Nontrivial Effective Lower Bounds for the Least Common Multiple of Some Quadratic Sequences

Sid Ali Bousla and Bakir Farhi Laboratoire de Mathématiques appliquées Faculté des Sciences Exactes Université de Bejaia 06000 Bejaia Algeria

                    bouslasidali@gmail.com
                     bakir.farhi@gmail.com

Abstract

This paper is devoted to studying the numbers

L_{c,m,n} := lcm (m^{2} + c, (m + 1)^{2} + c, …, n^{2} + c) , where c, m, n are positive integers such that m ≤ n. More precisely, we determine a nontrivial rational divisor of L_{c,m,n} and then we derive (as consequences) some nontrivial lower bounds for L_{c,m,n}. Our approach (focusing on commutative algebra) is new and different from those using previously by Farhi, Oon, and Hong.

1 Introduction and Notation

Throughout this paper, we let ℕ^{∗} denote the set ℕ \ {0} of positive integers. For t ∈ ℝ, we let ⌊t⌋ and ⌈t⌉ respectively denote the floor and the ceiling functions. We say that an integer a is a multiple of a non-zero rational number r (or equivalently, r is a divisor of a) if the quotient a/r is an integer. If m, n, c are positive integers such that m ≤ n, we set L_{c,m,n} := lcm {m^{2} + c, (m + 1)^{2} + c, …, n^{2} + c}. For a given polynomial P ∈ ℂ[X], we let \overline{P} denote the polynomial conjugate of P in ℂ[X], that is the polynomial we get by replacing each coefficient of P with its complex conjugate. It is well-known that the conjugation of polynomials in ℂ[X] is compatible with addition and multiplication, in the sense that for every P, Q ∈ ℂ[X], we have \overline{P + Q} = \overline{P} + \overline{Q} and \overline{P · Q} = \overline{P} · \overline{Q}. Further, we let I, E_{h} (h ∈ ℝ), and Δ denote the linear operators on ℂ[X] which respectively represent the identity, the shift operator with step h (E_{h}P (X) = P (X + h), ∀P ∈ ℂ[X]), and the forward difference (ΔP (X) = P (X + 1) − P (X), ∀P ∈ ℂ[X]). For n ∈ ℕ, the expression of Δ^{n} in terms of the E_{h}’s is easily obtained from the binomial formula, as follows

Δ^{n} = (E_{1} − I)^{n} = ∑_{m=0}^{n}(−1)^{n−m}\binom{n}{m}E_{1}^{m} = ∑_{m=0}^{n}(−1)^{n−m}\binom{n}{m}E_{m}.  (1)

For falling factorial powers, we use Knuth’s notation:

X\underline{n} := X (X − 1) (X − 2) ⋯ (X − n + 1) (∀n ∈ ℕ).

The study of the least common multiple of the first n consecutive positive integers (n ∈ ℕ^{∗}) began with Chebyshev’s work [2] in his attempts to prove the prime number theorem. The latter showed that the prime number theorem is equivalent to stating that log lcm (1, 2, …, n) ∼_{+∞} n. More recently, many authors are interested in the effective estimates of the least common multiple of consecutive terms of some integer sequences. In 1972, Hanson [6] showed (by using the expansion of the number 1 in Sylvester series) that lcm (1, 2, …, n) ≤ 3^{n} (∀n ∈ ℕ^{∗}). In 1982, investigating the integral ∫_{0}^{1} x^{n}(1 − x)^{n}dx, Nair [11] gave a simple proof that lcm (1, 2, …, n) ≥ 2^{n} (∀n ≥ 7). Later, the second author [5] obtained nontrivial lower bounds for the least common multiple of consecutive terms in an arithmetic progression. In particular, he proved that for any u_{0}, r, n ∈ ℕ^{∗} such that gcd (u_{0}, r) = 1, we have

lcm (u_{0}, u_{0} + r, …, u_{0} + nr) ≥ u_{0} (r + 1)^{n−1} ,  (2)

and conjectured that the exponent (n − 1) appearing in the right-hand side of (2) can be replaced by n, which is the optimal exponent that can be obtained. That conjecture was confirmed by Hong and Feng [7]. Furthermore, several authors obtained improvements of (2) for n sufficiently large in terms of u_{0} and r (see, e.g., [7, 8, 10]). The second author [5] also obtained nontrivial lower bounds for the least common multiple of some quadratic sequences. In particular, he proved that for any positive integer n, we have

lcm (1^{2} + 1, 2^{2} + 1, …, n^{2} + 1) ≥ 0.32(1.442)^{n}.  (3)

In 2013, Oon [12] managed to improve (3) by proving that for any positive integers c and n, we have

lcm (1^{2} + c, 2^{2} + c, …, n^{2} + c) ≥ 2^{n}.  (4)

Actually, we have something a little stronger:

Theorem 1 (Oon [12]). Let c, n, m be positive integers such that m ≤ ⌈\frac{n}{2}⌉. Then we have

L_{c,m,n} ≥ 2^{n}.

Later, Hong et al. [9] managed to generalize Theorem 1 for polynomial sequences (f(n))_{n≥1}, with f ∈ ℤ[X] and the coefficients of f are all nonnegative. In another direction, various asymptotic estimates have been obtained by several authors. For example, Bateman et al.

[1] proved that for any h, k ∈ ℤ with k > 0, h + k > 0, and gcd(h, k) = 1, we have

log lcm\{h + k, h + 2k, …, h + nk\} ∼_{+∞} (\frac{k}{φ(k)} ∑_{\begin{matrix}1≤m≤k \\ gcd(m,k)=1 \end{matrix}} \frac{1}{m}) n,  (5)

where φ denotes the Euler totient function. Another asymptotic estimate a little harder to prove is due to Cilleruelo [3] and states that for every irreducible quadratic polynomial f ∈ ℤ[X], we have

log lcm{f(1), …, f(n)} = n log n + Bn + o(n),  (6)

where B is a constant depending on f.

In this paper, we use arguments of commutative algebra and complex analysis to find a nontrivial rational divisor of L_{c,m,n} (c, m, n ∈ ℕ^{∗}). As a consequence, we derive some new nontrivial lower bounds for L_{c,m,n}. The rest of the paper is organized in five parts (subsections). In the first part, we give an algebraic lemma which allows us, on the one hand to re-demonstrate Theorem 1 of Oon by an easy and purely algebraic method, and on the other hand to reformulate the problem of bounding the number L_{c,m,n} from below. In this reformulation, we are led to introduce a vital arithmetic function, noted h_{c}, whose multiple provides a divisor for L_{c,m,n}. In the next two parts, we study the arithmetic function h_{c} and we find for h_{c} a simple multiple. In the fourth part, we use the obtained multiple of h_{c} to deduce a nontrivial divisor for L_{c,m,n}. Our new nontrivial lower bounds for L_{c,m,n} then follow from this divisor. We conclude the paper with a short last part which presents comparisons between our results and those of Oon (cf. Theorem 1).

2 The results and the proofs

2.1 An algebraic method

Although the method used by Oon [12] to obtain his result (i.e., Theorem 1) is analytic, the ingredients for its success are algebraic in depth, as we will show it below by applying the following fundamental algebraic lemma:

Lemma 2. Let A be an integral domain and n be a positive integer. Let also u_{0}, u_{1}, …, u_{n}, a, b be elements of A. Suppose that a and b satisfy the following conditions:

1. Each of the elements u_{0}, u_{1}, …, u_{n} of A divides a.

2. Each of the elements ∏\begin{matrix}0≤j≤n \\ j≠i \end{matrix} (u_{i} − u_{j}) (i = 0, 1, …, n) of A divides b.

Then the product ab is a multiple of the product u_{0}u_{1} ⋯ u_{n}.

Proof. If the elements u_{0}, u_{1}, …, u_{n} of A are not pairwise distinct, the result of the lemma is trivial, since by its second condition we have b = 0_{A}. Suppose in what follows that the u_{i}’s (i = 0, 1, …, n) are pairwise distinct. We use the well-known result that if a polynomial in one indeterminate, with coefficients in an integral domain, has a number of zeros (in that domain) greater than its degree then it is zero. Since a is a multiple of each of the elements u_{0}, u_{1}, …, u_{n} of A, there exist k_{0}, k_{1}, …, k_{n} ∈ A such that:

a = k_{0}u_{0} = k_{1}u_{1} = ⋯ = k_{n}u_{n}.  (7)

Similarly, since b is a multiple of each of the elements ∏\begin{matrix}0≤j≤n \\ j≠i \end{matrix} (u_{i} − u_{j}) (i = 0, 1, …, n), then

there exist ℓ_{0}, ℓ_{1}, …, ℓ_{n} ∈ A such that:

b = ℓ_{i} ∏_{\begin{matrix}0≤j≤n \\ j≠i \end{matrix}} (u_{i} − u_{j}) (∀i ∈ \{0, 1, …, n\}) .  (8)

Now, consider the following polynomial of A[X]:

P (X) := ∑_{i=0}^{n} (ℓ_{i} ∏_{\begin{matrix}0≤j≤n \\ j≠i \end{matrix}} (X − u_{j})) − b.

We have deg P ≤ n. On the other hand, we have (according to (8))

P (u_{i}) = 0 (∀i ∈ {0, 1, …, n}) ,

showing that the number of zeros of P in A is greater than its degree. So, according to the elementary result of commutative algebra announced above, the polynomial P is zero. In particular, we have P (0) = 0; that is

b = (−1)^{n} ∑_{i=0}^{n} ℓ_{i} ( ∏_{\begin{matrix}0≤j≤n \\ j≠i \end{matrix}} u_{j}) .

By multiplying the two sides of this last equality by a, we get (according to (7))

ab = (−1)^{n} ∑_{i=0}^{n} ℓ_{i}a ( ∏_{\begin{matrix}0≤j≤n \\ j≠i \end{matrix}} u_{j})

= (−1)^{n} ∑_{i=0}^{n} ℓ_{i}k_{i}u_{i} ( ∏_{\begin{matrix}0≤j≤n \\ j≠i \end{matrix}} u_{j})

= (−1)^{n} (∑_{i=0}^{n} ℓ_{i}k_{i}) u_{0}u_{1} ⋯ u_{n},

showing that ab is a multiple of u_{0}u_{1} ⋯ u_{n}, as required. This completes the proof.

Remark 3. Lemma 2 is inspired by Farhi’s result [5, Theorem 2] which becomes a special

case for A = ℤ, a = lcm(u_{0}, u_{1}, …, u_{n}), and b = lcm(∏\begin{matrix}0≤j≤n \\ j≠i \end{matrix}(u_{i} − u_{j}); i = 0, 1, …, n). It

was precisely this special case which led the second author [5] to establish the first nontrivial lower bounds for the least common multiples of arithmetic progressions.

Now, we use Lemma 2 to establish a new proof of Theorem 1, which is purely algebraic.

A new proof of Theorem 1. Since L_{c,m,n} is non-increasing relative to m, then it suffices to prove the result of the theorem for m = ⌈\frac{n}{2}⌉, that is L_{c,⌈\frac{n}{2}⌉,n} ≥ 2^{n}. For simplicity, put

m_{0} = ⌈\frac{n}{2}⌉. So, we have to show that L_{c,m_{0},n} ≥ 2^{n}. For n ∈ \{1, 2, …, 6\}, this can be easily checked by hand (as was done by Oon). Suppose in what follows that n ≥ 7. It is well-known and easily proved that for any integer r ≥ 7, we have ⌈\frac{r}{2}⌉ \binom{r}{⌈\frac{r}{2}⌉} ≥ 2^{r}. According to this

inequality for r = n, it suffices to show that L_{c,m_{0},n} ≥ m_{0}\binom{n}{m_{0}}. More generally, we shall show that:

L_{c,m′,n} ≥ m′\binom{n}{m′} (∀m′ ∈ ℕ^{∗}, m′ ≤ n).  (9)

Let m′ ∈ ℕ^{∗} such that m′ ≤ n. To prove (9), we apply Lemma 2 for A = ℤ[\sqrt{−c}] by taking for the u_{i}’s the elements m′ + \sqrt{−c}, m′ + 1 + \sqrt{−c}, …, n + \sqrt{−c} of A and for a and b the integers a = L_{c,m′,n} and b = (n − m′)!. For any k ∈ {m′, m′ + 1, …, n}, Since L_{c,m′,n} is a multiple (in ℤ, so also in A = ℤ[\sqrt{−c}]) of (k^{2} + c) and k^{2} + c = (k + \sqrt{−c}) (k − \sqrt{−c}) is a multiple (in ℤ[\sqrt{−c}]) of k + \sqrt{−c}, then L_{c,m′,n} is a multiple (in ℤ[\sqrt{−c}]) of k + \sqrt{−c}. This shows that the first condition of Lemma 2 is satisfied. On the other hand, we have for all k ∈ {m′, m′ + 1, …, n}:

∏_{\begin{matrix}m′≤ℓ≤n \\ ℓ≠k \end{matrix}} ((k + \sqrt{−c}) − (ℓ + \sqrt{−c})) = ∏_{\begin{matrix}m′≤ℓ≤n \\ ℓ≠k \end{matrix}}(k − ℓ) = (−1)^{n−k}(k − m′)!(n − k)!,

which divides (in ℤ, so also in ℤ[\sqrt{−c}]) the integer (n−m′)! (since \frac{(n−m′)!}{(k−m′)!(n−k)!} = \binom{n−m′}{k−m′} ∈ ℤ). This shows that the second condition of Lemma 2 is also satisfied. We thus deduce (by applying Lemma 2) that L_{c,m′,n}(n − m′)! is a multiple (in ℤ[\sqrt{−c}]) of ∏_{k=m′}^{n} (k + \sqrt{−c}). So, there exist x, y ∈ ℤ such that:

L_{c,m′,n}(n − m′)! = (x + y\sqrt{−c}) ∏_{k=m′}^{n} (k + \sqrt{−c}) .  (10)

Therefore, by taking the absolute value in ℂ on both sides, we get

L_{c,m′,n}(n − m′)! = \sqrt{x^{2} + cy^{2}} ∏_{k=m′}^{n} \sqrt{k^{2} + c}.

Next, since x^{2} + cy^{2} ∈ ℕ and x^{2} + cy^{2} ≠ 0 (because x^{2} + cy^{2} = 0 ⟹ L_{c,m′,n} = 0, which is false) then x^{2} + cy^{2} ≥ 1. Hence

L_{c,m′,n} = \frac{\sqrt{x^{2} + cy^{2}} ∏_{k=m′}^{n} \sqrt{k^{2} + c}}{(n − m′)!} ≥ \frac{∏_{k=m′}^{n} \sqrt{k^{2} + c}}{(n − m′)!} ≥ \frac{∏_{k=m′}^{n} k}{(n − m′)!} = m′\binom{n}{m′},

as required. This completes the proof of the theorem.

Naturally, we have the following question:

How could we improve the lower bound L_{c,m,n} ≥ ^{\frac{∏_{k=m}^{n} \sqrt{k^{2}+c}}{(n−m)!}}, obtained during the proof of Theorem 1 and initially established by Oon [12]?

To simplify, suppose that c = 1 and let m, n ∈ ℕ^{∗} such that m ≤ n. According to Formula (10), the positive integer L_{1,m,n}(n − m)! is a multiple (in ℤ[i]) of the Gauss integer ∏_{k=m}^{n}(k + i). Next, by taking the conjugates (in ℂ) of both sides of (10), we obtain that L_{1,m,n}(n − m)! is also a multiple (in ℤ[i]) of the Gauss integer ∏_{k=m}^{n}(k − i). It follows from those two facts that L_{1,m,n}(n − m)! is a multiple (in ℤ[i]) of^{1}

lcm_{ℤ[i]} ( ∏_{k=m}^{n}(k + i), ∏_{k=m}^{n}(k − i)) = \frac{∏_{k=m}^{n}(k + i) · ∏_{k=m}^{n}(k − i)}{gcd_{ℤ[i]} (∏_{k=m}^{n}(k + i), ∏_{k=m}^{n}(k − i))}

= \frac{∏_{k=m}^{n}(k^{2} + 1)}{gcd_{ℤ[i]} (∏_{k=m}^{n}(k + i), ∏_{k=m}^{n}(k − i))}.

unit.

1Let A be an integral domain and let a, b ∈ A. Then an element d of A is called a greatest common divisor of a and b (and denoted by gcd_{A}(a, b)) if d divides both a and b and if any other d′ ∈ A, which divides both a and b, also divides d. Similarly, an element m of A is called a least common multiple of a and b (and denoted by lcm_{A}(a, b)) if m is a multiple of both a and b and if any other m′ ∈ A, which is a multiple of both a and b, is also a multiple of m. Note that gcd_{A}(a, b) and lcm_{A}(a, b) exist at least when A is a unique factorization domain (which is the case of ℤ[i]) and they are unique up to a multiplication by a

Consequently

L_{1,m,n} ≥ \frac{∏_{k=m}^{n}(k^{2} + 1)}{(n − m)! |gcd_{ℤ[i]} (∏_{k=m}^{n}(k + i), ∏_{k=m}^{n}(k − i))|}.  (11)

Remarkably, the trivial upper bound

|gcd_{ℤ[i]} ( ∏_{k=m}^{n}(k + i), ∏_{k=m}^{n}(k − i))| ≤ | ∏_{k=m}^{n}(k + i)| ≤ ∏_{k=m}^{n} \sqrt{k^{2} + 1}

suffices to establish the Oon lower bound L_{1,m,n} ≥ ^{\frac{∏_{k=m}^{n} \sqrt{k^{2}+1}}{(n−m)!}}. So, a nontrivial upper bound for the number |gcd_{ℤ[i]} (∏_{k=m}^{n}(k + i), ∏_{k=m}^{n}(k − i))| certainly gives an improvement of the Oon theorem. On the other hand, for a, b ∈ ℤ such that (a, b) ≠ (0, 0), we can easily check that gcd_{ℤ[i]} (a + bi, a − bi) is not far from gcd_{ℤ}(a, b). Precisely, we have

gcd_{ℤ[i]} (a + bi, a − bi) = (σ + iτ) gcd_{ℤ}(a, b),

where σ, τ ∈ {−1, 0, 1} and (σ, τ) ≠ (0, 0). So, for the case c = 1, we are led to study the arithmetic function:

\begin{matrix}h : ℤ[i] \ \{0\} & ⟶ & ℕ^{∗} \\ a + bi & ⟼ & gcd(a, b) \end{matrix} ,

and precisely to find nontrivial upper bounds for the quantities h (∏_{k=m}^{n}(k + i)) (m, n ∈ ℕ^{∗}, m ≤ n). For the general case (c ∈ ℕ^{∗}), the arithmetic function we need to study is given by:

\begin{matrix}h_{c} : ℤ[\sqrt{−c}] \ \{0\} & ⟶ & ℕ^{∗} \\ a + b\sqrt{−c} & ⟼ & gcd(a, b) \end{matrix}

and the quantities we need to bound from above are h_{c} (∏_{k=m}^{n}(k + \sqrt{−c})) (m, n ∈ ℕ^{∗}, m ≤ n).

The following proposition has as objective to replace a specific arithmetic language of the ring ℤ[\sqrt{−c}] by its simpler analog in ℤ.

Proposition 4. Let c ∈ ℕ^{∗} and N, a, b ∈ ℤ, with (a, b) ≠ (0, 0). Then N is a multiple (in ℤ[\sqrt{−c}]) of (a + b\sqrt{−c}) if and only if N is a multiple (in ℤ) of \frac{a^{2}+cb^{2}}{gcd(a,b)}.

Proof. The result of the proposition is trivial for b = 0. Suppose in what follows that b ≠ 0.

Suppose that N is a multiple (in ℤ[\sqrt{−c}]) of (a + b\sqrt{−c}); that is there exist x, y ∈ ℤ such that:

N = (x + y\sqrt{−c}) (a + b\sqrt{−c}) .

By identifying the real and imaginary parts of the two hand-sides of this equality, we get

N = ax − byc,  (12)
0 = bx + ay.  (13)

Next, putting d := gcd(a, b), there exist a′, b′ ∈ ℤ, with b′ ≠ 0 and gcd(a′, b′) = 1, such that a = da′ and b = db′. By substituting these in (13), we obtain (after simplifying)

b′x = −a′y.  (14)

This last equality shows that b′ divides a′y. But since gcd(a′, b′) = 1, then (according to the Gauss lemma) b′ divides y. So there exists k ∈ ℤ such that y = kb′. By reporting this in (14), we get x = −ka′. Then by substituting x = −ka′ = −k \frac{a}{d} and y = kb′ = k \frac{b}{d} in (12), we finally obtain

N = −k \frac{a^{2} + cb^{2}}{d} = −k \frac{a^{2} + cb^{2}}{gcd(a, b)},

showing that N is a multiple (in ℤ) of \frac{a^{2}+cb^{2}}{gcd(a,b)}, as required.

Conversely, suppose that N is a multiple (in ℤ) of \frac{a^{2}+cb^{2}}{gcd(a,b)}. Then there exists k ∈ ℤ such that:

N = k \frac{a^{2} + cb^{2}}{gcd(a, b)} = k \frac{a − b\sqrt{−c}}{gcd(a, b)} (a + b\sqrt{−c}) = (k \frac{a}{gcd(a, b)} − k \frac{b}{gcd(a, b)}\sqrt{−c}) (a + b\sqrt{−c}) .

Since (k \frac{a}{gcd(a,b)} − k \frac{b}{gcd(a,b)} \sqrt{−c}) ∈ ℤ[\sqrt{−c}], the last equality shows that N is a multiple (in

ℤ[\sqrt{−c}]) of (a + b\sqrt{−c}), as required. This completes the proof of the proposition.

From Proposition 4, we derive the following corollary, which is the first key step to obtaining the results of this paper.

Corollary 5. Let c, m, n ∈ ℕ^{∗} such that m ≤ n. Then the positive integer L_{c,m,n}(n − m)! is a multiple (in ℤ) of the positive integer:

\frac{∏_{k=m}^{n} (k^{2} + c)}{h_{c} (∏_{k=m}^{n} (k + \sqrt{−c}))}.

Proof. Formula (10) (obtained during our new proof of Theorem 1) shows that L_{c,m,n}(n−m)! is a multiple (in ℤ[\sqrt{−c}]) of ∏_{k=m}^{n} (k + \sqrt{−c}). But, according to Proposition 4, this last property is equivalent to the statement of the corollary.

In view of Corollary 5, to bound from below L_{c,m,n} (c, m, n ∈ ℕ^{∗}, m ≤ n), it suffices to bound from above h_{c} (∏_{k=m}^{n}(k + \sqrt{−c})). Likewise, to find a nontrivial (rational) divisor of L_{c,m,n}, it suffices to find a nontrivial multiple of h_{c} (∏_{k=m}^{n}(k + \sqrt{−c})). This is what we will do in what follows.

2.2 An explicit Bézout identity

In the following, let c ∈ ℕ^{∗} and k ∈ ℕ be fixed and define

P_{k} (X) := (X + \sqrt{−c}) (X − 1 + \sqrt{−c}) ⋯ (X − k + \sqrt{−c}) := A_{k} (X) + B_{k} (X) \sqrt{−c},
\overline{P_{k}} (X) := (X − \sqrt{−c}) (X − 1 − \sqrt{−c}) ⋯ (X − k − \sqrt{−c}) := A_{k} (X) − B_{k} (X) \sqrt{−c},

where we understand that A_{k}, B_{k} ∈ ℤ[X]. In what follows, we find nontrivial multiples for the positive integers h_{c} (P_{k}(n)) = gcd (A_{k}(n), B_{k}(n)) (n ≥ 1). To do so, we look for two polynomial sequences (a_{k}(n))_{n} and (b_{k}(n))_{n} so that the polynomial sequence (a_{k}(n)A_{k}(n) + b_{k}(n)B_{k}(n))_{n} be independent on n. This leads to looking for two polynomials U_{k}, V_{k} ∈ ℚ[X] which satisfy the Bézout identity:

U_{k} (X) A_{k} (X) + V_{k} (X) B_{k} (X) = 1.

Next, since A_{k} = \frac{P_{k}+\overline{P_{k}}}{2} and B_{k} = \frac{P_{k}−\overline{P_{k}}}{2\sqrt{−c}}, the problem is equivalent to looking for σ_{k}, τ_{k} ∈ ℚ (\sqrt{−c}) [X] such that:

σ_{k} (X) P_{k} (X) + τ_{k} (X) \overline{P_{k}} (X) = 1.

Let us first justify the existence of such σ_{k} and τ_{k}. Denoting by Z (P) the set of all the complex zeros of a polynomial P ∈ ℂ[X], we have

Z (P_{k}) = \{−\sqrt{−c}, 1 − \sqrt{−c}, …, k − \sqrt{−c}\} and Z (\overline{P_{k}}) = \{\sqrt{−c}, 1 + \sqrt{−c}, …, k + \sqrt{−c}\} ,

showing that Z (P_{k}) ∩ Z (\overline{P_{k}}) = ∅; that is P_{k} and \overline{P_{k}} do not have a common zero in ℂ. This implies that P_{k} and \overline{P_{k}} are coprime in ℂ[X]; so coprime also in ℚ (\sqrt{−c}) [X] (since P_{k}, \overline{P_{k}} ∈ ℚ (\sqrt{−c}) [X]). It follows (according to Bézout’s theorem) that there exist σ_{k}, τ_{k} ∈ ℚ (\sqrt{−c}) [X] such that: σ_{k}P_{k} + τ_{k}\overline{P_{k}} = 1, as required.

Now, to find explicitly such σ_{k} and τ_{k}, we need the following more precise version of Bézout’s theorem:

Theorem 6. Let 𝕂 be a field and P and Q be two non-constant polynomials of 𝕂[X] such that gcd_{𝕂[X]} (P, Q) = 1. Then there exists a unique pair (U, V) of polynomials of 𝕂[X], with deg U < deg Q and deg V < deg P, such that:

P U + QV = 1.

Proof. Since gcd_{𝕂[X]} (P, Q) = 1, then (according to Bézout’s theorem) there exist U_{0}, V_{0} ∈ 𝕂[X] such that:

P U_{0} + QV_{0} = 1.

Next, consider in 𝕂[X] the euclidean division of U_{0} by Q and the euclidean division of V_{0} by (−P):

U_{0} = U_{1}Q + U,
V_{0} = V_{1} (−P) + V,

where U_{1}, V_{1}, U, V ∈ 𝕂[X], deg U < deg Q, and deg V < deg (−P) = deg P. So, we have

P U +QV = P (U_{0} − U_{1}Q)+Q (V_{0} + V_{1}P) = P Q (V_{1} − U_{1})+P U_{0} +QV_{0} = P Q (V_{1} − U_{1})+1.

If V_{1} − U_{1} ≠ 0, then the last equality implies that deg (P U + QV) ≥ deg (P Q), which is impossible, since deg U < deg Q and deg V < deg P. Thus V_{1} − U_{1} = 0, which gives P U + QV = 1. The existence of the pair (U, V) as required by the theorem is proved. It remains to prove the uniqueness of (U, V). Let (U_{∗}, V_{∗}) another pair of polynomials of 𝕂[X], with deg U_{∗} < deg Q, deg V_{∗} < deg P, and P U_{∗} + QV_{∗} = 1, and let us prove that (U_{∗}, V_{∗}) = (U, V). We have

P (UV_{∗} − U_{∗}V) = (P U) V_{∗} − (P U_{∗}) V = (1 − QV) V_{∗} − (1 − QV_{∗}) V = V_{∗} − V,

showing that the polynomial (V_{∗} − V) is a multiple of P in 𝕂[X]. But since deg (V_{∗} − V) < deg P (because deg V < deg P and deg V_{∗} < deg P), we have inevitably V_{∗} − V = 0; hence V_{∗} = V. Using this, we get P U_{∗} = 1 − QV_{∗} = 1 − QV = P U. Thus U_{∗} = U. Consequently, we have (U_{∗}, V_{∗}) = (U, V), as required. This completes the proof of the theorem.

In our context, the application of Theorem 6 gives the following corollary:

Corollary 7. There exists a unique polynomial σ_{k} ∈ ℂ[X], with degree ≤ k, such that:

σ_{k}P_{k} + \overline{σ_{k}}\overline{P_{k}} = 1.

Proof. According to Theorem 6 (applied for 𝕂 = ℂ and (P, Q) = (P_{k}, \overline{P_{k}})), there exists a unique pair (σ_{k}, τ_{k}) of polynomials of ℂ[X], with deg σ_{k} < deg \overline{P_{k}} = k + 1 and deg τ_{k} < deg P_{k} = k + 1, such that σ_{k}P_{k} + τ_{k}\overline{P_{k}} = 1. By taking the conjugates in ℂ[X] of both sides of the last equality, we derive that \overline{σ_{k}}\overline{P_{k}} + \overline{τ_{k}}P_{k} = 1, that is \overline{τ_{k}}P_{k} + \overline{σ_{k}}\overline{P_{k}} = 1. Since deg \overline{τ_{k}} = deg τ_{k} < k + 1 and deg \overline{σ_{k}} = deg σ_{k} < k + 1, this shows that the pair (\overline{τ_{k}}, \overline{σ_{k}}) satisfies the property that characterizes (σ_{k}, τ_{k}). Thus (\overline{τ_{k}}, \overline{σ_{k}}) = (σ_{k}, τ_{k}), that is τ_{k} = \overline{σ_{k}}. Consequently, we have σ_{k}P_{k} + \overline{σ_{k}}\overline{P_{k}} = 1. This completes the proof of the corollary.

Now, we are going to determine the explicit expression of the polynomial σ_{k} announced by Corollary 7. By replacing, in the identity σ_{k} (X) P_{k} (X) + \overline{σ_{k}} (X) \overline{P_{k}} (X) = 1, the indeterminate X by the numbers s + \sqrt{−c} (s = 0, 1, …, k), we get

σ_{k} (s + \sqrt{−c}) = \frac{1}{P_{k} (s + \sqrt{−c})} (∀s ∈ \{0, 1, …, k\}).  (15)

(since \overline{P_{k}} (s + \sqrt{−c}) = 0 for s = 0, 1, …, k). So the values of σ_{k} are known for (k + 1) equidistant points with distance 1. Since deg σ_{k} ≤ k, this is sufficient to determine the expression of σ_{k} (X) by using for example the Newton forward interpolation formula. Doing so, we obtain that:

σ_{k} (X) = ∑_{ℓ=0}^{k} \frac{(Δ^{ℓ}σ_{k}) (\sqrt{−c})}{ℓ!} (X − \sqrt{−c})\underline{ℓ} .

Then by using (1), we derive that:

σ_{k} (X) = ∑_{ℓ=0}^{k} ∑_{j=0}^{ℓ} \frac{(−1)^{ℓ−j}}{ℓ!} \binom{ℓ}{j}σ_{k} (j + \sqrt{−c}) (X − \sqrt{−c})\underline{ℓ}

= ∑_{ℓ=0}^{k} (\frac{1}{ℓ!} ∑_{j=0}^{ℓ}(−1)^{ℓ−j}\binom{ℓ}{j}σ_{k} (j + \sqrt{−c})) (X − \sqrt{−c})\underline{ℓ}

= ∑_{ℓ=0}^{k} (\frac{1}{ℓ!} ∑_{j=0}^{ℓ}(−1)^{ℓ−j}\binom{ℓ}{j}\frac{1}{P_{k} (j + \sqrt{−c})}) (X − \sqrt{−c})\underline{ℓ}

(according to (15)). So, by setting for all ℓ ∈ {0, 1, …, k}:

Θ_{k,ℓ} := \frac{1}{ℓ!} ∑_{j=0}^{ℓ}(−1)^{ℓ−j}\binom{ℓ}{j}\frac{1}{P_{k}(j + \sqrt{−c})},  (16)

we get

σ_{k} (X) = ∑_{ℓ=0}^{k} Θ_{k,ℓ} (X − \sqrt{−c})\underline{ℓ} .  (17)

It remains to simplify the expressions of the numbers Θ_{k,ℓ} (0 ≤ ℓ ≤ k). To do so, we introduce the functions R_{k,ℓ} (0 ≤ ℓ ≤ k), defined by:

R_{k,ℓ}(z) := \frac{1}{ℓ!} ∑_{j=0}^{ℓ}(−1)^{ℓ−j}\binom{ℓ}{j}\frac{1}{P_{k}(z + j + \sqrt{−c})},  (18)

so that we have

Θ_{k,ℓ} = R_{k,ℓ}(0) (∀ℓ ∈ {0, 1, …, k}).  (19)

Since the R_{k,ℓ}’s (0 ≤ ℓ ≤ k) are all rational functions, each of them is holomorphic on its domain of definition, that is each R_{k,ℓ} (0 ≤ ℓ ≤ k) is holomorphic on

D_{ℓ} := ℂ \ \{j − 2\sqrt{−c} ; j ∈ ℤ and − ℓ ≤ j ≤ k\} .

Consequently, the common domain of holomorphy of the functions R_{k,ℓ} (0 ≤ ℓ ≤ k) is the open connected region D of ℂ, given by:

D := ⋂_{0≤ℓ≤k} D_{ℓ} = ℂ \ \{j − 2\sqrt{−c} ; j ∈ ℤ and − k ≤ j ≤ k\}.

Using the principle of analytical continuation together with the theory of the gamma and beta functions, we can find another expression of R_{k,ℓ} (0 ≤ ℓ ≤ k), which is simpler than the above. We have the following proposition:

Proposition 8. For all ℓ ∈ ℕ, with ℓ ≤ k, and all z ∈ D, we have

R_{k,ℓ}(z) = \frac{(−1)^{k+ℓ}}{z + 2\sqrt{−c}}\binom{k + ℓ}{ℓ}\frac{1}{(k − 2\sqrt{−c} − z)\underline{k} (ℓ + 2\sqrt{−c} + z)\underline{ℓ}} .  (20)

Proof. Let ℓ ∈ ℕ such that ℓ ≤ k. According to the principle of analytical continuation, it suffices to prove Formula (20) for z ∈ ℂ, such that ℜ(z) > k. For a such z, we have

R_{k,ℓ}(z) := \frac{1}{ℓ!} ∑_{j=0}^{ℓ}(−1)^{ℓ−j}\binom{ℓ}{j}\frac{1}{P_{k}(z + j + \sqrt{−c})}

= \frac{1}{ℓ!} ∑_{j=0}^{ℓ}(−1)^{ℓ−j}\binom{ℓ}{j}\frac{1}{(z + j + 2\sqrt{−c}) (z + j − 1 + 2\sqrt{−c}) ⋯ (z + j − k + 2\sqrt{−c})}

= \frac{1}{ℓ!} ∑_{j=0}^{ℓ}(−1)^{ℓ−j}\binom{ℓ}{j}\frac{Γ (z + j − k + 2\sqrt{−c})}{Γ (z + j + 1 + 2\sqrt{−c})}

= \frac{1}{ℓ!} ∑_{j=0}^{ℓ}(−1)^{ℓ−j}\binom{ℓ}{j}\frac{1}{k!}β (z + j − k + 2\sqrt{−c}, k + 1)

= \frac{1}{k!ℓ!} ∑_{j=0}^{ℓ} ((−1)^{ℓ−j}\binom{ℓ}{j} ∫_{0}^{1} t^{z+j−k−1+2\sqrt{−c}}(1 − t)^{k}dt)

= \frac{1}{k!ℓ!} ∫_{0}^{1} t^{z−k−1+2\sqrt{−c}}(1 − t)^{k} (∑_{j=0}^{ℓ}(−1)^{ℓ−j}\binom{ℓ}{j}t^{j}) dt

= \frac{1}{k!ℓ!} ∫_{0}^{1} t^{z−k−1+2\sqrt{−c}}(1 − t)^{k} (t − 1)^{ℓ} dt

= \frac{(−1)^{ℓ}}{k!ℓ!} ∫_{0}^{1} t^{z−k−1+2\sqrt{−c}}(1 − t)^{k+ℓ}dt

= \frac{(−1)^{ℓ}}{k!ℓ!} β (z − k + 2\sqrt{−c}, k + ℓ + 1)

= \frac{(−1)^{ℓ}}{k!ℓ!} \frac{Γ (z − k + 2\sqrt{−c}) Γ (k + ℓ + 1)}{Γ (z + ℓ + 1 + 2\sqrt{−c})}

= (−1)^{ℓ}\binom{k + ℓ}{ℓ}\frac{1}{(z + ℓ + 2\sqrt{−c}) (z + ℓ − 1 + 2\sqrt{−c}) ⋯ (z − k + 2\sqrt{−c})}

= \frac{(−1)^{k+ℓ}}{z + 2\sqrt{−c}}\binom{k + ℓ}{ℓ}\frac{1}{(k − 2\sqrt{−c} − z)\underline{k} (ℓ + 2\sqrt{−c} + z)\underline{ℓ}} ,

as required. This completes the proof.

From Proposition 8, we immediately derive a simpler explicit expression of σ_{k} (X). We have the following corollary:

Corollary 9. We have

σ_{k} (X) = \frac{1}{2\sqrt{−c} (k − 2\sqrt{−c})\underline{k}} ∑_{ℓ=0}^{k} \frac{(−1)^{k+ℓ}\binom{k+ℓ}{ℓ}}{(ℓ + 2\sqrt{−c})\underline{ℓ}} (X − \sqrt{−c})\underline{ℓ} .

Proof. This immediately follows from Formulas (17), (19), and (20).

2.3 Nontrivial multiples of some values of h_{c}

In this subsection, we preserve the notation of Subsection 2.2. From Corollary 9, we derive the following theorem:

Theorem 10. For all c, n, m ∈ ℕ^{∗}, with m ≤ n, we have

h_{c} (∏_{ℓ=m}^{n} (ℓ + \sqrt{−c})) divides c ∏_{ℓ=1}^{n−m}(ℓ^{2} + 4c).

Proof. Let c, n, m ∈ ℕ^{∗}, with m ≤ n. Putting k := n−m ∈ ℕ and d := c ∏_{ℓ=1}^{n−m}(ℓ^{2}+4c) ∈ ℕ^{∗}, we have ∏_{ℓ=m}^{n} (ℓ + \sqrt{−c}) = P_{k}(n); so, we have to show that h_{c} (P_{k}(n)) divides d. By noting that 2d = \sqrt{−c} · 2\sqrt{−c} (k − 2\sqrt{−c})\underline{k} (k + 2\sqrt{−c})\underline{k}, we derive from Corollary 9 that 2dσ_{k} ∈ ℤ[\sqrt{−c}][X]. So, there exist r_{k}, s_{k} ∈ ℤ[X] such that:

2dσ_{k} (X) = r_{k} (X) + s_{k} (X) \sqrt{−c}.

Next, the identity of polynomials σ_{k}P_{k} + \overline{σ_{k}}\overline{P_{k}} = 1 (given by Corollary 7) implies that 2dσ_{k} · P_{k} + \overline{2dσ_{k}} · \overline{P_{k}} = 2d. By substituting in this last equality P_{k} by (A_{k} + B_{k}\sqrt{−c}) and 2dσ_{k} by (r_{k} + s_{k}\sqrt{−c}), we obtain (in particular) that:

r_{k}A_{k} − cs_{k}B_{k} = d,

implying that gcd_{ℤ[X]} (A_{k}, B_{k}) divides d. We then conclude that gcd_{ℤ} (A_{k}(n), B_{k}(n)) = h_{c} (P_{k}(n)) divides d, as required.

2.4 New estimates for the number L_{c,m,n}

We have the following theorem:

Theorem 11. Let c, m, n ∈ ℕ^{∗} such that m ≤ n. Then:

1. The positive integer L_{c,m,n} is a multiple of the rational number

\frac{∏_{k=m}^{n} (k^{2} + c)}{c · (n − m)! ∏_{k=1}^{n−m} (k^{2} + 4c)}.

2. We have

L_{c,m,n} ≥ λ_{1}(c) · m^{2} \frac{n!^{2}}{m!^{2}(n − m)!^{3}} ,

where λ_{1}(c) := e^{− \frac{2π^{2}}{3} c}/c.

Proof. The first point of the theorem is an immediate consequence of Corollary 5 and Theorem 10. Next, using the well-known inequality 1 + x ≤ e^{x} (∀x ∈ ℝ), we have

∏_{k=1}^{n−m} (k^{2} + 4c) = ∏_{k=1}^{n−m} k^{2} (1 + \frac{4c}{k^{2}})

= (n − m)!^{2} ∏_{k=1}^{n−m} (1 + \frac{4c}{k^{2}})

≤ (n − m)!^{2} ∏_{k=1}^{n−m} e^{\frac{4c}{k^{2}}}

≤ (n − m)!^{2} ∏_{k=1}^{+∞} e^{\frac{4c}{k^{2}}}
= (n − m)!^{2}e^{4c ∑_{k=1}^{+∞} \frac{1}{k^{2}}}
= (n − m)!^{2}e^{4c( \frac{π^{2}}{6} )}
= (n − m)!^{2}e^{\frac{2π^{2}}{3} c}.

Hence

\frac{∏_{k=m}^{n} (k^{2} + c)}{c · (n − m)! ∏_{k=1}^{n−m} (k^{2} + 4c)} ≥ \frac{∏_{k=m}^{n} k^{2}}{c · (n − m)! · (n − m)!^{2}e^{\frac{2π^{2}}{3} c}}

= \frac{m^{2} (\frac{n!}{m!})^{2}}{c · (n − m)!^{3}e^{\frac{2π^{2}}{3} c}}

= \frac{e^{− \frac{2π^{2}}{3} c}}{c} · m^{2} \frac{n!^{2}}{m!^{2}(n − m)!^{3}} .

The second point of the theorem then follows from the first one. This completes the proof.

We shall now impose conditions on m (in terms of n) in order to optimize (resp., simplify) the estimate of the second point of Theorem 11. To do so, we first need to get rid of the factorials in that estimate. We have the following:

Corollary 12. Let c, n, m ∈ ℕ^{∗} such that m < n. Then we have

L_{c,m,n} ≥ λ_{2}(c) · \frac{nm}{(n − m)^{3/2}} (\frac{m^{2}}{(n − m)^{3}} )^{n−m} e^{3(n−m)},  (21)

where λ_{2}(c) := \frac{e^{− \frac{2π^{2}}{3} c− \frac{5}{12}}}{(2π)^{3/2}c}.

Proof. Starting from the lower bound established by Theorem 11 for L_{c,m,n} and estimating each of its factorial terms by using the well-known double inequality:

k^{k}e^{−k}\sqrt{2πk} ≤ k! ≤ k^{k}e^{−k}\sqrt{2πk}e^{\frac{1}{12k}} (∀k ∈ ℕ^{∗})

(see, e.g., [4, Problem 1.15]), we get

L_{c,m,n} ≥ λ_{1}(c)(2π)^{−3/2} · \frac{nm}{(n − m)^{3/2}} · (\frac{n}{m})^{2n} · (\frac{m^{2}}{(n − m)^{3}} )^{n−m} e^{n−m} · e^{− \frac{1}{6m} − \frac{1}{4(n−m)}} .

Next, since e^{− \frac{1}{6m} − \frac{1}{4(n−m)}} ≥ e^{− \frac{1}{6} − \frac{1}{4}} = e^{− \frac{5}{12}} and (\frac{n}{m})^{2n} = e^{−2n log( \frac{m}{n})} ≥ e^{−2n( \frac{m}{n} −1)} = e^{2(n−m)}, then we deduce that:

L_{c,m,n} ≥ λ_{1}(c)(2π)^{−3/2}e^{−5/12} · \frac{nm}{(n − m)^{3/2}} (\frac{m^{2}}{(n − m)^{3}} )^{n−m} e^{3(n−m)},

as required.

In the context of Corollary 12, by supposing that n − m is of order of magnitude n^{α} for large n (where 0 < α < 1), then the dominant part of the lower bound (21) for L_{c,m,n} is

(\frac{m^{2}}{(n−m)^{3}} )^{n−m} and has order of magnitude n^{(2−3α)n^{α}}. So, to have an optimal estimate, we must take α less than but not too far from \frac{2}{3} (a study of the function α ⟼ (2 − 3α)n^{α} shows that the best value of α is α = \frac{2}{3} − \frac{1}{log n}). A concrete result specifying this heuristic reasoning is given by the following theorem:

Theorem 13. Let c, m, n ∈ ℕ^{∗} such that m ≤ n − \frac{1}{2}n^{2/3}. Then we have

L_{c,m,n} ≥ λ_{3}(c) · (n − \frac{1}{2}n^{2/3}) · (2e^{3})^{⌊\frac{1}{2} n^{2/3}⌋} ,

where λ_{3}(c) := ^{\frac{e^{− \frac{2π^{2}}{3} c− \frac{5}{12}}}{π^{3/2}c}}.

Proof. A simple calculation shows that the result of the theorem is true for n < 3. Suppose in what follows that n ≥ 3 and let m_{n} := n − ⌊\frac{1}{2}n^{2/3}⌋ < n; so m ≤ m_{n}. From Corollary 12, we have

L_{c,m_{n},n} ≥ λ_{2}(c)\frac{n (n − ⌊\frac{1}{2}n^{2/3}⌋)}{⌊\frac{1}{2}n^{2/3}⌋^{3/2}} (\begin{matrix}(n − ⌊\frac{1}{2}n^{2/3}⌋) & 2 \\ ⌊\frac{1}{2}n^{2/3}⌋ & 3 \end{matrix} )^{⌊\frac{1}{2} n^{2/3}⌋} e^{3⌊\frac{1}{2} n^{2/3}⌋}

≥ λ_{2}(c)\frac{n (n − \frac{1}{2}n^{2/3})}{(\frac{1}{2}n^{2/3})^{3/2}} (\begin{matrix}(n − \frac{1}{2}n^{2/3}) & 2 \\ (\frac{1}{2}n^{2/3}) & 3 \end{matrix} )^{⌊\frac{1}{2} n^{2/3}⌋} e^{3⌊\frac{1}{2} n^{2/3}⌋}

= 2^{3/2}λ_{2}(c) (n − \frac{1}{2}n^{2/3}) (8 (1 − \frac{1}{2n^{1/3}} )^{2})^{⌊\frac{1}{2} n^{2/3}⌋} e^{3⌊\frac{1}{2} n^{2/3}⌋}.

But since 1 − \frac{1}{2n^{1/3}} ≥ \frac{1}{2} (because n ≥ 1), we deduce that:

L_{c,m_{n},n} ≥ 2^{3/2}λ_{2}(c) (n − \frac{1}{2}n^{2/3}) (2e^{3})^{⌊\frac{1}{2} n^{2/3}⌋} .

The required result follows from the trivial fact that L_{c,m,n} ≥ L_{c,m_{n},n} (since m ≤ m_{n}).

In another direction, we derive from Corollary 12 the following theorem, which completes (in a way) Theorem 13 above.

Theorem 14. Let c, m, n ∈ ℕ^{∗} such that n − \frac{1}{2}n^{2/3} ≤ m ≤ n. Then we have

L_{c,m,n} ≥ λ_{2}(c) · ne^{3(n−m)},

where λ_{2}(c) is defined in Corollary 12.

Proof. The result of the theorem is trivial for m = n. Suppose in what follows that m < n; so we have n ≥ 2. Now, let f : [0, n] ⟶ ℝ be the function defined by f(x) = x^{2} − (n − x)^{3} (∀x ∈ [0, n]). Since f′(x) = 2x + 3(n − x)^{2} > 0 (∀x ∈ [0, n]), then f is increasing. Next, we have

f (n − \frac{1}{2}n^{2/3}) = (n − \frac{1}{2}n^{2/3})^{2} − (\frac{1}{2}n^{2/3})^{3}

= n^{2} − n^{5/3} + \frac{1}{4}n^{4/3} − \frac{1}{8}n^{2}

= \frac{7}{8}n^{2} − n^{5/3} + \frac{1}{4}n^{4/3}.

But since n^{2} ≥ \frac{8}{7}n^{5/3} (because n ≥ 2), it follows that f (n − \frac{1}{2}n^{2/3}) ≥ \frac{1}{4}n^{4/3} > 0. So, the increase of f ensures that f(m) > 0 (since m ≥ n − \frac{1}{2}n^{2/3} by hypothesis). Thus \frac{m^{2}}{(n−m)^{3}} > 1 and \frac{m}{(n−m)^{3/2}} > 1. By reporting these into (21), we then conclude that:

L_{c,m,n} ≥ λ_{2}(c) · ne^{3(n−m)},

as required. This completes the proof.

2.5 Comparison with Oon’s lower bound

In the application of Oon’s lower bound (i.e., Theorem 1), the number of the terms included in the least common multiple lcm(m^{2} + c, …, n^{2} + c) = L_{c,m,n} must be ≥ n − ⌈\frac{n}{2}⌉ + 1 > \frac{n}{2}; whereas when we put our theorems 13 and 14 together, this constraint is deleted. However, if the condition of application of Oon’s theorem holds then we obtain a lower bound for L_{c,m,n} stronger than those of our theorems. Further, our key result is rather the point 1 of Theorem 11 which provides a nontrivial rational divisor of L_{c,m,n}. Here, we have exploited this key result in a naive way. It is likely that “a smarter way” will provide better results.

3 Acknowledgments

The authors would like to thank the anonymous reviewer and the editor for their helpful comments and suggestions.

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2010 Mathematics Subject Classification: Primary 11A05; Secondary 11B83, 13G05, 30B40, 33B15. Keywords: least common multiple, quadratic sequence.

Received January 30 2020; revised versions received May 8 2020; May 14 2020. Published in Journal of Integer Sequences, June 11 2020.

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