amrit1968 wrote: Determine all polynomials P(x) with real coefficients such that [(x + 1)P(x − 1) − (x − 1)P(x)] is a constant polynomial. $(x+1)P(x-1)-(x-1)P(x)=c$ Let $P_1(x)=P(x)-\frac c2$ and equation is $(x+1)P_1(x-1)=(x-1)P_1(x)$ Setting there $x=1$, we get $P_1(0)=0$ and so : $P_1(x)=xP_2(x)$ and equation is $(x+1)(x-1)P_2(x-1)=(x-1)xP_2(x)$ Setting there $x=0$, we get $P_2(-1)=0$ and so : $P_2(x)=(x+1)P_3(x)$ and equation is $(x+1)(x-1)xP_3(x-1)=(x-1)x(x+1)P_3(x)$ And so $P_3(x)=a$ constant And so $\boxed{P(x)=ax^2+ax+b\quad\forall x}$ which indeed is a solution, whatever are $a,b\in\mathbb R$

The simplest solution: 1) We have $(x+1)P(x-1)-(x-1)P(x)=c$ (*) 2) If $x=1$, we get $P(0)=c/2$, if $x=-1$, we get $P(-1)=c/2$ 3) Thus we have $P(x)=x(x+1)Q(x)+c/2$ 4) Replacing in (*), we get: $Q(x-1)=Q(x)$, so $Q(x)=a$ 5) Thus $P(x)=ax(x+1)+c/2$. That's all!

$Q(x)=(x+1)P(x-1)-(x-1)P(x)$ First it is obviously that all constant polynomial are solution. Now suppose that $P(x)=ax^n+bx^{n-1}+r(x)$: The coefficient of $x^n$ in $Q(x)$ is $(n-2)a=0$ so $P(x)=ax^2+bx+c$ Setting these in the first we have $a=b$ so all the solution is $P(x)=ax^2+ax+c$ with $a,c$ real numbers.