Irreducibility of a Polynomial Shifted by a Power of Another Polynomial

Abstract

In this note, we show that, for any and any prime number , there exists for which the polynomial is irreducible over . For composite , this assertion is not true in general. However, it holds for any integer if is not of the form , where and are integers and .

1. Introduction

A polynomial in one or several variables with coefficients in a field is reducible over if it is a product of two nonconstant polynomials with coefficients in and irreducible otherwise. See, for instance, Schinzel’s book [1] for a systematic study of reducibility of polynomials.

Even in the case of univariate polynomials with coefficients in or in its ring of integers , there are very few criteria when the irreducibility of a given polynomial can be easily confirmed (Eisenstein’s criterion, Cohn’s criterion, and Newton polytopes method). However, usually a polynomial does not have a form for which any of the abovementioned methods can be applied. There are also some more special methods. For instance, reducibility of the polynomial when is irreducible and is chosen so that was recently studied in [2–4], whereas reducibility of has been considered in [5, 6]. In the latter case, it was shown that, for any coprime polynomials , and for all but finitely many prime numbers , the polynomial is irreducible.

In this note, instead of , we consider and show the following.

Theorem 1. Let be a prime number. Then, for each there exists such that the polynomial is irreducible over .

In the case when is a composite number, the assertion of Theorem 1 is not true. Indeed, suppose that , where are integers. Take, for instance, . Then, for any , we have

The degree of is if is a constant and otherwise it is . The degree of the factor is 1 if is a constant and otherwise it is . So, in both cases, is a factor of of degree at least 1 and at most . Hence, is reducible over .

We also state a sufficient condition for under which the assertion of Theorem 1 is true for composite .

Theorem 2. Let be an integer, and let be a polynomial which is not of the form with integers , , and . Then, there exists such that the polynomial is irreducible over .

Since can be expressed as , we can formulate Theorem 1 in the following equivalent form: for any prime number each polynomial in is expressible by the sum of a th power of a polynomial in and an irreducible over polynomial in .

In particular, selecting (or ), we can claim that each polynomial in is the sum of a square (resp. cube) in and an irreducible over integer polynomial. A corresponding problem for integers asserts that each sufficiently large integer is either a square (resp. cube) in or the sum of a square (resp. cube) in and a prime number (see the paper of Hardy and Littlewood (p. 49 in [][][7]) and (p. 51 in [][][7]). Both these problems are wide open, see, e.g., [8–11] for some progress on the representations of integers by the sum of a square and a prime number.

It is not surprising at all that an additive problem in integer polynomials involving irreducible polynomials is much easier than the corresponding problem in integers involving prime numbers, since “almost all” integer polynomials are irreducible (see [12], for a precise statement), whereas “almost none” integer is a prime number. The same happens with Goldbach-type problems in polynomials with integer coefficients when much more is known compared to classical Goldbach problems for integers. There is a considerable literature concerning this, see, for instance, [13–22].

Throughout, without loss of generality, we may assume that is nonconstant. Indeed, for , it suffices to take any constant polynomial . Then, for each , the polynomial is a constant, so it is irreducible over .

In Section 2, we give some auxiliary results. Then, in Section 3, we complete the proofs of the theorems.

2. Auxiliary Results

We first recall the simplest version of Hilbert’s irreducibility theorem (see p. 298 in [1]).

Lemma 1. Let be an irreducible over polynomial. Then, there are infinitely many for which the polynomial is irreducible over .

The next lemma follows from the result of Davenport et al. [23].

Lemma 2. Let be an integer and let be a nonconstant polynomial such that, for each , there is for which . Then, there exists such that .

Here is a more special version of the above result due to Perelli and Zannier [24].

Lemma 3. Let be a nonconstant polynomial such that, for each , there are and satisfying . If the prime divisors of all belong to a finite set , then there are integers , , and a polynomial such that .

Next, we recall a theorem of Capelli, which was generalized by Kneser, see p. 92 in [1].

Lemma 4. Let be a field and let be an integer. The polynomial , where , is irreducible over except when, for some , either and or with some prime .

We conclude this section with several simple lemmas.

Lemma 5. Let be a field and let be an integer. Suppose that is a polynomial of degree in with coefficient for . If is reducible over , then is reducible over for each .

Proof. Since is reducible, there are satisfying , satisfying , and such thatwhere is of degree at most in and is of degree at most in . Hence, for any , the degrees of the polynomials and are and , respectively. In particular, these polynomials are both nonconstant. This implies that their product is reducible over .
Here is a simple corollary of Lemma 4:

Lemma 6. The polynomial , where and is a prime number, is irreducible over except when for some .

Proof. Suppose that is reducible. Then, for each , by Lemma 5 with , , and , the polynomial is reducible over . Thus, by Lemma 4 with and , we must have for some . Moreover, from , it follows that . Therefore, by Lemma 2, we conclude that there is a polynomial such that .
We also have the following.

Lemma 7. Let be a nonconstant polynomial and let be a prime number. Then, the polynomialis irreducible over .

Proof. Denote the polynomial by . Suppose is reducible over . Fix any for which . From Lemma 5, it follows that must be reducible over . Since , the polynomialmust be reducible over as well. However, by Eisenstein’s criterion, this is not the case. Hence, is irreducible over .

3. Proof of Theorems 1 and 2

Proof of Theorem 1. Suppose first that the polynomial is irreducible over . Then, by Lemma 1, for some the polynomial is irreducible over , so we can simply take the constant polynomial .
The only alternative is indicated by Lemma 6. Then, , where . Consider with some to be chosen later. It is clear thatBy Lemma 7 combined with Lemma 1, there is an integer for which is irreducible over . Hence, so is the polynomial too, which is the desired conclusion.

Proof of Theorem 2. If is irreducible over , then the argument is the same as that in the Proof of Theorem 1. Suppose is reducible over . Then, by Lemma 5, for each , the polynomial is reducible over . By Lemma 4, for each , we have , where , and is prime divisor of or . Thus, by Lemma 3, we must have for some integer and some polynomial . This is not the case by the assumption of the theorem, which completes the proof.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This research was funded by European Social Fund (Project no. 09.3.3-LMT-K-712-01-0037) under grant agreement with the Research Council of Lithuania (LMTLT).