The Least Common Multiple of a Quadratic Sequence

The least common multiple of a quadratic sequence

Javier Cilleruelo

Abstract For any irreducible quadratic polynomial f(x) in Z[x] we obtain the estimate log l.c.m. (f(1), . . . , f(n)) = n log n + Bn + o(n) where B is a constant depending on f.

1. Introduction

The problem of estimating the least common multiple of the firstPn positive integers was first investigated by Chebyshev [Che52] when he introduced the function Ψ(n) = pm6n log p = log l.c.m.(1, . . . , n) in his study on the distribution of prime numbers. The prime number theorem asserts that Ψ(n) ∼ n, so the asymptotic estimate log l.c.m.(1, . . . , n) ∼ n is equivalent to the prime number theorem. The analogous asymptotic estimate for any linear polynomial f(x) = ax + b is also known [Bat02] and it is a consequence of the prime number theorem for arithmetic progressions: q X 1 log l.c.m.(f(1), . . . , f(n)) ∼ n , (1) φ(q) k 16k6q (k,q)=1 where q = a/(a, b). We address here the problem of estimating log l.c.m.(f(1), . . . , f(n)) when f is an irreducible quadratic polynomial in Z[x]. When f is a reducible quadratic polynomial the asymptotic estimate is similar to that we obtain for linear polynomials. This case is studied in section §4 with considerably less effort than the irreducible case. We state our main theorem.

Theorem 1. For any irreducible quadratic polynomial f(x) = ax2 + bx + c in Z[x] we have log l.c.m. (f(1), . . . , f(n)) = n log n + Bn + o(n) where B = Bf is deﬁned by the formula X (d/p) log p 1 X ³ r ´ B = γ − 1 − 2 log 2 − + log 1 + (2) f p − 1 φ(q) q p 16r6q (r,q)=1 X ³1 + (d/p) X s(f, pk)´ + log a + log p − . p − 1 pk p|2aD k>1 In this formula γ is the Euler constant, D = b2 − 4ac = dl2, where d is a fundamental discriminant, (d/p) is the Kronecker symbol, q = a/(a, b) and s(f, pk) is the number of solutions of f(x) ≡ 0 (mod pk) which can be easily calculated using Lemma 2.

2 For the simplest case, f(x) = x + 1, the constant Bf in Theorem 1 can be written as log 2 X (−1/p) log p B = γ − 1 − − , (3) f 2 p − 1 p6=2

2000 Mathematics Subject Classiﬁcation 2000 Mathematics Subject Classiﬁcation: 11N37. Keywords: least common multiple, quadratic sequences, equidistribution of roots of quadratic congruences This work was supported by Grant MTM 2008-03880 of MYCIT (Spain) Javier Cilleruelo

p−1 where (−1/p) is the Kronecker symbol (or Legendre symbol since p is odd) deﬁned by (−1/p) = (−1) 2 when p is odd.

In section §3 we give an alternative expression for the constant Bf which is more convenient for numerical computations. As an example we will see that the constant Bf in (3) can be written as log 2 X∞ ζ0(2k) X∞ L0(2k, χ ) X∞ log 2 B = γ − 1 − + + −4 − f k k 2k 2 ζ(2 ) L(2 , χ−4) 2 − 1 k=1 k=0 k=1 = −0.066275634213060706383563177025...

It would be interesting to extend our estimates to irreducible polynomials of higher degree, but we have found a serious obstruction in our argument. Some heuristic arguments and computations allow us to conjecture that the asymptotic estimate log l.c.m. (f(1), . . . , f(n)) ∼ (deg(f) − 1)n log n (4) holds for any irreducible polynomial f in Z[x] of degree deg(f) > 3. In subsection 2.4 we explain the obstruction to prove this conjecture. There we also proof that log l.c.m. (f(1), . . . , f(n)) ∼ n log n (5) holds for any irreducible quadratic polynomial f(x). Although this estimate is weaker than Theorem 1, the proof is easier. To obtain the linear term in Theorem 1 we need a more involved argument. An important ingredient in this part of the proof is a deep result about the distribution of the solutions of the quadratic congruences f(x) ≡ 0 (mod p) when p runs over all the primes. It was proved by Duke, Friedlander and Iwaniec [DFI95] (for D < 0), and by Toth (for D > 0). Actually we need a more general statement of this result, due to Toth.

Theorem 2 [Tot00]. For any irreducible quadratic polynomial f in Z[x], the sequence {ν/p, 0 6 ν < p 6 x, p ∈ S, f(ν) ≡ 0 (mod p)} is well distributed in [0, 1) as x tends to inﬁnity for any arithmetic progression S containing inﬁnitely many primes p for which the congruence f(x) ≡ 0 (mod p) has solutions.

Acknowledgment. We thank to Arpad Toth for clarifying the statement of Theorem 1.2 in [Tot00], to Guoyou Qian for detecting a mistake in a former version of Lemma 2, to Adolfo Quirósfor conversations on some algebraic aspects of the problem, to Enrique GonzálezJiménezfor the calculations of some constants and to Fernando Chamizo for some suggestions and a carefully reading of the paper. Finally we thank to the anonymous referee for some valuable suggestions which have improved the quality of the presentation of this paper.

2. Proof of Theorem 1

2.1 Preliminaries For f(x) = ax2 + bx + c we deﬁne D = b2 − 4ac and

Ln(f) = l.c.m.(f(1), . . . , f(n)).

Since Ln(f) = Ln(−f) we can assume that a > 0. Also we can assume that b and c are nonnegative integers. If this is not the case, we consider a polynomial fk(x) = f(k + x) for a k such that fk(x) has nonnegative coeﬃcients. Then we observe that log Ln(f) = log Ln(fk) + Ok(log n) and that this error term is negligible for the statement of Theorem 1.

We deﬁne the numbers βp(n) by the formula Y βp(n) Ln(f) = p (6) p

2 The least common multiple of a quadratic sequence where the product runs over all the primes p. The primes involved in this product are those for which the congruence f(x) ≡ 0 (mod p) has some solution. Except for some special primes (those such that p | 2aD) the congruence f(x) ≡ 0 (mod p) has 0 or 2 solutions. We will discus this in detail in Lemma 2.

We denote by Pf the set of non special primes for which the congruence f(x) ≡ 0 (mod p) has exactly two solutions. More concretely Pf = {p : p - 2aD, (D/p) = 1} where (D/p) is the Kronecker symbol. This symbol is just the Legendre symbol when p is an odd prime.

The quadratic reciprocity law shows that the set Pf is the set of the primes lying in exactly ϕ(4D)/2 of the ϕ(4D) arithmetic progressions modulo 4D, coprimes with 4D. As a consequence of the prime number theorem for arithmetic progressions we have x #{p 6 x : p ∈ P } ∼ f 2 log x or equivalently, X x 1 ∼ . log x 06ν

 2/3 Small primes : p < n . (  2 2/3 bad primes: p | f(i) for some i 6 n. – p ∈ Pf : Medium primes: n 6 p < Cn :  good primes: p2 - f(i) for any i 6 n.  Large primes: Cn 6 p 6 f(n). We will use diﬀerent strategies to deal with each class.

2.2 Large primes

We consider Pn(f) and the numbers αp(n) deﬁned by Yn Y αp(n) Pn(f) = f(i) = p . (7) i=1 p Next lemma allow us to analyze the large primes involved in (6).

Lemma 1. If p > 2an + b then αp(n) = βp(n).

Since C = 2a + b we can write X log Ln(f) = log Pn(f) + (βp(n) − αp(n)) log p. (8) p

Indeed we can take C to be any constant greater than 2a+b. As we will see, the ﬁnal estimate of log Ln(f) will not depend on C.

The estimate of log Pn(f) is easy: Yn Yn ³ b c ´ log P (f) = log f(k) = log ak2 1 + + (9) n ka k2a k=1 k=1 Xn ³ b c ´ = n log a + log(n!)2 + log 1 + + ka k2a k=1 = 2n log n + n(log a − 2) + O(log n)

3 Javier Cilleruelo and we obtain X log Ln(f) = 2n log n + n(log a − 2) + (βp(n) − αp(n)) log p + O(log n). (10) p

2.3 The number of solutions of f(x) ≡ 0 (mod pk) and the special primes The number of solutions of the congruence f(x) ≡ 0 (mod pk) will play an important role in the proof of Theorem 1. We write s(f, pk) to denote this quantity. The lemma below resumes all the casuistic for s(f, pk). We observe that except for a ﬁnite number of primes, those dividing 2aD, we have that s(f; pk) = 2 or 0 depending on (D/p) = 1 or −1. Lemma 2. Let f(x) = ax2 + bx + c be an irreducible polynomial and D = b2 − 4ac. l (i) If p - 2a, D = p Dp and (Dp, p) = 1 then  bk/2c p , k 6 l s(f, pk) = 0, k > l, l odd or (D /p) = −1  p  l/2 2p , k > l, l even (Dp/p) = 1. ( 0, if p | b (ii) If p | a, p 6= 2 then s(f, pk) = 1, if p - b.  1 if a is even (iii) If b is odd then, for all k > 2, s(f, 2k) = s(f, 2) = 0 if a is odd and c is odd  2 if a is odd and c is even. (iv) If b is even and a is even then s(f, 2k) = 0 for any k > 1. (v) If b is even and a is odd, let D = 4lD0,D0 6≡ 0 (mod 4). k bk/2c (a) If k 6 2l − 1, s(f; 2 ) =( 2 2l,D0 ≡ 1 (mod 4) (b) If k = 2l, s(f; 2k) = 0,D0 6≡ 1 (mod 4). ( 2l+1,D0 ≡ 1 (mod 8) (c) If k > 2l + 1, s(f; 2k) = 0,D0 6≡ 1 (mod 8). Proof. The proof is a consequence of elementary manipulations and Hensel’s lemma. When the modulo is an odd prime p and p - a, the congruence ax2 +bx+c ≡ 0 (mod p) is equivalent to the congruence y2 ≡ D (mod p). Hensel’s lemma (see for example Theorem 123 in [HW08]) provides a method to obtain all the solutions of the congruence y2 ≡ D (mod pk+1) from the solutions of y2 ≡ D (mod pk). In this way we obtain all the distinct cases contained in part (i) of the lemma. Part (ii) is trivial and parts (iii),(iv) and (v) correspond to the case p = 2 which can be analyzed easily. Corollary 1. If p - 2aD then s(f, pk) = 1 + (D/p). k Proof. In this case, l = 0 and Dp = D in Lemma 2. Thus s(f, p ) = 0 = 1 + (D/p) if (D/p) = −1 and s(f, pk) = 2 = 1 + (D/p) if (D/p) = 1. Lemma 3. µ ¶ X s(f, pk) log n α (n) = n + O . (11) p pk log p k>1 where s(f; pk) denotes the number of solutions of f(x) ≡ 0 (mod pk), 0 6 x < pk.

αp,i Proof.P We observe that the maximum exponent αp,i such that p | f(i) can be written as αp,i = k 1. Thus k>1, p |f(i) X X X X X αp(n) = αp,i = 1 = 1. (12) i6n i6n k>1 k>1 i6n pk|f(i) pk|f(i)

4 The least common multiple of a quadratic sequence

The trivial estimate · ¸ µ· ¸ ¶ n X n s(f; pk) 6 1 6 s(f; pk) + 1 pk pk i6n, pk|f(i) gives X s(f; pk) 1 = n + O(s(f; pk)). (13) pk i6n pk|f(i) Putting (13) in (12) and observing that k 6 log f(n)/ log p and that s(f, pk) ¿ 1, we get X s(f, pk) ³log n´ α (n) = n + O . p pk log p k>1

Since pβp(n) 6 f(n) we have always the trivial estimate

βp(n) 6 log f(n)/ log p ¿ log n/ log p. (14) When we substitute (14) and (11) in (10) for the special primes we obtain   X X s(f, pk) log p log L (f) = 2n log n + n log a − 2 −  (15) n pk p|2aD k>1 X + (βp(n) − αp(n)) log p + O(log n). p

Lemma 4. For any p - 2aD we have 1 + (D/p) ³log n´ α (n) = n + O . (16) p p − 1 log p Proof. It is a consequence of Lemma 3 and Corollary 1.

2.4 The asymptotic estimate This subsection is a break in the proof of Theorem 1 to prove, in an easy way, that the weaker estimate log l.c.m.(f(1), . . . , f(n)) ∼ n log n (17) holds for any irreducible quadratic polynomial f. We substitute (14) and (16) in (15) to obtain X log Ln(f) = 2n log n + (βp(n) − αp(n)) log p + O(n) (18) p

P log p P (D/p) log p Now we get (17) using that p6x p−1 ∼ log x and that the sum p p−1 is a convergent sum. This is the moment to explain the main obstruction to obtain the analogous estimate for polynomials of degree deg(f) > 3. For example we consider the polynomial f(x) = x3 + 2. Using the same approach used in the quadratic case we get X log Ln(f) = 3n log n + (βp(n) − αp(n)) log p + O(n). (19) p<3n2

5 Javier Cilleruelo

We observe that the primes involved in the sum have the quadratic bound 3n2 instead the linear bound we have in the case of quadratic polynomials. The reason is that if p | k3 + 2 and p | j3 + 2 with j < k 6 n we only can say that p 6 3n2. sp It is easy to check that βp(n) ¿ log n/ log p and αp(n) = n p−1 +O(log n/ log p) where sp is the number of solutions of x3 + 2 ≡ 0 (mod p). Then we obtain X s log p X log L (f) = 3n log n − n p + O(n) + (β (n) − α (n)) log p. (20) n p − 1 p p p

P sp log p The Frobenius density theorem [LS96] implies that sp = 1 in average, so p6x p−1 ∼ log x. Then, in the case f(x) = x3 + 2 we have X log Ln(f) = 2n log n(1 + o(1)) + (βp(n) − αp(n)) log p. (21) n

3 3 We observe that βp(n) = αp(n) unless there exist j < k 6 n such that p | k + 2, p | j + 2. In that case we have the trivial bound |βp(n) − αp(n)| ¿ 1. Thus, to obtain the asymptotic log Ln(f) ∼ 2n log n for f(x) = x3 + 2 we should prove that |{p : n 6 p 6 3n2, p | k3 + 2, p | j3 + 2 for some 1 6 j < k 6 n}| = o(n) In general, when f(x) is an irreducible polynomial, the asymptotic estimate log l.c.m(f(1), . . . , f(n)) ∼ (deg(f) − 1)n log n would follow from the estimate |{p : n 6 p ¿ ndeg(f)−1, p | f(k), p | f(j) for some 1 6 j < k 6 n}| = o(n). (22) This is obviously true when deg(f) = 2 but we do not know how to prove it when deg(f) > 3. We come back to the proof of Theorem 1.

2.5 Medium primes These primes can be also classiﬁed in bad and good primes. Bad primes are those p such that p2 | f(i) for some i 6 n. Good primes are those are not bad primes.

As we have seen in the previous section, for any prime p ∈ Pf the congruence f(x) ≡ 0 (mod p) has exactly two solutions, say 0 6 νp,1, νp,2 < p.

If p is a good prime, we have that αp(n) is just the number of integers i 6 n such that p | f(i). All these integers have this form · ¸ n − ν ν + kp, 0 6 k 6 p,1 (23) p,1 p · ¸ n − ν ν + kp, 0 6 k 6 p,2 . (24) p,2 p

Also it is clear that if p is a good prime then βp(n) 6 1. These observations motivate the following deﬁnition:

Definition 1. For any p ∈ Pf we deﬁne · ¸ · ¸ n − ν n − ν α∗(n) = p,1 + p,2 + 2 (25) p p p ( 1, if β (n) > 1 β∗(n) = p (26) p 0, otherwise.

Lemma 5. For any p ∈ Pf we have ∗ 2n i) αp(n) − αp(n) = p(p−1) + O(log n/ log p) ∗ ∗ 2 ii) αp(n) = αp(n) and βp(n) = βp (n) if p - f(i) for any i 6 n.

6 The least common multiple of a quadratic sequence

2n Proof. i) Lemma 4 implies that αp(n) = p−1 + O(log n/ log p) when p ∈ Pf . On the other hand we have ∗ 2n ∗ 2n that αp(n) = p + O(1). Thus, αp(n) − αp(n) = p(p−1) + O(log n/ log p). ii) The ﬁrst assertion has been explained at the beginning of the subsection. For the second, if p - f(i) ∗ ∗ for any i 6 n then βp(n) = βp (n) = 0. And if p | f(i) for some i 6 n we have that βp (n) = βp(n) = 1 since p2 - f(i).

By substituting (16) and (14) in (15) for small primes we obtain   X X s(f, pk) log p log L (f) = 2n log n + n log a − 2 −  (27) n pk p|2aD k>1 X (1 + (D/p)) log p X − + (β (n) − α (n)) log p + O(n2/3). p − 1 p p p

∗ ∗ To estimate S1(n) we observe that Lemma 5 ii) implies that βp(n)−βp (n)−αp(n)+αp(n) = 0 for any good ∗ ∗ prime p. On the other hand, Lemma 5 i) and (14) implies that |βp(n)−βp (n)−αp(n)+αp(n)| ¿ log n/ log p. Thus,

2/3 |S1(n)| ¿ log n |{p : n < p < Cn, p bad}|. (29)

Lemma 6. The number of bad primes p - D,Q 6 p < 2Q is ¿ n2/Q2.

2 2 Proof. Let Pr the set of all primes p such that f(i) = ai + bi + c = rp for some i 6 n. For p ∈ Pr we 2 2 2ai+b √ 1 1 have (2ai + b) − 4arp = D and then, | p − 2 ra| ¿ p2 ¿ Q2 . We observe that all the fractions 2ai+b 0 0 p , 1 6 i 6 n, Q 6 p < 2Q are pairwise diﬀerent. Otherwise (2ai + b)p = (2ai + b)p and then 2 2 ¯p | 2ai + b. But¯ it would imply that p | (2ai + b) − 4arp = D, which is not possible. On the other hand, ¯ 2ai+b 2ai0+b ¯ 1 1 ¯ p − p0 ¯ > pp0 À Q2 . Thus, the number of primes p ∈ Pr lying in [Q, 2Q] is ¿ 1. We ﬁnish the proof by observing that r 6 f(n)/Q2 ¿ n2/Q2.

Now, if we split the interval [n2/3, Cn] in dyadic intervals and apply lemma above to each interval we 2/3 obtain |S1(n)| ¿ n log n. P ∗ To estimate S (n) = 2/3 α (n) we start by writing 3 n

7 Javier Cilleruelo

Thus X (1 + (D/p)) log p S (n) = n (30) 3 p n2/3

Substituting this in (28) and then in (27) we obtain   X X s(f, pk) log p log L (f) = 2n log n + n log a − 2 −  (34) n pk p|2aD k>1 X (1 + (D/p)) log p − + S (n) − T (n) − T (n) + O(n2/3 log n) p − 1 2 1 2 p

Sums T1(n) and T2(n) will be o(n) as a consequence of Theorem 2. But this is not completely obvious and we will provide a detailed proof in the next subsection. First we will obtain in the next lemma a simpliﬁed expression for (34).

Lemma 7.

2/3 log Ln(f) = n log n + cn + S2(n) − T1(n) − T2(n) + O(n log n), (38) where X (d/p) log p X ³ 1 X s(f, pk)´ c = log a − log C − 2 + γ − + log p − p − 1 p − 1 pk p-2aD p|2aD k>1 and S2(n),T1(n) and T2(n) are as in (35), (36) and (37).

Proof. Let D = l2d where d is a fundamental discriminant. First we observe that (D/p) = (l/p)2(d/p) and that if p - D then (D/p) = (d/p). As a consequence of the prime number theorem on arithmetic P (d/p) log p progressions we know that the sum p p−1 is convergent. On the other hand, the well known estimate

8 The least common multiple of a quadratic sequence

P log p p6x p−1 = log x − γ + o(1) where γ is the Euler constant, implies that

X (1 + (D/p)) log p X log p = log n + log C − γ − (39) p − 1 p − 1 p

Finally we substitute (39) in (34).

2.6 Equidistribution of the roots (mod p) of a quadratic polynomial

Now we develop a method to prove that T1(n), T2(n) and other similar sums which will appear in the estimate of S2(n) are all o(n). These sums are all of the form X a(ν, p, x) log p (40) 06ν

Hence, to prove that the sums (40) are o(x) we must show that X a(ν, p, x) = o(x/ log x). 06ν

Theorem 2 implies, in particular, that for any arithmetic progression S and for any piecewise continuos R 1 function g in [0, 1] such that 0 g = 0 we have that X g(ν/p) = o(x/ log x). (43) 06ν

Lemma 8. Let f be an irreducible polynomial in Z[x]. We have that the sums T1(n) and T2(n) deﬁned in (36) and (37) are both o(n).

Proof. To prove that T1(n) = o(n) we apply (43) to the function g(x) = x − 1/2.

To prove that T2(n) = o(n), the strategy is splitting the range of the primes in small intervals such that n/p are almost constant in each interval. We take H a large, but ﬁxed number and we divide the h−1 h interval [1, Cn] in H intervals Lh = ( H Cn, H Cn], h = 1,...,H. Now we write X ³nn − ν o 1´ − = Σ + Σ + Σ + O(n/(H1/3 log n)) (44) p 2 31 32 33 06ν

9 Javier Cilleruelo where X X ³nH ν o 1´ Σ = − − 31 h p 2 H2/36h6H 06ν

Σ31 = o(Hn/ log n) = o(n log n) (45) since H is a constant. ν H H To bound Σ32 we observe that if p ∈ Lh and p 6∈ [ h , h−1 ], then nn ν o nH ν o n H H 0 6 − − − = − 6 . p p h p p h h(h − 1) Thus X X H X X 1 π(n) n |Σ | ¿ ¿ ¿ ¿ . (46) 32 h2 H1/3 H1/3 H1/3 log n H2/36h

To bound Σ33 ﬁrst we observe that X X Σ33 ¿ 1

H2/36h

10 The least common multiple of a quadratic sequence

To present Lemma 10 we need some previous considerations.

For primes p ∈ Pf the congruence f(x) ≡ 0 (mod p) has exactly two solutions, say 0 6 νp,1, νp,2 < p. In some parts of the proof of Theorem 1 we will need to estimate some quantities depending on min(νp,1, νp,2). For this reason it is convenient to know how they are related. 2 If f(x) = ax + bx + c and p ∈ Pf then νp,1 + νp,2 ≡ −b/a (mod p). The next lemma will give more information when the prime p belongs to some particular arithmetic progression. Lemma 9. Let q = a/(a, b), l = b/(a, b). For any r, (r, q) = 1 and for any prime p ≡ lr−1 (mod q) and p ∈ Pf we have ν ν r l p,1 + p,2 ≡ − (mod 1). (48) p p q pq

Proof. To avoid confusions we denote by qp and pq the inverses of q (mod p) and p (mod q) respectively. qp pq 1 From the obvious congruence qqp + ppq ≡ 1 (mod pq) we deduce that p + q − pq ∈ Z. Since p ≡ lrq qp 1 rlq (mod q) we obtain p ≡ pq − q (mod 1). Thus µ ¶ ν ν −lq 1 rl r l p,1 + p,2 ≡ p ≡ −l − q ≡ − (mod 1). p p p pq q q pq

h Since the two roots are symmetric respect to r − l , necessarily one of then lies in r − l , 1 + ´ 2q 2pq 2q 2pq 2 r l 2q − 2pq (mod 1) and the other in the complementary set.

−1 Definition 2. For (r, q) = 1, 1 6 r 6 q, p ≡ lr (mod q) and p ∈ Pf we deﬁne νp,1 the root of f(x) ≡ (mod p) such that ν h r l 1 r l ´ p,1 ∈ T = − , + − (mod 1), p rp 2q 2pq 2 2q 2pq νp,2 and we deﬁne νp,2 the root of f(x) ≡ 0 (mod p) such that p ∈ [0, 1) \ Trp.

Lemma 10. Assume the notation above. Let α1, α2, β1, β2, c1, c2 be constants and g1(x), g2(x) two linear functions satisfying that · µ ¶ µ ¶ ¸ n c n c J (p) = g + 1 , g + 2 ⊂ T n 1 p p 2 p p rp for any prime p ∈ Kn = [α1n + β1, α2n + β2]. We have X ³ ³ν ´ ´ χ p,1 − 2|J (p)| log p = o(n) (49) Jn(p) p n p∈Kn∩Pf p≡lr−1 (mod q) where χI is the characteristic function of the set I.

Proof. Since Jn(p) ⊂ Trp then ν2/p 6∈ Jn(p) and we can write X ³ν ´ X ³ν ´ χ p,1 log p = χ log p Jn(p) p Jn(p) p p∈Kn∩Pf 16ν6p∈Kn, p≡lr−1 (mod q) f(ν)≡0 (mod p) p≡lr−1 (mod q) and X X 2|Jn(p)| log p = |Jn(p)| log p.

p∈Kn∩Pf 16ν6p∈Kn, p≡lr−1 (mod q) f(ν)≡0 (mod p) p≡lr−1 (mod q) Thus, X ³ ³ν ´ ´ X ³ ³ν ´ ´ χ p,1 − 2|J (p)| log p = χ − |J (p)| log p. Jn(p) p n Jn(p) p n p∈Kn∩Pf 16ν6p∈Kn, p≡lr−1 (mod q) f(ν)≡0 (mod p) p≡lr−1 (mod q)

11 Javier Cilleruelo

The proof will be accomplished by showing that X ³ ³ν ´ ´ χ − |J (p)| = o(n/ log n). (50) Jn(p) p n 16ν6p∈Kn, f(ν)≡0 (mod p) p≡lr−1 (mod q)

h−1 h We split Kn in intervals Lh = ( H n, H n] of length n/H and two extra intervals I,F (the initial and the ﬁnal intervals) of length 6 n/H. Here h runs over a suitable set of consecutive integers H of cardinality ¿ (α2 − α1)H.

Let Ih denote the interval [g1(H/h) + c1H/(nh), g2(H/h) + c2H/(nh)]. We write

X ³ ³ν ´ ´ χ − |J (p)| = Σ + Σ + Σ + Σ (51) Jn(p) p n 1 2 3 4 16ν6p∈Kn, f(ν)≡0 (mod p) p≡lr−1 (mod q) where X X ³ ³ν ´ ´ Σ = χ − |I | 1 Ih p h h∈H 06ν

h∈H 06ν

The inner sum in Σ1 can be estimated as we did in Lemma 8, (with the function g(x) = χI (x) − |I| instead of g(x) = x − 1/2), and we get again that Σ1 = o(n/ log n).

To estimate Σ2 and Σ3 we observe that if p ∈ Lh then Jn(p) and Ih are almost equal. Actually, comparing the end points of both intervals and because g is a linear function, we have that and χJn(p)(x) =

χIh (x) except for an interval (or union of two intervals) Eh of measure

2 |Eh| ¿ min(1, H/h ).

2 In particular, the estimate ||Jn(p)| − |Ih|| ¿ min(1, H/h ) holds. Thus, we have X X X X X X 1 Σ ¿ min(1, H/h2) ¿ 1 + 2 H1/3 h∈H p∈Ln h6H2/3 p∈Lh H2/3

12 The least common multiple of a quadratic sequence

To bound Σ3 ﬁrst we observe that X X

Σ3 ¿ χEh (ν/p)

h∈H 06ν

= (χEh (ν/p) − |Eh|) + |Eh|.

h∈H 06ν

(χEh (ν/p) − |Eh|) = o(n/ log n).

06ν

1/3 Thus, Σ3 ¿ o(n/ log n) + n/(H log n).

Finally we estimate Σ4. We observe that X X |Σ4| 6 1 + 1 ¿ n/(H log n) p∈I p∈F as a consequence of the prime number theorem. Then 1/3 Σ1 + Σ2 + Σ3 + Σ4 = O(n/(H log n)) + O(n/(H log n)) + o(n/ log n) ﬁnishing the proof because we can take H arbitrarily large.

2.7 Estimate of S2(n) and end of the proof Lemma 11. ³ 1 X r ´ S (n) = n 1 + log C − log 4 + log(1 + ) + o(n) (52) 2 φ(q) q (r,q)=1

Proof. Following the notation of Lemma 9 we split X X X ∗ S2(n) = S2r(n) + βp (n) log p = S2r(n) + O(1) (r,q)=1 p6l (r,q)=1 16r6q 16r6q where X ∗ S2r(n) = βp (n) log p. (53) l

−1 νp,1 νp,2 r l Since p ≡ lr (mod q), Lemma 9 implies that p + p ≡ q − pq (mod 1). We observe also that, r l since p > l we have that 0 < q − pq 6 1.

13 Javier Cilleruelo

Now we will check that  1, if n > 1 + r − l  p 2 2q 2pq  r l n 1 r l χ r l n (νp,1/p), if − < < + − ∗ [ 2q − 2pq , p ] q pq p 2 2q 2pq βp (n) = r l n r l χ[ r − l , r − l ](νp,1/p), if − 6 6 −  2q 2pq q pq 2q 2pq p q pq  n r l χ r l n r l (νp,1/p) if < − [ q − pq − p , q − pq ] p 2q 2pq

∗ νp,1 n νp,2 n We observe that βp (n) = 1 if and only if p 6 p or p 6 p . We remind that

r l ν 1 r l − 6 p,1 < + − (54) 2q 2pq p 2 2q 2pq

Also we observe that Lemma 9 implies that ( ν r − l − νp,1 if νp,1 6 r − l p,2 = q pq p p q pq (55) p r l νp,1 νp,1 r l q − pq − p + 1 if p > q − pq . ³ ´ n 1 r l 1 r l ∗ – Assume p > 2 + 2q − 2pq . Then νp,1 < p 2 + 2q − 2pq < n, so βp (n) = 1

r l n 1 r l – Assume q − pq < p < 2 + 2q − 2pq .

∗ ∗ If χ r l n (νp,1/p) = 1 then νp,1 6 n, so β (n) = 1. [ 2q − 2pq , p ] p νp,1 n r l νp,2 ∗ If χ r l n (νp,1/p) = 0 then > > − . Relations (54) and (55) imply that = [ 2q − 2pq , p ] p p q pq p r l νp,1 1 r l n ∗ 1 + q − pq − p > 2 + 2q − 2pq > p . Since νp,1 > n and νp,2 > n we get βp (n) = 0.

r l n r l – Assume 2q − 2pq 6 p 6 q − pq .

νp,2 r l ∗ If χ r l r l (νp,1/p) = 1 then (55) imply that 0 < 6 − , which implies that [ 2q − 2pq , q − pq ] p 2q 2pq ∗ νp,2 6 n, so βp (n) = 1. νp,1 r l n νp,2 ∗ If χ r l r l (νp,1/p) = 0 then > − > and relation (55) imply that = [ 2q − 2pq , q − pq ] p q pq p p r l νp,1 r l n ∗ q − pq − p + 1 > q − pq > p . Since νp,1 > n and νp,2 > n we get βp (n) = 0.

n r l – Assume p < 2q − 2pq .

νp,1 r l νp,2 r l ∗ If χ r l n r l (νp,1/p) = 1 then 6 − and relation (55) implies that = − − [ q − pq − p , q − pq ] p q pq p q pq νp,1 r l r l n n ∗ p 6 q − pq − ( q − pq − p ) = p , so βp (n) = 1 ∗ If χ r l n r l (νp,1/p) = 0 we distinguish two cases: [ q − pq − p , q − pq ]

· If r − l 6 νp,1 < r − l − n then ν1,p > r − l > n , and also we have that νp,2 = 2q 2q p q ³pq p ´ p 2q 2q p p r l νp,1 r l r l n n ∗ q − pq − p > q − pq − q − pq − p = p . Thus βp (n) = 0 ³ ´ · If r − l < νp,1 < 1 + r − l then νp,1 > 1 r − l > n . On the other hand, νp,2 = q pq p 2 2q ³2pq p ´ 2 q pq p p r l νp,1 r l 1 r l 1 r l n q − pq − p + 1 > q − pq − 2 + 2q − 2pq + 1 = 2 + 2q − 2pq > p . Thus, again we have ∗ that βp (n) = 0.

14 The least common multiple of a quadratic sequence

P4 Now we split S2r(n) = i=1 S2ri(n) according the ranges of the primes involved in the lemma above. X S2r1(n) = log p n+l/(2q) l

−1 Since (q, D) = 1 and the primes are odd numbers, the primes p ≡ lr (mod q), p ∈ Pf lie in a set of φ(4qD)/(2φ(q)) arithmetic progressions modulo 4qD. The prime number theorem for arithmetic progressions implies that

X x log p ∼ (56) 2φ(q) p6x −1 p≡lr (mod q), p∈Pf and

X log p log(b/a) = + o(1) (57) p 2φ(q) ax

We will use these estimates and lema 10 to estimate S2ri(n), i = 1, 2, 3, 4. By (56) we have

n q S (n) = + o(n). (58) 2r1 φ(q) q + r

To estimate S2r2 we write X S2r2(n) = χ r l n (νp,1/p) log p [ 2q − 2pq , p ] n+l/(2q) n+l/q 1/2+r/(2q)

15 Javier Cilleruelo

Lemma 10 implies that the last sum is o(n). Thus, X ³2n r ´ S = − log p + o(n) 2r2 p q n+l/(2q) n+l/q 1/2+r/(2q)

To estimate S2r3(n) we write X ³r l ´ S (n) = − log p 2r3 q pq q l 2q l r (n+ q )6p6 r (n+ q ) p≡lr−1 (mod q) p∈Pf X ³ ³r l ´´ + χ[ r − l , r − l ](νp,1/p) − − log p 2q 2pq q pq q pq q l 2q l r (n+ q )6p6 r (n+ q ) p≡lr−1 (mod q) p∈Pf n = + o(n) 2φ(q) by (56) and lema 10.

To estimate S2r4(n) we write X ³2n 2l ´ S (n) = + log p 2r4 p pq 2q l r (n+ 2q )

S2r(n) = S2r1(n) + S2r2(n) + S2r3(n) + S2r4(n) + O(1) n q = + o(n) φ(q) q + r n ³1 q ´ n ³1 r ´ + log + − − + o(n) φ(q) 2 2r φ(q) 2 q + r n + + o(n) 2φ(q) n ³ ´ + log C − log(2q/r) + o(n) φ(q) n ³ ´ = 1 + log C − log 4 + log(1 + r/q) + o(n). φ(q)

16 The least common multiple of a quadratic sequence

Now sum in all r 6 q, (r, q) = 1 to ﬁnish the estimate of S2(n).

Finally we substitute (52) in (38) to conclude the proof of Theorem 1.

3. Computation of the constant Bf

P (d/p) log p The sum p p−1 , appearing in the formula of the constant Bf converges very slowly. The next lemma gives an alternative expression for this sum, more convenient in order to obtain a fast computation.

Lemma 12. X X∞ 0 k X∞ 0 k X (d/p) log p ζ (2 ) L (2 , χd) = k − k + sp. (59) p − 1 ζ(2 ) L(2 , χd) p k=1 k=0 p|d

P∞ log p where sp = k . k=1 p2 −1

³ ´(d/p) Q 1 Proof. For s > 1 we consider the function Gd(s) = p 1− ps . Taking the derivative of the logarithm of Gd(s) we obtain that

G0 (s) X (d/p) log p d = . (60) G (s) ps − 1 d p

³ ´−1 Q (d/p) s Since L(s, χd) = p 1 − p we have

Y³ 1 ´(d/p)³ (d/p)´−1 G (s)L(s, χ ) = 1 − 1 − (61) d d ps ps p Y ³ 1 ´−1 = 1 − (62) p2s (d/p)=−1 ³ ´ (d/p)−1 ³ ´ Y 1 2 Y 1 1/2 = 1 − 1 − (63) p2s p2s p p|d 1/2 1/2 1/2 = Gd (2s)ζ (2s)T (2s) (64) ³ ´ Q 1 where T (s) = p|d 1 − ps . The derivative of the logarithm gives G0 (s) G0 (2s) ζ0(2s) T 0 (2s) L0(s, χ ) d − d = + d − d . Gd(s) Gd(2s) ζ(2s) Td(2s) L(s, χd) Thus µ ¶ 0 0 m mX−1 0 k 0 k+1 Gd(s) Gd(2 s) Gd(2 s) Gd(2 s) − m = k − k+1 (65) Gd(s) Gd(2 s) Gd(2 s) Gd(2 s) k=0 Xm 0 k Xm 0 k mX−1 0 k ζ (2 s) Td(2 s) L (2 s, χd) = k + k − k . (66) ζ(2 s) Td(2 s) L(2 s, χd) k=1 k=1 k=0

17 Javier Cilleruelo

By (60) we have that for s > 2,

ζ0(s) X Λ(n) log 2 X log n | | 6 6 + ζ(s) ns − 1 2s − 1 ns − 1 n>2 n>3 Z 4 log 2 9 X log n 4 log 2 9 ∞ log x 6 + 6 + dx 3 2s 8 ns 3 2s 8 xs n>3 2 4 log 2 9³ log 2 1 ´ = + + 3 2s 8 2s−1(s − 1) 2s−1(s − 1)2 1 ³20 log 2 + 8´ 5 2−s 6 6 · . 2s(s − 1) 9 2 s − 1

0 k −2k 0 k 0 k 0 k ζ (2 ) 5 2 Gd(2 ) Td(2 ) L (2 ,χd) Thus, | k | 6 · k . The same estimate holds for | k |, | k | and | k |. When m → ∞ and ζ(2 ) 2 2 −1 Gd(2 ) Td(2 ) L(2 ,χd) then s → 1 we get

X X∞ 0 k X∞ 0 k X∞ 0 k (d/p) log p ζ (2 ) L (2 , χd) Td(2 ) = k − k + k . (67) p − 1 ζ(2 ) L(2 , χd) Td(2 ) p k=1 k=0 k=1

0 k P P 0 k P Td(2 ) log p ∞ Td(2 ) Finally we observe that k = k , so k = sp. Td(2 ) p|d p2 −1 k=1 Td(2 ) p|d

The advantage of the lemma above is that the series involved converge very fast. For example,

X∞ 0 k X6 0 k L (2 , χd) L (2 , χd) k = k + Error L(2 , χd) L(2 , χd) k=0 k=0 with |Error| 6 10−40.

Hence we can write Bf = C0 + Cd + C(f) where C0 is an universal constant, Cd depends only on d, and C(f) depends on f. More precisely,

X∞ ζ0(2k) C = γ − 1 − 2 log 2 − = −1.1725471674190148508587521528364 ... 0 ζ(2k) k=1 X∞ 0 k X L (2 , χd) Cd = k − sp L(2 , χd) k=0 p|d 1 X ³ r ´ X ³1 + (d/p) X s(f, pk)´ C(f) = log 1 + + log a + log p − . φ(q) q p − 1 pk 16r6q p|2aD k>1 (r,q)=1

P 0 k k The values of sp and k>0 L (2 , χd)/L(2 , χd), can be calculated with MAGMA with high precision. We include some of the values of Cd and C(f):

C−4 = + 0.346538435736895987549 − s2 = + 0.066550762366036180349 ...

C−8 = − 0.076694093066485311184 − s2 = − 0.356681766437345118384 ...

C−3 = + 0.586272400297149523649 − s3 = + 0.435045713698422447292 ...

C−7 = − 0.070022837990444988815 − s7 = − 0.111373766208260107471 ...

C−15 = − 0.486320692903261758405 − s3 − s5 = − 0.707190640126000030028 ...

18 The least common multiple of a quadratic sequence

C(x2 + 1) = (3 log 2)/2 = 1.039720770839917964125 ... C(x2 + 2) = (3 log 2)/2 = 1.039720770839917964125 ... C(x2 + x + 1) = log 2 + (log 3)/6 = 0.876249228671296924649 ... C(x2 + x + 2) = log 2 + (log 7)/(42) = 0.739478374585071816681 ... C(2x2 + 1) = 3 log 2 = 2.079441541679835928251 ... C(2x2 + x + 1) = 2 log 2 + log 3 + (log 7)/(42) = 1.838090663253181508076 ... C(2x2 + x + 2) = log 2 + (7 log 3)6 + (log 5)/(20) = 2.055333412961111634775 ... C(2x2 + 2x + 1) = 3 log 2 = 2.079441541679835928251 ...

2 Table below contains the constant B = Bf for all irreducible quadratic polynomial f(x) = ax +bx+c with 0 6 a, b, c 6 2. When f1, f2 are irreducible quadratic polynomials such that f1(x) = f2(x + k) for some k, we only include one of them since Ln(f1) = Ln(f2) + O(log n).

f(x) d q Bf x2 + 1 -4 1 −0.06627563421306070638. . . x2 + 2 -8 1 −0.48950816301644200511. . . x2 + x + 1 -3 1 +0.13874777495070452108. . . x2 + x + 2 -7 1 −0.54444255904220314164. . . 2x2 + 1 -8 1 +0.55021260782347595900. . . 2x2 + x + 1 -7 2 +0.55416972962590654974. . . 2x2 + x + 2 -15 2 +0.17559560541609675388. . . 2x2 + 2x + 1 -4 1 +0.97344513662685725774. . .

Table below shows the error term Ef (n) = log Ln(f) − n log n − Bf n for the polynomials above and some values of n. 2 3 4 5 6 7 f(x) Ef (10 ) Ef (10 ) Ef (10 ) Ef (10 ) Ef (10 ) Ef (10 ) x2 + 1 −18 +6 −111 +34 −2634 −1557 x2 + 2 −36 −11 −263 −761 −1462 −8457 x2 + x + 1 −6 −9 +17 −654 −2528 −1685 x2 + x + 2 +9 −20 −218 −2120 +687 −686 2x2 + 1 −15 −1 −301 −251 +1084 −14821 2x2 + x + 1 −1 +6 +18 −1289 +235 −2553 2x2 + x + 2 −34 +4 −295 +27 +1169 +1958 2x2 + 2x + 1 −9 −89 +9 −232 −2876 −10624

4. Quadratic reducible polynomials To complete the problem of estimating the least common multiple of quadratic polynomials we will study here the case of reducible quadratic polynomials. Being this case much easier than the irreducible case, we will give a complete description for the sake of the completeness. 2 0 If f(x) = ax + bx + c with g = (a, b, c) > 1, it is easy to check that log Ln(f) = log Ln(f ) + O(1) where f 0(x) = a0x2 + b0x + c0 with a0 = a/g, b0 = b/g, c0 = c/g. 2 2 2 2 2 If f(x) = (ax + b) with (a, b) = 1 then, since (m , n ) = (m, n) , we have that Ln((ax + b) ) = 2 Ln(ax + b) and we can apply (1) to get a X 1 log l.c.m.{(a + b)2,..., (an + b)2} ∼ 2n . (68) φ(a) k 16k6a (k,a)=1

19 The least common multiple of a quadratic sequence

Now we consider the more general case f(x) = (ax + b)(cx + d), (a, b) = (c, d) = 1.

Theorem 3. Let f(x) = (ax + b)(cx + d) with (a, b) = (c, d) = 1 and ad 6= bd. Let q = ac/(a, c). We have n X ³ a c ´ log l.c.m.(f(1), . . . , f(n)) ∼ max , . (69) ϕ(q) (br)a (dr)c 16r6q, (r,q)=1

p6Mn p>Mn p6Mn p

βp(n) where ²p(n) = 1 if p | f(i) for some i 6 n and ²p(n) = 0 otherwise. Since p 6 f(n) we have that βp(n) ¿ log n/ log p and then X √ (βp(n) − ²p(n)) log p ¿ (log n)π(Mn) ¿ n. (71)

p6Mn Thus, X √ log Ln(f) = log p + O( n). (72) p|f(i) for some i6n −1 Let q = ac/(a, c). Suppose that p ≡ r (mod q), (r, q) = 1. Let k = (br)a the least positive integer such that k ≡ br (mod a). Then p | (ai + b) for some i 6 n if and only if kp 6 an + b. Similarly, let j = (dr)c be the least positive integer such that j ≡ dr (mod c). Again, p | (ci + d) for some 6 i 6 n if jp 6 cn + d. −1 an+b cn+d Thus, the primes p ≡ r (mod ac) counted in the sum above are those such that p 6 max( k , j ). n a c The prime number theorem for arithmetic progressions implies that there are ∼ ϕ(q) max( k , j ) of such primes. We ﬁnish the proof summing up in all 1 6 r 6 q, (r, q) = 1.

References Bat02 P. Bateman, A limit involving Least Common Multiples: 10797, American Mathematical Monthly 109 (2002), no. 4, 393-394. Che52 P.L. Chebyshev, Memoire sur les nombres premiers. J. Math. pures et appl. 17 (1852), 366-390. DFI95 W. Duke, J. Friedlander and H. Iwaniec, Equidistribution of roots of a quadratic congruence to prime moduli, Ann. of Math. 141 (1995), no. 2, 423–441. HW08 G.H. Hardy and E.M. Wright, An Introduction to the Theory of Numbers (Sixth edition), Oxford: Oxford University Press, (2008). LS96 H. Lenstra and P. Stevenhagen, Chebotar¨evand his density theorem, The Mathematical Inteligencer 18(1996), no. 2, 26–37. Tot00 A. Toth, Root of quadratic congruences, Internat. Math. Res. Notices 14 (2000), 719–739.

Javier Cilleruelo [email protected] Instituto de Ciencias Matemáticas(CSIC-UAM-UC3M-UCM) and Departamento de Matemáticas, Uni- versidad Autónomade Madrid , 28049 Madrid, Spain