substituting in x=0, P(0)=0 or 1
Suppose P(0)=1
then let P(x)=xQ(x)+1
substituting into our functional equation
xQ^2(x)+2Q(x)=xQ(x²)
which implies that x|Q(x)
then we can replace Q(x)=xR(x)
x³R²(x)+2xR(x)=x³R(x²)
which implies that x²| R(x)
this process continues on forever. which means Q(x)=0
so we have a solution P(x)=1
Suppose P(0)=0
since its a non-zero polynomial, then let P have k zeros of the form x=0, k>0
P(x)=x^k Q(x)
hence x^2kQ(2x)=x^2kQ(x)²
Q(2x)=Q(x)², with Q(0)!=0. But Q also satisfy the functional equation, with Q(0)!=0, so Q(x)=1 (first case). then
hence P(x)=x^k, k>0 There are n solutions here.

total of n + 1 solutions.

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