Research Article Irreducibility of a Polynomial Shifted by a Power of Another Polynomial

is properly cited. In this note, we show that, for any f ∈ Z [ x ] and any prime number p , there exists g ∈ Z [ x ] for which the polynomial f ( x ) − g ( x ) p 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 ah ( x ) k , where a ≠ 0 and k ≥ 2 are integers and h ∈ Z [


Introduction
A polynomial in one or several variables with coefficients in a field K 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 in K � 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.
ere are also some more special methods. For instance, reducibility of the polynomial f(g(x)) when f ∈ K[x] is irreducible and g ∈ K[x] is chosen so that degg < deg f was recently studied in [2][3][4], whereas reducibility of f(x) − pg(x) has been considered in [5,6]. In the latter case, it was shown that, for any coprime polynomials f, g ∈ Z[x], and for all but finitely many prime numbers p, the polynomial f(x) − pg(x) is irreducible.
In this note, instead of f(x) − pg(x), we consider f(x) − g(x) p and show the following.
In the case when m � p ≥ 2 is a composite number, the assertion of eorem 1 is not true. Indeed, suppose that m � qℓ, where q, ℓ ≥ 2 are integers. Take, for instance, f(x) � x q . en, for any g ∈ Z[x], we have (1) e degree of w is q if g is a constant and otherwise it is ℓqdegg. e degree of the factor x − g(x) ℓ is 1 if g is a constant and otherwise it is ℓdegg. So, in both cases, is a factor of w of degree at least 1 and at most q − 1 degw. Hence, w is reducible over Q.
We also state a sufficient condition for f under which the assertion of eorem 1 is true for composite m � p. Since f(x) can be expressed as g(x) p + f(x) − g(x) p , we can formulate eorem 1 in the following equivalent form: for any prime number p each polynomial in Z[x] is expressible by the sum of a pth power of a polynomial in Z[x] and an irreducible over Q polynomial in Z[x].
In particular, selecting p � 2 (or p � 3), we can claim that each polynomial in Z[x] is the sum of a square (resp. cube) in Z[x] 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 [7]). Both these problems are wide open, see, e.g., [8][9][10][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. e same happens with Goldbach-type problems in polynomials with integer coefficients when much more is known compared to classical Goldbach problems for integers. ere is a considerable literature concerning this, see, for instance, [13][14][15][16][17][18][19][20][21][22].
roughout, without loss of generality, we may assume that f is nonconstant. Indeed, for f(x) � a ∈ Z, it suffices to take any constant polynomial g(x) � b ∈ Z. en, for each m ∈ N, the polynomial f(x) − g(x) m � a − b m is a constant, so it is irreducible over Q.
In Section 2, we give some auxiliary results. en, in Section 3, we complete the proofs of the theorems.

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

Lemma 1.
Let F(x, y) ∈ Z[x, y] be an irreducible over Q polynomial. en, there are infinitely many y 0 ∈ Z for which the polynomial F(x, y 0 ) ∈ Z[x] is irreducible over Q. e next lemma follows from the result of Davenport et al. [23].

Lemma 2. Let k ≥ 2 be an integer and let f ∈ Z[x] be a nonconstant polynomial such that, for each
Here is a more special version of the above result due to Perelli and Zannier [24]. Next, we recall a theorem of Capelli, which was generalized by Kneser, see p. 92 in [1].

Lemma 4.
Let K be a field and let m ≥ 2 be an integer. e polynomial x m − a, where a ∈ K, is irreducible over K except when, for some b ∈ K, either a � − 4b 4 and 4|m or a � b p with some prime p|m.
We conclude this section with several simple lemmas.

Lemma 5.
Let K be a field and let m ≥ 2 be an integer.
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 x 0 ∈ K, the degrees of the polynomials c 1 y n + u(y, x 0 ) and c 2 y k + v(y, x 0 ) are n and k, respectively. In particular, these polynomials are both nonconstant. is implies that their product F(x 0 , y) ∈ K[y] is reducible over K.
Here is a simple corollary of Lemma 4: Proof. Suppose that y p − f(x) is reducible. en, for each x 0 ∈ Z, by Lemma 5 with K � Q, c � 1, and m � p, the polynomial y p − f(x 0 ) ∈ Z[y] is reducible over Q. us, by Lemma 4 with K � Q and m � p, we must have f(x 0 ) � b p for some b ∈ Q. Moreover, from f(x 0 ) ∈ Z, it follows that b ∈ Z. erefore, by Lemma 2, we conclude that there is a We also have the following.
is irreducible over Q.
Proof. Denote the polynomial by u(x, y). Suppose u is reducible over Q. Fix any x 0 ∈ Z for which h(x 0 ) ≠ 0. From Lemma 5, it follows that u(x 0 , y) ∈ Z[y] must be reducible over Q. Since h(x 0 ) ∈ Z, the polynomial 2 Journal of Mathematics must be reducible over Q as well. However, by Eisenstein's criterion, this is not the case. Hence, u is irreducible over Q.

Proof of Theorems 1 and 2
Proof of eorem 1. Suppose first that the polynomial f(x) − y p ∈ Z[x, y] is irreducible over Q. en, by Lemma 1, for some y 0 ∈ Z the polynomial f(x) − y p 0 ∈ Z[x] is irreducible over Q, so we can simply take the constant polynomial g(x) � y 0 . e only alternative is indicated by Lemma 6. en, . Consider g(x) � h(x) + y with some y ∈ Z to be chosen later. It is clear that By Lemma 7 combined with Lemma 1, there is an integer □ Proof of eorem 2. If f(x) − y m ∈ Z[x, y] is irreducible over Q, then the argument is the same as that in the Proof of eorem 1. Suppose f(x) − y m is reducible over Q. en, by Lemma 5, for each x ∈ Z, the polynomial y m − f(x) ∈ Z[y] is reducible over Q. By Lemma 4, for each x ∈ Z, we have f(x) � a(x)ℓ(x) k(x) , where a(x) ∈ 1, − 4 { }, ℓ(x) ∈ Z and k(x) is prime divisor of m or k(x) � 4. us, by Lemma 3, we must have f(x) � ah(x) k for some integer a ≠ 0 and some polynomial h ∈ Z[x]. is is not the case by the assumption of the theorem, which completes the proof. □

Data Availability
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Conflicts of Interest
e authors declare that they have no conflicts of interest.