F(x) is a polynomial with real coefficients such that:

F(0) = 1, and:
F(2) + F(3) = 125, and:
F(x)*F(2x²) = F(2x³ + x)

Find the value of F(5).

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(2x²) = F(2x³ + 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 2014-08-24 12:46:23