#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#
