non costant

I do not know but it looks interesting. I will be interested in the answer when it comes  

it's pretty clear P(x) can't be an odd degree so we'll try a 2nd degree polynomial

$\text{Letting }P(x) = c_0 + c_1 x + c_2 x^2\\ \text{then grinding out the expansions of each side and equating like power of x coefficients}\\ \text{we get two solutions}\\ P(x) = 0\\ P(x) = x+x^2$