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Find all polynomials $P(x)$ such that $P(x^2+1)=P(x)^2+1$ and $P(0)=0$.

It's almost the same question as this: Find all polynomials $P$ such that $P(x^2+1)=P(x)^2+1$.

But here I have the condition $P(0)=0$.

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    Referring to the other question: It follows easily by induction that, for all $P(x)$ of positive even degree, $P(0)>0$. The constant solutions are not zero, so the only polynomial for your condition is the one of odd degree. – Empy2 Jun 16 '15 at 15:08
  • Yup, based on the answers in the previous post that you linked, and @Michael's point, you already have your answer. – Robert Short Jun 16 '15 at 15:09
  • If $P(0)=0$ then $P(0^2+1)=P(1)=P(0)^2+1=1$.

    Then, $P(1^2+1)=P(2)=1+1=2$.

    Then, $P(2^2+1)=P(5)=2+1=3$.

    Then, $P(5^2+1)=P(26)=4$.

    And so on, so, if there exists such polynomial I believe it should be unique.

     – hjhjhj57 Jun 16 '15 at 15:11

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Assume coefficients $c_0,c_1, \cdots$ such that: $$ P(x) = \sum_{i=0}^n c_i x^i$$ Coefficients of $P(x^2+1)$ and $P(x)^2+1$ are easy to compute in terms of $\{c_k\}$. Do that and then set them to be equal for each coefficient of the equation. This way $P(0)=0$ just tells us $c_0=0$. Left hand side will obviously have all odd coeffs = 0 which will likely make it help starting with those ones.

Also, start with lower exponents, as fewer coefficients will be involved there.

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