Let x=1 then the relation reduces to P(0) = 0. So then let P(x) = x*Q(x).
Substituting this and simplifying yields (x+1)^2*Q(x-1) - (x-1)^2*Q(x+1) = 4x*Q(x)
Evaluating this at x=1, x=0, and x=-1 eventually leads to Q(-1)=Q(0)=Q(1).
This suggests Q(x) is a constant function, call the constant k. Then P(x) = kx is a solution.
But what if there is a larger Q(x)? Then it would look like (x-1)*x*(x+1)*R(x)+k. I tried substituting this but could not make any more decent progress.
Answer
