P(x) = x^n for any even power works
Thus includes the constant function P(x)=1
Also the zero polynomial P(x)=0
It's easy to show x^n does not work for odd powers.
If the polynomial has more terms the LHS will have terms with powers lacking on the RHS. These come from P(x)^2 compared to P(x^2). Furthermore, these extra terms are of higher degree than P(x) so there's no way to fix this with the difference between P(-x) and P(x).
Posted by Jer
on 2020-11-04 10:45:16