Since F(0)=1 there can be no negative exponents and the coefficient of the x^0 term is 1.
I tried Ax + 1, then Ax^2 + Bx + 1, then Ax^3 + Bx^2 + Cx + 1.
I found F(x)=1 is a trivial solution to the equation, F(x)*F(2x
^{2}) = F(2x
^{3} + x), but the boundary condition of F(2) + F(3) = 2 instead of 125.
I found F(x)=x^2 + 1 also works for the equation, but F(2) + F(3) = 50 instead of 125.

Posted by Larry
on 20140824 12:46:23 