commutive polyomials ? A(B) = B(A) ??

[tommy1729]

Hold on Bo.

Polynomials have many fixpoints.

And there is also the 1-periodic theta mapping.

So just because the regular koenigs function based on a selected fixpoint does not result in another polynomial that commutes with the given one does not disprove its existance.

Secondly suppose we have a given polynomial A(x) of degree j.

Then proving that no other polynomial commutes with it requires checking it for all degrees ...

That is a supertask, since there are infinitely many integers.

That adresses the issue with finding a polynomial B(x) such that A(B(x)) = B(A(x) WHEN we are given an A(x).

The question in the link is more subtle :

A,B,C are distinct polynomials where none are given.

We want to FIND these 3 polynomials.

And they must satisfy 

A(B(x)) = B(A(x))

A(C(x)) = C(A(x))

B(C(x)) = C(B(x))

and the polynomials A,B,C must have rational coefficients and have distinct degrees.

Now read the question in the link again.

I think that should clarify all.

In the comment mick conjectured x^7  + x + 1 does not commute with polynomials of even degree.
( which seems close to what you referred to )

regards

tommy1729