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Determine, with proof, all polynomials $P(x)$ such that, $P(x)P(x+1)=P(x^2)$

Put $x=1$ into this equation, we can get $P(2)=1$ and put $x=0$, we can get $P(1)=1$ $x=-1$, $P(-1)P(0)=P(1)$

Then what should I do, just simply calculate the product of $P(x)$ and $P(x+1)$? So confused.

How to do the multiplication for these 2 polynomials and determine the $P(x)$.

Clark Zhang
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    Multiplying those gets you a really complicated expression, which is almost of no use unless you bring the degree of the polynomials down. Note that apart from literally multiplying them, you can also substitute different values of $x$. – Trebor Jul 25 '21 at 08:34
  • Special values of $x$ that are worth considering are 0, 1, -1, and the zero-points of the polynomials. – Trebor Jul 25 '21 at 08:35
  • Yeah, I realised this, but how to determine the value of P(0)? – Clark Zhang Jul 25 '21 at 08:39
  • You don't need to determine $P(0)$, although it may be a rational step to try when solving this problem – Trebor Jul 25 '21 at 08:41
  • You find that if $P(0) \ne 0$, $P(1) = 1, P(2) = 1, P(0)=1/P(-1), P(3) = P(4), \dots$ by plugging several values. Maybe at some point this gives the desired system. Note that the expression of a $n$ degree polynomial is entirely determined by $n+1$ points. – Rundasice Jul 25 '21 at 08:47
  • I find the description in the question confusing. Why is it working with three different lists of coefficients, $$a_n,a_{n-1},...\b_n,b_{n-1},...\c_n,c_{n-1},...$$ All other parts of both question and comments seem to imply there should only be one set so that $P$ denotes the same polynomial throughout, right? – String Jul 25 '21 at 08:53
  • Yes, just P(X). – Clark Zhang Jul 25 '21 at 08:55
  • What field is your polynomial defined over? – Sebastian Monnet Jul 25 '21 at 09:45
  • All Real Numbers! – Clark Zhang Jul 25 '21 at 10:34
  • Sorry, I should edit this. – Clark Zhang Jul 25 '21 at 10:34

1 Answers1

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Hint. The given identity holds also for any complex number. Note that if $z$ is a complex root then also $z^2$ is a complex root, which implies that $z=0$ or $|z|=1$ (otherwise we have infinite distinct roots). In a similar way, if $z$ is a complex root then also $(z-1)^2$ is a complex root, which implies that $z-1=0$ or $|z-1|=1$.

So what are the possible roots of $P$? The answer will give you a factorization of $P$.

Robert Z
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