P(2x-1) / P(x) = P(2x^2 -1 ) / P(x^2) = ... = P(2.x^2^n - 1) / P( x^2^n) = lim P(2t-1)/P(t) as t tends to infinity So we have P(2x-1)/P(x) = 2^n where n is deg of P(x) Denote g(x) = P(x+1) we have 2^n .g(x) = 2^n . P(x+1) = P(2 (x+1) -1 ) = P(2x +1) = g(2x) We could prove that g(x) = a. x^n for all x Hence P(x) = a. (x-1)^n for all x

A very nice problem. Proof Let $P(2x-1)=2^n.P(x)+R(x)$ for $R(x)$ is a polynomial st deg $R(x)=m<degP(x)=n$ After that we have $P(2x^2-1)=2^n.P(x^2)+R(x^2)$ so the original equation become: $P(x).[2^n.P(x^2)+R(x^2)]=P(x^2).[2^n.P(x)+R(x)]$ so $P(x).R(x^2)=P(x^2).R(x)$ Compare degree of two side of equation above we have: $n=m$ Contracdiction! It implies that $R\equiv 0$ so $P(2x-1)=P(x).2^n$ Let $Q(x)=P(x+1)$ so $Q(2x)=2^n.Q(x)$ So $Q(x)=a_n.x^n$ so $P(x)=a_n.(x-1)^n$ With $a_n \neq 0$

Let $D(x)= gcd(P(x), P(2x-1))$. Then $P(x)=D(x)F(x)$ and $P(2x-1)=D(x)G(x)$, where $F$ and $G$ are relatively prime polynomials. One also has $P(x^2 )=D(x^2 )F(x^2 )$ and $P(2x^2 -1 )=D(x^2 )G(x^2 )$ .From the given condition we obtain $D(x) F(x) D(x^2 ) G(x^2 ) = D(x) G(x)D(x^2 ) F(x^2 )$, therefore it follows that $F(x)G(x)=G(x) F(x^2 )$. Since $F(x)$ and $G(x)$ are relatively prime polynomials, the same are $F(x)$ and $G(x^2 )$, thus $F(x^2 )$ divides ${F(x}$ and $G(x^2 )$ divides $G(x)$. It follows that $F$ and $G$ are constant, whence $P(2x - 1) '= 2^np(x).Now it's easy to see $P(x)=c,c(1-x)^n$