A polynomial with unique root in every $ \mathbb Z _ n $

113 Views Asked by At

Let $ p ( x ) \in \mathbb N [ x ] $ be a polynomial with nonnegative integer coefficients, and $ a \in \mathbb Z $ be a given integer constant. If for all positive integers $ n $, $ p ( x ) + a $ has a unique root in the finite ring $ \mathbb Z _ n $, can we conclude that $ p ( x ) $ must be linear?

This problem comes from another one that I'm working on, and can be considered as an special case weaker than the original problem. While working, I realized that I have no method for solving this special case either, and I need to become sure that it holds, first.

The intuition behind why I think $ p ( x ) $ must be linear comes from Schur's conjecture (it is now a theorem in fact, yet still widely referred to by this name). If instead of having a unique root, $ p ( x ) + a $ were supposed to be a permutation polynomial, then by the conjecture it should have been the composition of linear functions and Dickson polynomials (of the first kind). But since all the coefficients of $ p ( x ) $ are nonnegative, and Dickson polynomials have a certain pattern of alternating signs in coefficients, it seems that $ p ( x ) $ must be linear (I haven't checked this in full detail, but it seems to be true). That made me more confident in believing that the nonnegativity of the coefficients of $ p ( x ) $ might be a strong enough condition to add to the weaker assumption of $ p ( x ) + a $ having a unique root, to get the same result. Also note that while the assumption of Schur's conjecture is that the polynomial is a permutation polynomial in the finite field $ \mathbb Z _ p $ for infinitely many prime numbers $ p $, here we have the unique root property in $ \mathbb Z _ n $ for all positive integers $ n $, which seems to be a very strong restriction.

The above mentioned theory of permutation polynomials was the closest theory to my problem in the literature that I could find searching online. I couldn't find much information on polynomials having unique roots when considered over several finite fields or rings. This may come from my unfamiliarity with the subject, of course. I find it very helpful if you could give references to parts of the literature more directly related to the problem at hand.

1

There are 1 best solutions below

7
On BEST ANSWER

The positivity assumptions are irrelevant. So more generally, suppose $q\in\mathbb{Z}[x]$ has exactly one root mod $n$ for all $n$. Let $f$ be any irreducible factor of $q$. By the Chebotarev density theorem, there are infinitely many primes $p$ such that $f$ splits as a product of distinct linear factors mod $p$. By our assumption on $q$, this means $f$ must be linear.

So, $q$ must actually be a product of linear factors over $\mathbb{Z}$, i.e. all its roots are rational. If $q$ has two distinct rational roots, there is some $p$ such that those roots exist and remain distinct mod $p$ (just take $p$ that does not divide their denominators or the numerator of their difference). So, $q$ can only have one rational root, so $q(x)=a(bx+c)^n$ for some $a,b,c\in\mathbb{Z}$ and some $n\in\mathbb{N}$ with $b$ and $c$ relatively prime. Clearly $a$ must be $\pm 1$ and then $b$ must also be $\pm 1$ (otherwise there would be no root mod $b$). Replacing $x$ with $bx+c$ (which is an invertible change of variables), we may thus assume $q(x)=x^n$. But now if $n>1$, $q$ has more than one root mod $p^n$ for any prime $p$ (namely both $x=0$ and $x=p$ are roots), so $n$ can only be $1$.