# Irreducibility of a Polynomial Shifted by a Power of Another Polynomial

Authors: Artūras Dubickas
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this note, we show that, for anyf∈ℤx and any prime number p, there exists g∈ℤx for which the polynomial fx−gxp is irreducible over ℚ. For composite p≥2, this assertion is not true in general. However, it holds for any integer p≥2 if f is not of the form ahxk, where a≠0 and k≥2 are integers and h∈ℤx.


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## Body

## 1. Introduction

A polynomial in one or several variables with coefficients in a fieldK is reducible over K if it is a product of two nonconstant polynomials with coefficients in K 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 inK=ℚ or in its ring of integers ℤ, there are very few criteria when the irreducibility of a given polynomial f 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 fgx when f∈Kx is irreducible and g∈Kx is chosen so that degg<degf was recently studied in [2–4], whereas reducibility of fx−pgx has been considered in [5, 6]. In the latter case, it was shown that, for any coprime polynomials f,g∈ℤx, and for all but finitely many prime numbers p, the polynomial fx−pgx is irreducible.In this note, instead offx−pgx, we consider fx−gxp and show the following.Theorem 1.
Letp≥2 be a prime number. Then, for each f∈ℤx there exists g∈ℤx such that the polynomial fx−gxp is irreducible over ℚ.In the case whenm=p≥2 is a composite number, the assertion of Theorem 1 is not true. Indeed, suppose that m=qℓ, where q,ℓ≥2 are integers. Take, for instance, fx=xq. Then, for any g∈ℤx, we have(1)wx=fx−gxm=xq−gxℓq=x−gxℓxq−1+xq−2gxℓ+⋯+gxℓq−1.The degree ofw is q if g is a constant and otherwise it is ℓqdegg. The degree of the factor x−gxℓ is 1 if g is a constant and otherwise it is ℓdegg. So, in both cases, x−gxℓ∈ℤx is a factor of w of degree at least 1 and at most q−1degw. Hence, w is reducible over ℚ.We also state a sufficient condition forf under which the assertion of Theorem 1 is true for composite m=p.Theorem 2.
Letm≥2 be an integer, and let f∈ℤx be a polynomial which is not of the form ahxk with integers a≠0, k≥2, and h∈ℤx. Then, there exists g∈ℤx such that the polynomial fx−gxm is irreducible over ℚ.Sincefx can be expressed as gxp+fx−gxp, we can formulate Theorem 1 in the following equivalent form: for any prime number p each polynomial in ℤx is expressible by the sum of a pth power of a polynomial in ℤx and an irreducible over ℚ polynomial in ℤx.In particular, selectingp=2 (or p=3), we can claim that each polynomial in ℤx is the sum of a square (resp. cube) in ℤx and an irreducible over ℚ integer polynomial. A corresponding problem for integers asserts that each sufficiently large integer is either a square (resp. cube) in ℤ or the sum of a square (resp. cube) in ℤ 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 thatf is nonconstant. Indeed, for fx=a∈ℤ, it suffices to take any constant polynomial gx=b∈ℤ. Then, for each m∈ℕ, the polynomial fx−gxm=a−bm is a constant, so it is irreducible over ℚ.In Section2, 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.
LetFx,y∈ℤx,y be an irreducible over ℚ polynomial. Then, there are infinitely many y0∈ℤ for which the polynomial Fx,y0∈ℤx is irreducible over ℚ.The next lemma follows from the result of Davenport et al. [23].Lemma 2.
Letk≥2 be an integer and let f∈ℤx be a nonconstant polynomial such that, for each x∈ℤ, there is ℓx∈ℤ for which fx=ℓxk. Then, there exists h∈ℤx such that fx=hxk.Here is a more special version of the above result due to Perelli and Zannier [24].Lemma 3.
Letf∈ℤx be a nonconstant polynomial such that, for each x∈ℤ, there are ax,ℓx∈ℤ and kx∈ℕ∖1 satisfying fx=axℓxkx. If the prime divisors of all ax belong to a finite set S, then there are integers a≠0, k≥2, and a polynomial h∈ℤx such that fx=ahxk.Next, we recall a theorem of Capelli, which was generalized by Kneser, see p. 92 in [1].Lemma 4.
LetK be a field and let m≥2 be an integer. The polynomial xm−a, where a∈K, is irreducible over K except when, for some b∈K, either a=−4b4 and 4|m or a=bp with some prime p|m.We conclude this section with several simple lemmas.Lemma 5.
LetK be a field and let m≥2 be an integer. Suppose that Fx,y∈Kx,y is a polynomial of degree m in y with coefficient c∈K∖0 for ym. If Fx,y is reducible over K, then Fx0,y∈Ky is reducible over K for each x0∈K.Proof.
SinceF is reducible, there are c1,c2∈K∖0 satisfying c1c2=c, n,k∈ℕ satisfying n+k=m, and u,v∈Kx,y such that(2)Fx,y=c1yn+ux,yc2yk+vx,y,where u is of degree at most n−1 in y and v is of degree at most k−1 in y. Hence, for any x0∈K, the degrees of the polynomials c1yn+uy,x0 and c2yk+vy,x0 are n and k, respectively. In particular, these polynomials are both nonconstant. This implies that their product Fx0,y∈Ky is reducible over K.
Here is a simple corollary of Lemma4:Lemma 6.
The polynomialyp−fx, where f∈ℤx and p≥2 is a prime number, is irreducible over ℚ except when fx=hxp for some h∈ℤx.Proof.
Suppose thatyp−fx is reducible. Then, for each x0∈ℤ, by Lemma 5 with K=ℚ, c=1, and m=p, the polynomial yp−fx0∈ℤy is reducible over ℚ. Thus, by Lemma 4 with K=ℚ and m=p, we must have fx0=bp for some b∈ℚ. Moreover, from fx0∈ℤ, it follows that b∈ℤ. Therefore, by Lemma 2, we conclude that there is a polynomial h∈ℤx such that fx=hxp.
We also have the following.Lemma 7.
Leth∈ℤx be a nonconstant polynomial and let p be a prime number. Then, the polynomial(3)hx+yp−hxpy=∑j=1ppjyj−1hxp−j∈ℤx,y,is irreducible over ℚ.Proof.
Denote the polynomial byux,y. Suppose u is reducible over ℚ. Fix any x0∈ℤ for which hx0≠0. From Lemma 5, it follows that ux0,y∈ℤy must be reducible over ℚ. Since hx0∈ℤ, the polynomial(4)hx01−pux0,yhx0=1+yp−1y=∑j=1ppjyj−1∈ℤy,must be reducible over ℚ as well. However, by Eisenstein’s criterion, this is not the case. Hence, u is irreducible over ℚ.


## 3. Proof of Theorems1 and 2

Proof of Theorem 1.
Suppose first that the polynomialfx−yp∈ℤx,y is irreducible over ℚ. Then, by Lemma 1, for some y0∈ℤ the polynomial fx−y0p∈ℤx is irreducible over ℚ, so we can simply take the constant polynomial gx=y0.
The only alternative is indicated by Lemma6. Then, fx=hxp, where h∈ℤx. Consider gx=hx+y with some y∈ℤ to be chosen later. It is clear that(5)fx−gxp=hxp−hx+yp=−yhx+yp−hxpy.
By Lemma7 combined with Lemma 1, there is an integer y0≠0 for which vx=hx+y0p−hxp/y0 is irreducible over ℚ. Hence, so is the polynomial −y0vx=fx−hx+y0p too, which is the desired conclusion.Proof of Theorem 2.
Iffx−ym∈ℤx,y is irreducible over ℚ, then the argument is the same as that in the Proof of Theorem 1. Suppose fx−ym is reducible over ℚ. Then, by Lemma 5, for each x∈ℤ, the polynomial ym−fx∈ℤy is reducible over ℚ. By Lemma 4, for each x∈ℤ, we have fx=axℓxkx, where ax∈1,−4, ℓx∈ℤ and kx is prime divisor of m or kx=4. Thus, by Lemma 3, we must have fx=ahxk for some integer a≠0 and some polynomial h∈ℤx. This is not the case by the assumption of the theorem, which completes the proof.



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*Source: 8869499-2020-12-02.xml*