...+Q(0)+P(0)$
$= \displaystyle\sum^{n-1}_{i=1}3i(i+1) + nQ(0)+P(0)$
$= \displaystyle\sum^{n-1}_{i=1}3i(i+1) + nP(1)- nP(0)+P(0)$
$= \displaystyle\sum^{n-1}_{i=1}3i(i+1) + nP(1)-(n-1)P(0)$
$= 3\displaystyle\sum^{n-1}_{i=1}(i^2+i) + nP(1)-(n-1)P(0)$
$= n^3-n + nP(1)-(n-1)P(0)\;\; \forall x \in \mathbb{N}$
Since $P(x)-x^3-x(P(1)-1)+(n-1)P(0)$ has infinitely many roots so 
$P(x)-x^3-x(P(1)-1)+(n-1)P(0) = 0$, we get
$P(x)=x^3+x(P(1)-1)-(n-1)P(0)\;\; \forall x \in \mathbb{R}$
As $P(1)-1$ and $(n-1)P(0)$ are in $\mathbb{R}$, we obtain
$P(x)=x^3+cx+d,\; \forall x \in \mathbb{R}$ and $c,d$ are real constants.

REPLY [3 votes]: I did in a pretty simpler way than you did.
$p(x)=ax^2+bx+c$ doesn't work because $p(x-1)+p(x+1)-2 p(x)=2a$ is a constant
so I thought to $3rd$ degree $p(x)=ax^3+bx^2+cx+d$
computed $p(x-1)+p(x+1)-2 p(x)=2 b + 6 a x$ and concluded that it must be $a=1;\;b=0$
So all third degree polynomial which satisfy the request are of the form $p(x)=x^3+cx+d$ for any $c,\;d\in\mathbb{R}$
Fourth degree or higher like $p(x)=a x^4+b x^3+c x^2+d x+e$ can't satisfy the request because $p(x-1)+p(x+1)-2 p(x)=12 a x^2+2 a+6 b x+2 c$ and the condition to eliminate $x^2$ or other powers greater than is to put $a=0$ and the polynomial is no more $4-$th degree.