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$P(x)$ is a polynomial with real coefficients satisfying $$P(x)P(x+1)=P(x^2+2)$$

I did find some solutions for $P(x)P(x+1)=P(x^2)$ and I wonder if I can do the same for this problem

I tried making $P(x)=a_nx^n+\cdots+a_1x+a_0$, and got that $a_n=1$, but I don't know how to make progress

P/s: I do found that $(x^2-x+2)^n$ is a solution. The solution is here: https://artofproblemsolving.com/community/c6h392562p2239953.

Kii
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  • $P(x)=0$ and $P(x)=1$ are two possible solutions. – Gonçalo Feb 03 '24 at 03:21
  • Oh yes, I had done that too. If P(x) = c(const) ... – Kii Feb 03 '24 at 03:22
  • Welcome to Math SE. By "Find P(x) with real coefficients", I assume the "real coefficients" are referring to that where $P(x)$ is a polynomial. Please clarify that. – John Omielan Feb 03 '24 at 03:34
  • @JohnOmielan thank you very much! Actually, I study math in Vietnamese so there are many math terms that I don't understand. – Kii Feb 03 '24 at 03:37
  • You're welcome. FYI, using an Approach0 search, there's the AoPS threads P(x) that satisfies ..., Polynomials! and Polynomial equation (with this one giving a source of "Swiss TST $2015$. Problem $6$"). – John Omielan Feb 03 '24 at 03:42
  • @JohnOmielan Thank you. This is the problem my teacher gave me yesterday to prepare for VMO. – Kii Feb 03 '24 at 03:43
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    Thanks for the additional information. Please edit your question with some of the material you have in the comments (I received from teacher / I already know that constants are solutions). If the question is online by any chance and not in your textbook then you can attach the link to it. Thanks. Also, what method was used to solve the similar equation $P(x)P(x+1) = P(x^2)$? – Sarvesh Ravichandran Iyer Feb 03 '24 at 03:45
  • @SarveshRavichandranIyer For that problem, if $z$ is a root of $P$, then so are $z^2$ and $(z-1)^2$. There are a lot of values of $z$ for which if $z$ is a root, then $P$ would have infinitely many roots. After you rule these out, there are not too many possible roots. – JimmyK4542 Feb 03 '24 at 03:57
  • @JimmyK4542 Thanks for the clarification. I think a similar method can be used here but it will be a little more complicated. – Sarvesh Ravichandran Iyer Feb 03 '24 at 04:00
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    @Kii if you seem to have found the solution, please post your answer below. – D S Feb 03 '24 at 04:16

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