P(2x^2)=(P(2x^3+x))/(P(x)) is an even function so
a) P(2x^3+x) and P(x) are odd functions
b) P(2x^3+x) and P(x) are even functions
We have also
P(0)P(0)=P(0) so
P(0)=0 or P(0)=1
If P(0)=0 then P(x) must be odd or even. If P(x) is even
let a_(2n)x^(2n) the minimum x power in it's composition.
Then for that term,
(P_b(x)+a_(2n)x^(2n))(P_b(2x^2)+a_(2n)(2x^2)^(2n))=P_b(2x^3+x)+a_(2n)(2x^3+x)^(2n)
considering the lower order powers to the left and to the rigth
a_(2n)x^(2n)a_(2n)(2x^2)^(2n)=a_(2n)(2x^3)^(2n)+cdots+a_(2n)x^(2n) which implies
a_(2n)=0
In the same line of reasoning for the case P(0)=0 and P(x) odd, we can prove that P(x) such that P(0)=0 cannot be also odd. So
P(0)=1 and P(x) is even, then
P_n(x)=1+sum_(k=1)^na_(2n)x^(2n)
Now taking P_1(x) = 1+a_2x^2 we have
(1+a_2x^2)(1+a_2(2x^2)^2)=1+a_2(2x^3+x)^2
This equality is verified for a_2=1 so
P_1(x) = 1 + x^2
now considering
P_2(x)=1+a_2x^2+a_4x^4
after
P_2(x)P_2(2x^2)=P_2(2x^3+x)
solving for a_2, a_4 we obtain
P_2(x) = 1 + 2x^2+x^4 = P_1(x)^2
Finally we can verify that making
P_n(x)=(1+x^2)^n
then it is true
P_n(x)P_n(2x^2)=P_n(2x^3+x)
which is equivalent to
(1+x^2)^n(1+4x^4)^n=(1+x^2(2x^2+1)^2)^n
and also to
(1+x^2)(1+4x^4)=1+x^2(2x^2+1)^2
So the solutions are
P_n(x) = (1+x^2)^n for n=0,1,2,3,cdots
