Find all polynomial function $P$ of real coefficients such that for all $x \in \mathbb{R}$ $$P(x)P(x+1)=P(x^2+2)$$
Problem
Source: Swiss TST 2015. Problem 6
Tags: algebra, polynomial, function, functional equation
pco
29.07.2017 15:56
tchebytchev wrote: Find all polynomial function $P$ of real coefficients such that for all $x \in \mathbb{R}$ $$P(x)P(x+1)=P(x^2+2)$$ Posted (and solved) many many times (2006, 2009, 2011, 2015, 2016 twice) Dont hesitate to use the search function (see here ). Set (for example copy/paste) in the "search term" field the exact following string : +"P(x)P(x+1)" +"P(x^2+2)" You'll get in the five first results (excluded your own post and this post itself) all the help you are requesting for.
I'm sorry not providing you the direct link to this result but I encountered users who never tried the search function, thinking quite easier to have other users make the search for them. So now I prefer to point to the search function and to give the appropriate search term (I checked that it indeed will give you the expected result) instead of the link itself (wink)
tchebytchev
29.07.2017 16:11
pco wrote: tchebytchev wrote: Find all polynomial function $P$ of real coefficients such that for all $x \in \mathbb{R}$ $$P(x)P(x+1)=P(x^2+2)$$ Posted (and solved) many many times (2006, 2009, 2011, 2015, 2016 twice) Dont hesitate to use the search function (see here ). Set (for example copy/paste) in the "search term" field the exact following string : +"P(x)P(x+1)" +"P(x^2+2)" You'll get in the five first results (excluded your own post and this post itself) all the help you are requesting for.
I'm sorry not providing you the direct link to this result but I encountered users who never tried the search function, thinking quite easier to have other users make the search for them. So now I prefer to point to the search function and to give the appropriate search term (I checked that it indeed will give you the expected result) instead of the link itself (wink)
I'm not requesting for a solution. I create Swiss TST. I'm very thankful if other people already noticed the existence of such problems here. It will be taken into account when creating the whole contest. Anyway, thank you for noticing that this problem is already in the forum.