we take $ (a_n)_{n\in\mathbb{N}}: \ a_0=0,a_{n+1}=(a_n+1)^3$ let the sequence $ (b_n)_{n\in\mathbb{N}}: \ b_n=P(a_n)$ we have $ b_0=a_0$ and $ b_{n+1}=(b_n+1)^3$ so $ \forall n: \ a_n=b_n$ so $ \forall n\in\mathbb{N}: \ P(a_n)-a_n=0$ and the set $ \{a_n: \ n\in\mathbb{N}\}$ is infinit so $ P(X)-X=0$

Very nice solution any other ideas?

aviateurpilot wrote: we take $ (a_n)_{n\in\mathbb{N}}: \ a_0 = 0,a_{n + 1} = (a_n + 1)^3$ let the sequence $ (b_n)_{n\in\mathbb{N}}: \ b_n = P(a_n)$ we have $ b_0 = a_0$ and $ b_{n + 1} = (b_n + 1)^3$ so $ \forall n: \ a_n = b_n$ so $ \forall n\in\mathbb{N}: \ P(a_n) - a_n = 0$ and the set $ \{a_n: \ n\in\mathbb{N}\}$ is infinit so $ P(X) - X = 0$ can someone prove why if the set is infinite then P(x)=x?

Because a polynomial of degree n has at most n roots (except we have a constant polynomial) If we have that a polynomial has infinite roots it must be the constant polynomial equal to zero. Since this polynomial is P(x)-x we obtain P(x)=x as desired The theorem he used is the fundamental theorem of algebra and is very well known, if you search you will find tons of proofs Daniel

To be sure, the fundamental theorem of algebra is much deeper and harder to prove than the fact used here. Here, we used that a polynomial cannot have more roots than its degree. The fundamental theorem of algebra says that a polynomial cannot have less roots than its degree (at least over the complex numbers). We really don't need this here. Here is a slightly more interesting problem: Can you find all polynomials $p$ with $p((x+1)^3)=(p(x)+1)^3$ (i.e. we remove the condition $p(0)=0$)?