Stéphane R Louboutin

Upper bounds on residues of Dedekind zeta functions of non-normal totally real cubic fields by

Keywords: 2020 Mathematics Subject Classification: Primary 11R42; Secondary 11M20, 11R11, 11R16 L-function, zeta function, quadratic and cubic number fields, Euler-Kronecker constants

Introduction. Various bounds on the absolute values of L-functions

of number fields at s = 1 and on residues at s = 1 of Dedekind zeta functions of a number field L are known. Also, better bounds depending on the splitting behavior of given prime ideals of L of small norms are known. These bounds involve a term ν L in the series expansion at s = 1 of the Dedekind zeta function of L. We explain why one should expect to have bounds on ν L which also depend on the splitting behavior in L of given prime integers. We explicitly do that for L a real quadratic number field. We deduce very good upper bounds on the residue at s = 1 of the Dedekind zeta function of a non-normal totally real cubic number field K, bounds depending on the splitting behavior of the prime p = 2 in K. We have almost reached in Theorem 1.7 an explicit bound

Res s=1 (ζ K (s)) ≤ Q 2 8 (log d L + c 2 )(log(d 2 K /d L ) + c 2 ) (1) ≤ Q 2 8 (log d K + (c 2 + c 2 )/2) 2
for the residue at s = 1 of the Dedekind zeta function of a non-normal totally real cubic number field K of discriminant of absolute value d K , where d L is the absolute value of the discriminant of the real quadratic subfield

L = Q( √ d K )
of the normal closure of K. Here, Q 2 ∈ {1, 1/2, 1/3, 1/4, 1/7} defined in (15) depends on the splitting behavior of the prime p = 2 in K and c 2 and c 2 would also depend explicitly on the splitting behavior of the prime p = 2 in K. Unfortunately, the sizes of the error terms in Theorem 1.1 and Proposition 2.1 prevent us from getting such a simple bound and we only get the bound

Res s=1 (ζ K (s)) ≤ Q 2 + o(1) 8 (log d L )(log(d 2 K /d L )) ≤ Q 2 + o(1) 8 (log d K ) 2
as d L tends to infinity and d K /d L ≥ log 2 d K . We hope somebody could deal nicely with these error terms or could find a completely different approach to get an explicit bound as close as possible to (1). Combined with the results obtained in [START_REF] Louboutin | Real zeros of Dedekind zeta functions[END_REF], this would have great consequences on the solution of the class number one problem for the non-normal sextic CM-fields (see [START_REF] Louboutin | Some explicit upper bounds for residues of zeta functions of number fields taking into account the behavior of the prime 2[END_REF] Remark 10]).

1.1. Notation. Let us set the notation we will be using throughout the paper.

Let K be a number field, of degree m = r 1 +2r 2 , absolute discriminant d K and Dedekind zeta function ζ K (s). Set

Z K (s) = d K 4 r 2 π m s/2 Γ r 1 (s/2)Γ r 2 (s)ζ K (s).
In particular, Z Q (s) = π -s/2 Γ (s/2)ζ(s). Then Z K (s) is a meromorphic function satisfying the functional equation Z K (1 -s) = Z K (s), with only two poles, at s = 0 and s = 1, both simple. We define (see [START_REF] Louboutin | Upper bounds on L(1, χ) and applications[END_REF](22)] for the positivity)

0 < λ K := Res s=1 (Z K (s)) = √ d K (2π) r 2 Res s=1 (ζ K (s)), 0 < µ K := lim s→1 1 λ K Z K (s) - 1 s(s -1) , 0 < ν K := µ K Res s=1 (ζ K (s)).
In particular, λ Q = 1 and

(2) obtaining explicit bounds on µ K is not easy. In fact, recalling that the Euler-Kronecker constant γ K of K is defined by

µ Q = (2 + γ -log( 4π
ζ K (s) = Res s=1 (ζ K (s)) 1 s -1 + γ K + O(s -1) ,
we have

µ K = γ K + 1 + 1 2 (log d K -m(γ + log(4π
)) + r 2 log 4) and only conditional upper bounds (under the assumption of the Generalized Riemann Hypothesis) for γ K are known:

γ K log log d K ,
where the implied constants are absolute (see [IM, Theorem 3 and Corollary 3.3.2]).

1.2. Bounds on L(1, χ). In [START_REF] Louboutin | Majorations explicites de |L(1, χ)[END_REF] (see also [START_REF] Ramaré | Approximate formulae for L(1, χ)[END_REF], [Edd] and [PE]), we obtained the explicit bound (4)

|L(1, χ)| ≤ 1 2 log f χ + µ Q for the value at s = 1 of the L-function associated with a primitive and even Dirichlet character χ, of conductor f χ > 1 (see also [GS] and [Pin] for better non-explicit bounds).

This was considerably generalized in [Lou04, Theorem 1] and [Ram04, Corollary 1], leading to better explicit bounds taking into account the behavior of χ at given prime integers (see also [Toy] for better non-explicit bounds). Our present approach based on Lemma 3.1 will give a simpler proof of these latter bounds (see the first item of Theorem 1.3).

In [Lou00, Theorem 7], we generalized (4) to the case of L-functions associated with primitive characters on ray class groups of a given number field L. Focussing on characters unramified at all the infinite places of L, we obtained

|L(1, χ)| ≤ Res s=1 (ζ L (s)) × 1 2 log f χ + µ L + o(1) (5) = 1 2 Res s=1 (ζ L (s)) log f χ + ν L + o(1)
, where f χ = N (F χ ) is the norm of the finite part F χ of the conductor of χ. It took us some time to generalize (5), i.e. to obtain better bounds taking into account the behavior of χ at given prime ideals of L. In 2018, we eventually obtained a generalization of [Lou04, Theorem 1] and [Ram04, Corollary 1] (see Proposition 2.1 below for the size of the present error terms):

Theorem 1.1 (see [START_REF] Louboutin | Upper bounds on L(1, χ) taking into account a finite set of prime ideals[END_REF]Theorem 2]). Let P be a given finite set of prime ideals of a given number field L of degree m. Let N (P) = N L/Q (P) be the norm of P ∈ P . Let χ range over the non-trivial and primitive characters on ray class groups for L which are unramified at all the infinite real places of L. Set P 1 (χ) = {P ∈ P ; P F χ }. Then

|L(1, χ)| ≤ B P (χ) × Res s=1 (ζ L (s)) × 1 2 log f χ + µ L + S P (χ) + O (log(ed L )) m-1 (log(ed L f χ )) f χ ,
where the implied constant depends only on m and the norms of the prime ideals in P and where

0 ≤ B P (χ) = P∈P N (P) -1 N (P) -χ(P) ≤ 1, and
0 ≤ S P (χ) = log 2 |P 1 (χ)| + P∈P log N (P) N (P) -1 .
Hence, by (3), whenever f χ ≥ (log(ed L )) 2 we have

(6) |L(1, χ)| ≤ B P (χ) × 1 2 Res s=1 (ζ L (s)) log f χ + ν L + O(log m-1 (ed L )
). 1.3. Main results: bounds on ν L for real quadratic number fields and on Res s=1 (ζ K (s)) for non-normal totally real cubic number fields. Let K be a non-normal totally real cubic number field. The best explicit known bound is [START_REF] Louboutin | Upper bounds for residues of Dedekind zeta functions and class numbers of cubic and quartic number fields[END_REF], where it is said to hold true for d K ≥ 146, but according to [START_REF] Cohen | A Course in Computational Algebraic Number Theory[END_REF]Appendix B4] there are only two totally real cubic number fields of discriminant less than 146, both of them being abelian). In [Lou08, Theorem 8] we obtained a better bound depending on the behavior of the prime 2 in K:

(7) Res s=1 (ζ K (s)) ≤ 1 8 (log d K + λ 3 ) 2 , where λ 3 := 2 + 2γ -2 log(4π) = -1.90761 . . . (see
(8) Res s=1 (ζ K (s)) ≤ Q 2 8 (log d K + 5) 2 ,
where Q 2 ∈ {1, 1/2, 1/3, 1/4, 1/7} defined in (15) below depends explicitly on the splitting behavior of the prime p = 2 in K. In Theorem 1.7, we will obtain still better bounds. Let N be the normal closure of K. Hence N is a totally real sextic field with Galois group isomorphic to the symmetric group of order 6. Let L be the unique (real) quadratic subfield of N. The cubic extension N/L is cyclic. Let χ N/L denote any of the two complex conjugate cubic characters associated with

N/L. Let f N/L = N L/Q (F N/L ) be the norm of the conductor F N/L of χ N/L . We have (9) (ζ K (s)/ζ(s)) 2 = ζ N (s)/ζ L (s) = L(s, χ N/L )L(s, χN/L ), which gives d 2 K = d N /d L = d 2 L f 2 N/L , i.e. (10) f N/L = d K /d L and Res s=1 (ζ K (s)) = |L(1, χ N/L )|.
Now, let L be a real quadratic number field, hence associated with a primitive even Dirichlet character χ L of conductor d L . By (4), we have

(11) Res s=1 (ζ L (s)) = L(1, χ L ) ≤ 1 2 log d L + µ Q .
While we have no explicit upper bounds on µ L in (5), we do have one on [START_REF] Louboutin | Majorations explicites du résidu au point 1 des fonctions zêta de certains corps de nombres[END_REF]Proposition 6] or [Lou01, Theorem 2], a better bound than the one in (3) for m = 2.

(12) ν L := µ L Res s=1 (ζ L (s)) ≤ 1 8 log 2 d L , by
Remark 1.2. Let L be a real quadratic number field of discriminant d L . Then

ν L = L(1, χ L ) µ Q + Λ (1, χ L ) Λ(1, χ L ) = L(1, χ L ) 1 2 log d L + 1 -log(4π) + L (1, χ L ), by (25)
. Starting from this formula it would be difficult to end up with the factor 1/8 in (12), as by (11) it would be difficult to have anything better than a factor 1/4. Using (10), (5), (11), and (12) we obtain a bound on Res s=1 (ζ K (s)) which is better than the one in (7

) if d K ≥ d 2 L [Lou18, p. 255].
In this paper we improve upon (12) by taking into account the behavior of small prime integers in L: see ( 14) and Corollary 1.5. Then, using (10), (5), (13) for χ = χ L , and (14) we obtain a bound on Res s=1 (ζ K (s)) which is better than the one in (8

) if d K ≥ d L log 2 d L -see Theorem 1.7.
In other words, the aim of this paper is to prove Theorem 1.3 below. Not only does our present approach based on Lemma 3.1 give a simpler proof of (13) than the ones in [START_REF] Louboutin | Explicit upper bounds for |L(1, χ)| for primitive characters χ[END_REF] and [START_REF] Ramaré | Approximate formulae for L(1, χ). II[END_REF]; it also enables us to get (13) and (14) in one stroke.

Theorem 1.3. Let d 1 , d 2 ≥ 1 be coprime square-free positive integers.

(i) Let χ range over the primitive and even Dirichlet characters of conductor

f χ > 1 for which gcd(d 1 , f χ ) = 1 and d 2 | f χ . Setting c 1 (d 1 , d 2 ) = µ Q + ω(d 1 ) log 2 + p|d 1 d 2 log p
p-1 , we have an effective bound with a negative error term for ω(d 2 ) ≤ 1 and f χ large enough:

(13) |L(1, χ)| ≤ 1 2 p|d 1 d 2 p -1 p -χ(p) × {log f χ + 2c 1 (d 1 , d 2 )} + O log f χ f χ .
(ii) Let L range over the real quadratic fields for which

p | d 1 ⇒ χ L (p) = -1 and p | d 2 ⇒ χ L (p) = 0. For c 2 (d 1 , d 2 ) = 2µ Q -1 + p|d 1 2p log p p 2 -1 + p|d 2 log p
p-1 and c 3 (d 1 , d 2 ) some explicit constant depending only on d 1 and d 2 given in Section 3.1, we have an effective bound with a negative error term for ω(d 2 ) ≤ 2 and d L large enough:

(14) ν L ≤ 1 8 p|d 1 d 2 p -1 p -χ L (p) × {(log d L + 2c 2 (d 1 , d 2 )) 2 + 4c 3 (d 1 , d 2 )} + O log 2 d L √ d L .
From the effectiveness of these bounds we obtain Corollary 1.4 (compare with (4)). For any real quadratic field L we have

Res s=1 (ζ L (s)) ≤      (log d L + 2 + γ -log(4π))/2 in all cases, (log d L + 2 + γ -log π)/4 if 2 is ramified in L, (log d L + 2 + γ -log(π/4))/6 if 2 is inert in L.
Corollary 1.5 (compare with (12)). Set κ = 4µ Q -2 + (8 log 2)/3 = -0.05922 . . . and κ = 4κ 2 = 19.52508 as defined in (39). For any real quadratic field L we have

ν L ≤          (log d L ) 2 /8 in all cases, (log d L ) 2 /16 if 2 is ramified in L, ((log d L + κ) 2 + κ )/24 if 2 is inert in L, (log d L ) 2 /24 if 2 is inert in L and d L ≥ 5 • 10 71 .
Remark 1.6. Numerical computations of ν L based on [Lou98b, Section 4.2] suggest that the fourth bound (which follows from the third one) holds true whenever 2 is inert in L and d L ≥ 5. Indeed, with a 24h30mn long high precision computation on a 3GHz PC using UBASIC, we checked that max{24ν L /log 2 d L ; d L ≡ 5 (mod 8) and d L ≤ 10 5 } = 0.9055 . . . for d L = 62269. Notice that χ L (p) = 1 for p ∈ {3, 5, 7, 11, 13, 17, 19, 23, 29} and d L = 62269.

Let p ≥ 2 be a given prime. Take P equal to the set of prime ideals of L dividing p. To simplify the presentation, we assume that d L goes to infinity and that f N/L = d K /d L log 2 d L , to be allowed to use (6). By (10), ( 6) and Theorem 1.3, we have

Res s=1 (ζ K (s)) ≤ B P (χ N/L ) × 1 2 Res s=1 (ζ L (s)) log(d K /d L ) + ν L + O(log d L ) with Res s=1 (ζ L (s)) ≤ B {p} (χ L ) × 1 + o(1) 2 log d L and ν L ≤ B {p} (χ L ) × 1 + o(1) 8 log 2 d L .
Hence,

Res s=1 (ζ K (s)) ≤ Q p + o(1) 8 (log d L )(log(d 2 K /d L )) + O(log d L ),
where

Q p := B {p} (χ L ) × B {P; P|(p) in L} (χ N/L ).
From (9) and the uniqueness of the Euler product of Dirichlet series, we have

1 - 1 p s P|(p) in K 1 - 1 N K/Q (P) s -1 2 = P|(p) in L 1 - χ N/L (P) N L/Q (P) -1 1 - χN/L (P) N L/Q (P) -1
.

Hence,

Q p = 1 -1 p 3 P|(p) in K 1 - 1 N K/Q (P) (15) =                  1 if (p) = P 1 P 2 P 3 splits in K, p-1 p if (p) = P 1 P 2 2 is partially ramified in K, p-1 p+1 if (p) = P 1 P 2 in K, (p-1) 2 p 2 if (p) = P 3 is totally ramified in K, (p-1) 2 p 2 +p+1 if (p) = P is inert in K.
Putting everything together and using XY ≤ ((X + Y )/2) 2 for X, Y ≥ 0, we obtain Theorem 1.7 (compare with (7) and [Lou08, Theorem 8]). Let p ≥ 2 be a given prime integer. Let L be the real quadratic subfield of the normal closure N of a non-normal totally real cubic number field K. Let Q p be as in (15). Then

Res s=1 (ζ K (s)) ≤ Q p + o(1) 8 (log d L )(log(d 2 K /d L )) ≤ Q p + o(1) 8 log 2 d K ,
where the effective error terms tend to 0 as d L → ∞, and

d K /d L ≥ log 2 d L .
2. Proof of the size of the error term in Theorem 1.1

Proposition 2.1 (see [Lou18, Proposition 9]). Let L be a totally real number field of degree m. Let P 1 and P 2 be two disjoint finite sets of prime ideals of L. For a > 0 and X > 0, set

Π L (s) = a 1-s s -1 P∈P 1 1 - 1 N (P) s + a s s P∈P 1 1 - 2 N (P) + 1 N (P) 1-s × P∈P 2 1 - 1 N (P) s and F L (s, X) = Π L (s)X s-1 Z L (s)/ d L ,
a meromorphic functions with at most two poles, a double one at s = 1 and an at most double one at s = 0. Define

I L (X) := 1 2πi (s)=α F L (s, X) ds (16) = Res s=1 (F L (s, X)) + Res s=0 (F L (s, X)) + R L (X)
(which does not depend on α > 1), where

R L (X) := 1 2πi (s)=1-α F L (s, X) ds = 1 2πi (s)=α F L (1 -s, X) ds (17) = 1 2πi (s)=α Π L (1 -s)Z L (s) X s √ d L ds = O log m (ed L ) X .
Let χ be a non-trivial, primitive and character on a ray class group for L, unramified at all the infinite places of K. Assume that P ∈ P 1 implies P F χ and P ∈ P 2 implies P | F χ . Then

(18) P∈P 1 - χ(P) N (P) L(1, χ) ≤ I L ( f χ ) = Res s=1 (F L (s, X)) + O log m (ed L ) X .
The implied constants in (17)-( 18) depend only on a, m and the norms of the prime ideals in P .

Proof. Only the sizes of the error terms in (17) and (18) require proof.

For 1 < σ = (s) = 1 + ≤ 4 we have |ζ L (s)| ≤ ζ L (σ) ≤ ζ(σ) m , ζ(σ) = 1 σ -1 + n≥1 1 n σ - n+1 n dx x σ ≤ 1 σ -1 + n≥1 1 n σ - 1 (n + 1) σ = 1 σ -1 + 1 = σ σ -1 ≤ 4 and |Z L (s)/(X s √ d L )| ≤ -m d /2 L |Γ (σ/2)| m (4/ √ π) m /X
. By [Lou00, Lemma 10], taking = 2m/log(ed L ) we have ≤ 3 and (17) follows.

The last assertion follows from ( 16), (17) and the formula (see [Lou18, Lemma 10])

Res s=0 (F L (s, X)) = - Res s=1 (ζ L (s)) X × P∈P 1 1 - 1 N (P) ×            log X -µ L + 1 + log a -a if P 2 = ∅ and P 1 = ∅, log X -µ L + 1 + log a + P∈P log N (P) N (P)-1 if P 2 = ∅ and P 1 = ∅, log N (P 0 ) if P 2 = {P 0 }, 0 otherwise. which yields Res s=0 (F L (s, X)) (log m-1 (ed L ))(log(ed L X))/X, by (3)
. Theorem 1.1 follows from the last assertion and the fact that

Res s=1 (F L (s, X)) = P∈P 1 - 1 N (P) × Res s=1 (ζ L (s)) × log X + µ L -1 -log a + 2 |P 1 | a + P∈P log N (P) N (P) -1 is minimal for a = 1/2 |P 1 | .
3. Proof of Theorem 1.3. In this section we use Lemma 3.1 below to prove Theorem 1.3.

Notice that the present proof of (13) drastically shortens the one of [Lou04, Theorem 1] and that the use of Lemma 3.1 would also drastically shorten the proof of [Lou18, Theorem 2].

First, in (24) and (25) we give integral representations on the vertical line (s) = α > 1 of L(1, χ) and ν L , respectively. Second, in Lemma 3.2 we prove the positivity of some inverse Mellin transforms appearing in these integral representations. Third, in (29) we deduce bounds for |L(1, χ)| and ν L as inverse Mellin transforms of some G 1 (s) = Π 1 (s)Z Q (s) and G 2 (s) = Π 2 (s)Z Q (s) for some explicit factors Π 1 (s) and Π 2 (s) depending on both d 1 and d 2 . Fourth, in (32) we reduce the proof of Theorem 1.3 to the computation of Res s=1 (G 1 (s)) and Res s=1 (G 2 (s)). Finally, in (33) and (34) we compute these residues. By Stirling's formula, in any given vertical strip we have

(19) Γ (σ + it) = O(e -π|t|/2 |t| σ-1/2 ) for α 1 ≤ σ ≤ α 2 and |t| ≥ 1.
Let χ be a primitive and even Dirichlet character of conductor f > 1. Then

Λ(s, χ) = (π/f ) -s/2 Γ (s/2)L(s, χ)
is an entire function of order one satisfying (20)

Λ(σ + it, χ) = O(e -π|t|/4 |t| M ) for α 1 ≤ σ ≤ α 2 and |t| ≥ 1,
for some M = M (α 1 , α 2 ) (e.g. see [START_REF] Davenport | Multiplicative Number Theory[END_REF]Chapter 12] and [Lou01, Sec-tion 4.1]). Recall also the functional equation

(21) Λ(1 -s, χ) = W χ Λ(s, χ)
for some complex root number W χ of absolute value equal to one (see e.g. [START_REF] Davenport | Multiplicative Number Theory[END_REF]Chapter 9]). The bounds ( 19) and ( 20) show that all the integrals on vertical lines of the complex plane below are absolutely convergent and that we are allowed to move these lines to the left, provided that we take into account the residues at the poles encountered.

Lemma 3.1. Let d 1 ≥ 1 be a square-free integer. Set

P (s, χ, d 1 ) = p|d 1 1 - χ(p) p s .
Let χ be a primitive and even Dirichlet character of conductor f > 1. Assume that gcd(d

1 , f ) = 1. Set L(s, χ, d 1 ) := n≥1 gcd(n,d 1 )=1 χ(n) n s = P (s, χ, d 1 )L(s, χ)
(the L-function associated with the character modulo d 1 f induced by χ) and Λ(s, χ, d 1 ) = (π/f ) -s/2 Γ (s/2)L(s, χ, d 1 ) = P (s, χ, d 1 )Λ(s, χ).

For a > 0 and α > 1, we have

Λ(1, χ, d 1 ) = 1 2πi (s)=α Λ(s, χ, d 1 )a 1-s s -1 + Λ(1 -s, χ, d 1 )a s s ds, (22)
Λ (1, χ, d 1 ) = 1 2πi (s)=α Λ(s, χ, d 1 ) (s -1) 2 - Λ(1 -s, χ, d 1 ) s 2 ds, (23)
where these integrals are absolutely convergent and do not depend on α, by (20). Moreover,

(24) p|d 1 1 - χ(p) p L(1, χ) = Λ(1, χ, d 1 )/ f
and for a real quadratic number field L we have

(25) p|d 1 1 - χ L (p) p ν L = µ Q -p|d 1 χ L (p) log p p-χ L (p) Λ(1, χ L , d 1 ) + Λ (1, χ L , d 1 ) √ d L ,
where χ L is the real primitive quadratic Dirichlet character of conductor d L associated with L.

Proof. Set

G(s) = Λ(s, χ, d 1 )a 1-s s -1 + Λ(1 -s, χ, d 1 )a s s .
The bound (20) allows us to move the vertical line of integration (s) = α > 1 to the left to (s) = 1 -α < 0. Noticing that we pick up residues at s = 1 and s = 0, both simple and of residue Λ(1, d 1 , χ), we obtain

1 2πi (s)=α G(s) ds = Res s=1 (G(s)) + Res s=0 (G(s)) + 1 2πi (s)=1-α G(s) ds = 2Λ(1, d 1 , χ) + 1 2πi (s)=α G(1 -s) ds = 2Λ(1, d 1 , χ) - 1 2πi (s)=α G(s) ds,
by making the change of variables s → 1-s to come back to the line (s) = α and then using G(1 -s) = -G(s). Identity (22) follows. In the same way, we have

Λ (1, χ, d 1 ) -(log a)Λ(1, χ, d 1 ) = 1 2πi (s)=α Λ(s, χ, d 1 )a 1-s (s -1) 2 - Λ(1 -s, χ, d 1 )a s s 2 ds,
which for a = 1 gives (23). Now, let L be a real quadratic number field . Then

W L = 1, Λ(1 -s, χ L ) = Λ(s, χ L ), (26) Λ(s, χ L ) = Z L (s)/Z Q (s) = λ L λ Q (1 + (µ L -µ Q )(s -1) + O((s -1) 2 )), and
Λ(s, d 1 , χ L ) = P (s, χ, d 1 )Λ(s, χ L ) = λ L λ Q P (1, χ, d 1 ) 1 + µ L -µ Q + P (1, χ, d 1 ) P (1, χ, d 1 ) (s -1) + O((s -1) 2 ) .
Then (25) follows by using λ Q = 1 and ν L = λ L µ L / √ d L , for any totally real number field L. Lemma 3.2. For X > 0, c real, m ≥ 1 an integer and k ∈ {1, 2}, set

(27) F k (X, c, m) := 1 2πi (s)=α X -s Γ m (s/2) ds (s -c) k (α > max(0, c)),
an absolutely convergent integral not depending on α, by (19). Then

F k (X, c, m) > 0, ∂F 1 ∂c (X, c, m) = F 2 (X, c, m) > 0, ∂F 2 ∂c (X, c) > 0 and X → F k (X, c, m) is decreasing. Proof. Clearly, (t 1 , . . . , t m , t) → e -(t 1 +•••+tm) (t 1 • • • t m ) (α+it)/2-1 X -(α+it) (α -c + it) 2 is in L 1 (R m + × R, C
) for α > 0 and α = c. Now, for α > c we have

1 2πi (s)=α u s (s -c) 2 ds = 0 if u < 1, u c log u if u > 1.
(move the line of integration (s) = α of this absolutely convergent integral to the right to infinity for 0 < u < 1 and to the left to infinity for u > 1, in which case you pick up a residue at the pole s = c). Therefore, by Fubini's theorem, we have

F 2 (X, c, m) = J m (X, c) > 0,
where

J m (X, c) := • • • t 1 ,...,tm>0 t 1 •••tm>X 2 e -(t 1 +•••+tm) √ t 1 • • • t m X c log √ t 1 • • • t m X dt 1 • • • dt m t 1 • • • t m = 1 2 • • • t 1 ,...,tm>0 t 1 •••tm>1 e -X 2/m (t 1 +•••+tm) (t 1 • • • t m ) c/2 log(t 1 • • • t m ) dt 1 • • • dt m t 1 • • • t m .
Finally, fix X > 0 and α > 0. Let c range in (-∞, α). Clearly,

c → I m (X, c) := • • • t 1 ,...,tm>0 t 1 •••tm>1 e -X 2/m (t 1 +•••+tm) (t 1 • • • t m ) c/2 dt 1 • • • dt m t 1 • • • t m is differentiable on (-∞, α), with derivative c → J m (X, c) = F 2 (X, c, m). Us- ing Stirling's formula (19), it is clear that c → F 1 (X, c, m) is differentiable on (-∞, α), with derivative c → F 2 (X, c, m). Hence, c → I m (X, c)-F 1 (X, c, m) is constant on (-∞, α). Since lim c→-∞ I m (X, c) = lim c→-∞ F 1 (X, c, m) = 0, by
Lebesgue's dominated convergence theorem we obtain

F 1 (X, c, m) = I m (X, c) > 0.
Finally, these two integral representations of F 1 (X, c, m) and F 2 (X, c, m) show that they are decreasing functions of X > 0.

Lemma 3.3 (see [Lou18, Lemma 7]). Set φ(δ) = p|δ (p -1) and φ(δ) = p|δ (p -2) for δ ≥ 1 a square-free integer (by convention, empty products are 1). Let n≥1 a n be absolutely convergent. Then for d 1 ≥ 1 we have

δ|d 1 µ(δ)δ n≥1 δ|n a n = δ|d 1 µ(δ)φ(δ) n≥1 gcd(n,d 1 )=δ a n and δ|d 1 φ(δ) n≥1 gcd(n,d 1 )=δ a n = δ|d 1 φ(δ) n≥1 δ|n a n . Theorem 3.4. For d ≥ 1, set P (s, d) := p|d 1 - 1 p s and Q(s, d) := p|d 1 1 - 2 p + 1 p 1-s
(empty products are 1). For d 1 and d 2 coprime square-free positive integers, k ∈ {1, 2}, a > 0, X > 0, and α > 1, set

Π k (s, a, d 1 , d 2 ) := a 1-s (s -1) k P (s, d 1 ) + a s s k Q(s, d 1 ) P (s, d 2 ), G k (s, a, d 1 , d 2 , X) := Π k (s, a, d 1 , d 2 )X s-1 Z Q (s) and (28) I k (a, d 1 , d 2 , X) := 1 2πi (s)=α G k (s, a, d 1 , d 2 , X) ds.
Then, for any primitive and even Dirichlet character χ of conductor f > 1 for which gcd(d 1 , f ) = 1 and d 2 | f , we have

(29) |Λ(1, χ, d 1 )|/ f ≤ I 1 (a, d 1 , d 2 , f ), |Λ (1, χ, d 1 )|/ f ≤ I 2 (1, d 1 , d 2 , f ).
Proof. Let the notation be as in Lemma 3.1 and (27). First, by the positivity of F k (X, c, 1) (Lemma 3.2), the absolute value of the first integral in ( 22) and ( 23)

I 1 := 1 2πi (s)=α Λ(s, χ, d 1 )a 1-s (s -1) k ds = n≥1 gcd(n,d 1 d 2 )=1 χ(n)aF k an π f , 1, 1
(where we have used χ(n) = 0 for gcd(n, d 2 ) > 1) is less than or equal to

n≥1 gcd(n,d 1 d 2 )=1 aF k an π f , 1, 1 = 1 2πi (s)=α a 1-s (s -1) k f s/2 P (s, d 1 d 2 )Z Q (s) ds.
Second, let us deal with the second integral in ( 22) and (23):

I 2 := 1 2πi (s)=α Λ(1 -s, χ, d 1 )a s s k ds.
We have

p|d 1 1 - χ(p) p s = δ|d 1 µ(δ) χ(δ) δ s = µ(d 1 )χ(d 1 ) d s 1 δ|d 1 µ(δ) χ(δ)δ s ,
by making the change of variables δ → d 1 /δ. Using (21), we obtain

Λ(1 -s, χ, d 1 ) = W χ µ(d 1 )χ(d 1 )d s-1 1 Λ(s, χ) δ|d 1 µ(δ) χ(δ)δ 1-s .
By the first assertion in Lemma 3.3, for (s) > 1 we have

L(s, χ) δ|d 1 µ(δ) χ(δ)δ 1-s = δ|d 1 µ(δ)δ n≥1 δ|n χ(n)n -s = δ|d 1 µ(δ)φ(δ) n≥1 gcd(n,d 1 )=δ χ(n)n -s .
By the second assertion in Lemma 3.3 the absolute value of

I 2 = W χ µ(d 1 )χ(d 1 ) d 1 δ|d 1 µ(δ)φ(δ) n≥1 gcd(n,d 1 )=δ χ(n)F k n ad 1 π f , 0, 1
is less than or equal to

1 d 1 δ|d 1 φ(δ) n≥1 gcd(n,d 1 )=δ gcd(n,d 2 )=1 F k n ad 1 π f , 0, 1 = 1 d 1 δ|d 1 φ(δ) n≥1 gcd(n,d 2 )=1 F k δn ad 1 π f , 0, 1 = 1 2πi (s)=α a s s k f s/2 Q(s, d 1 )P (s, d 2 )Z Q (s) ds,
where we have used χ(n) = 0 for gcd(n, d 2 ) > 1 and the positivity of F k (X, 0, 1) (Lemma 3.2). Third, the bounds in (29) with I k (a, d 1 , d 2 , X) as in (28) follow.

Lemma 3.5. Let α > 1 and let the notation be as in Theorem 3.4. Then

(30) I k (a, d 1 , d 2 , X) = ρ k (s, a, d 1 , d 2 , X) + R k (a, d 1 , d 2 , X),
where

ρ k (s, a, d 1 , d 2 , X) = Res s=1 (G k (s, a, d 1 , d 2 , X)) + Res s=0 (G k (s, a, d 1 , d 2 , X))
and

(31) R k (a, d 1 , d 2 , X) := 1 2πi (s)=α Π k (1 -s, a, d 1 , d 2 )X -s Z Q (s) ds = O(X -α ). Since Res s=0 (G k (s, a, d 1 , d 2 , X)) = O((log k X)/X) as X → +∞,
we have

(32) I k (a, d 1 , d 2 , X) = Res s=1 (G k (s, a, d 1 , d 2 , X)) + O((log k X)/X).
Proof. Moving the line (s) = α > 1 in (28) to the left to (s) = 1-α < 0 we pick up residues at s = 1 and s = 0 and using

Z Q (1 -s) = Z Q (s), we obtain R k (a, d 1 , d 2 , X) := 1 2πi (s)=1-α G k (s, a, d 1 , d 2 , X) ds = 1 2πi (s)=α G k (1 -s, a, d 1 , d 2 , X) ds = 1 2πi (s)=α Π k (1 -s, a, d 1 , d 2 )X -s Z Q (s) ds, and (30) follows.
Lemma 3.6. Let ω(d) := p|d 1 denote the number of distinct prime divisors of d. Set (where empty products are 1 and empty sums are 0)

P (d) := p|d 1 - 1 p , S 1 (d) := p|d log p p -1 , S 2 (d) := p|d p log 2 p (p -1) 2 .
Let µ Q be as in (2). Set

µ Q := π 2 8 -γ 2 -2γ(1) = 1.04615 . . . , where
ζ(s) = 1 s -1 + γ -γ(1)(s -1) + O((s -1) 2 ).
Let the notation be as in Theorem 3.4. Then

(33) Res s=1 (G 1 (s, a, d 1 , d 2 , X)) = P (d 1 d 2 ) × {log X + µ Q -1 -log a + 2 ω(d 1 ) a + S 1 (d 1 d 2 )}, which is minimal for a = 1/2 ω(d 1 ) , (34)
Res s=1 (G 2 (s, 1, d 1 , d 2 , X)) = P (d 1 d 2 ) 2 × log X + µ Q -1 + S 1 (d 1 d 2 ) 2 + µ Q + 2 ω(d 1 )+1 -S 2 (d 1 d 2 ) , Res s=0 (G 1 (s, a, d 1 , d 2 , X)) = - P (d 1 ) X ×          log X -µ Q + 1 + log a -a if d 2 = d 1 = 1, log X -µ Q + 1 + log a + S 1 (d 1 ) if d 2 = 1 and d 1 > 1, log d 2 if d 2 is prime, 0 otherwise;
and

Res s=0 (G 2 (s, a, d 1 , d 2 , X)) = - P (d 1 ) 2X ×                              (log X -µ Q + 1 + log a) 2 + µ Q + 2a if d 2 = d 1 = 1, (log X -µ Q + 1 + log a + S 1 (d 1 )) 2 + µ Q + p|d 1 (p -2) log 2 p (p -1) 2 if d 2 = 1 and d 1 > 1, (log d 2 )(2 log X -2µ Q + 2 + 2 log a + 2S 1 (d 1 ) -log d 2 ) if d 2 is prime, 2(log p)(log q) if d 2 = pq, 0 otherwise.
In particular, by (31

), for k ∈ {1, 2} we have Res s=0 (G k (s, a, d 1 , d 2 , X)) + R k (a, d 1 , d 2 , X) ≤ 0 and I k (a, d 1 , d 2 , X) ≤ Res s=1 (G k (s, a, d 1 , d 2 , X)) for ω(d 2 )
≤ k and X large enough.

3.1. Proof of Theorem 1.3. For the first bound, use (24), (29), (32), (33) for a = 1/2 ω(d 1 ) , and the last assertion of Lemma 3.6. For the second bound, using (25), (29), (32), (33) for a = 1/2 ω(d 1 ) , and (34), we obtain

p|d 1 1 + 1 p ν L = µ Q + p|d 1 log p p+1 Λ(1, χ L , d 1 ) + Λ (1, χ L , d 1 ) √ d L ≤ A × I 1 (2 -ω(d 1 ) , d 1 , d 2 , f χ ) + I 2 (1, d 1 , d 2 , d L ) = A × Res s=1 (G 1 (s, 2 -ω(d 1 ) , d 1 , d 2 , d L )) + Res s=1 (G 2 (s, 1, d 1 , d 2 , d L )) + O log 2 d L √ d L = 1 8 p|d 1 d 2 1 - 1 p {(log d L + 2c 2 (d 1 , d 2 )) 2 + 4c 3 (d 1 , d 2 )} + O log 2 d L √ d L ,
where

A = µ Q + p|d 1 log p p+1 , B = µ Q + log 2 ω(d 1 ) + S 1 (d 1 d 2 ), C = µ Q -1 + S 1 (d 1 d 2 ), D = µ Q + 2 ω(d 1 )+1 -S 2 (d 1 d 2 ), c 2 (d 1 , d 2 ) = A + C and c 3 (d 1 , d 2 ) = D + A(2B -2C -A). The error terms O log 2 d L √ d L
are negative for d L large enough, by the last assertion of Lemma 3.6.

Proof of Corollary 1.4

Remark 4.1. Take c ∈ {0, 1}, k ∈ {1, 2} and X > 0. Taking α = 2 in (27) and using |ζ(2

+ it)| ≤ ζ(2) = π 2 /6, we have 0 < Z k (X, c) := 1 2πi (s)=α X -s Z Q (s) ds (s -c) k ≤ M k (c) 12X 2 , where M k (c) := (s)=2 Γ (s/2) (s -c) k dt = ∞ -∞ πt/2 sinh(πt/2) dt ((2 -c) 2 + t 2 ) k/2 .
Moreover, M 1 (0) = 1.88826 . . . , M 1 (1) = 2.94137 . . . , M 2 (0) = 0.77941 . . . and M 2 (1) = 2.10564 . . . . Moreover, X → Z k (X, c) is decreasing and Z k (X, 0) ≤ Z k (X, 1), by Lemma 3.2. Theorem 4.2. Let χ be a primitive even Dirichlet character of conductor f > 1. Then

(35) |L(1, χ)| ≤ 1 2 log f + µ Q . Proof.
We apply (24), ( 29), (30), (31) and Lemma 3.6 with a = d 1 = d 2 = 1 and Π 1 (s, 1, 1, 1) = 1 s-1 + 1 s . We obtain

|L(1, χ)| ≤ I( √ f ), where I(X) = ρ(X) + R(X) with ρ(X) = (log X + µ Q ) -(log X -µ Q )/X, R(X) = - 1 2πi (s)=α 1 s + 1 s -1 X -s Z Q (s) ds = -Z 1 (X, 0) -Z 1 (X, 1) < 0.
Since χ is a primitive even Dirichlet character of conductor f > 1, we have

X = √ f ≥ √ 5 ≥ exp(µ Q ).
Theorem 4.3. Let p ≥ 2 be a prime. Let χ be a primitive even Dirichlet character of conductor f divisible by p. Assume that f ≥ 5p 12 log p 2 , which is the case for 2 ≤ p ≤ 139. Then

(36) |L(1, χ)| ≤ 1 - 1 p 1 2 log f + µ Q + log p p -1 .
Proof. We apply (24), (29), (30), (31) and Lemma 3.6 with a = d 1 = 1,

d 2 = p and Π 1 (s, 1, 1, p) = 1 -1 p s 1 s-1 + 1 s . We obtain |L(1, χ)| ≤ I( √ f ), where I(X) = ρ(X) + R(X) with ρ(X) = 1 - 1 p log X + µ Q + log p p -1 - log p X , R(X) = 1 2πi (s)=α (p s-1 -1) 1 s + 1 s -1 X -s Z Q (s) ds = Z 1 (X/p, 0) + Z 1 (X/p, 1) p -Z 1 (X, 0) -Z 1 (X, 1) ≤ Z 1 (X/p, 0) + Z 1 (X/p,
(37) |L(1, χ)| ≤ p -1 p -χ(p) × 1 2 log f + µ Q + log 2 + log p p -1 .
Proof. We apply (24), ( 29), ( 30), (31) and Lemma 3.6 with a = 1/2, d 1 = p and d 2 = 1. We obtain

1 - χ(p) p L(1, χ) ≤ I( f ),
where

I(X) = ρ(X) + R(X) with ρ(X) = 1 - 1 p log X + µ Q + log 2 + log p p -1 - log X + log π-γ 2 + log p p-1 X , R(X) = 1 2πi (s)=α 2 -s s (p s-1 -1) - 2 s-1 s -1 1 - 2 p + 1 p s X -s Z Q (s) ds = 1 p Z 1 (2X/p, 0) -Z 1 (2X, 0) - p -2 2p Z 1 (X/2, 1) -1 2 Z 1 (pX/2, 1) ≤ 1 p Z 1 (2X/p, 0) ≤ p 24X 2 , by Remark 4.1. Hence, -1 - 1 p log √ f + log π-γ 2 + log p p-1 √ f + R( f ) ≤ p -12 √ f 1 -1 p log f + log π -γ + 2 log p p-1 24f ≤ p -11 √ f (log f + log π -γ) 24f if p ≥ 13.
The first bound is negative for f ≥ 5 and 2 ≤ p ≤ 61. The second bound is negative for √ f log(πe -γ f ) ≥ p/11, hence for f ≥ p 6 log p 2 and p ≥ 67.

5. Proof of Corollary 1.5 5.1. Proof of the first and second bounds in Corollary 1.5. First assume that d 1 = 1. Then Λ(s, χ, d 1 ) = Λ(s, χ). Lemma 3.1 with a = 1 and

d 1 = 1 gives Λ(1, χ L ) = 1 2πi (s)=α Λ(s, χ L ) 1 s -1 + 1 s ds (α > 1), Λ (1, χ L ) = 1 2πi (s)=α Λ(s, χ L ) 1 (s -1) 2 - 1 s 2 ds (α > 1).
By Lemma 3.2, for X > 0 we have

1 2πi (s)=α X -s Γ (s/2) 1 s -1 + 1 s ds = F 1 (X, 1, 1) + F 1 (X, 0, 1) > 0, 1 2πi (s)=α X -s Γ (s/2) 1 (s -1) 2 - 1 s 2 ds = F 2 (X, 1, 1) -F 2 (X, 0, 1) > 0.
Hence, by (25), we have ν

L ≤ I(d 2 , √ d L ), where I(d 2 , X) := 1 2πi (s)=α G(s, d 2 , X) ds = Res s=1 (G(s, d 2 , X)) + Res s=0 (G(s, d 2 , X)) + R(d 2 , X) with G(s, d 2 , X) := p|d 2 1- 1 p s 1 (s -1) 2 - 1 s 2 +µ Q 1 s -1 + 1 s X s-1 Z Q (s), Res s=1 (G(s, d 2 , X)) = P (d 2 ) 2 × log X + 2µ Q -1 + S 1 (d 2 ) 2 -µ Q -S 2 (d 2 ) , Res s=0 (G(s, d 2 , X)) = 1 2X ×          (log X -2µ Q + 1) 2 -µ Q if d 2 = 1, (log p)(2 log X -4µ Q +2-log p) if d 2 = p, 2(log p)(log q) if d 2 = pq, 0 otherwise,
where P (d), S 1 (d) and S 2 (d) are as in Lemma 3.6 and where

µ Q = 1 + (1 -µ Q ) 2 -µ Q = 1 + (1 -µ Q ) 2 -π 2 /8 + γ 2 + 2γ(1) = 0.90818 . . .

and (by taking

α = 2) R(d 2 , X) := 1 2πi (s)=1-α G(s, d 2 , X) ds = 1 2πi (s)=α G(1 -s, d 2 , X) ds = O(X -2 ).
Hence we get a neat asymptotic and effective bound which we will make explicit in Theorems 5.2 and 5.3:

Theorem 5.1. Let d 2 ≥ 1 be a square-free integer. Let L range over the real quadratic fields of discriminants divisible by d 2 . Then for d L effectively large enough we have

ν L ≤ 1 8 p|d 2 1 - 1 p × log d L + 4µ Q -2 + 2 p|d 2 log p p -1 2 .
In particular, for

d 1 = d 2 = 1, we have G(1-s, 1, X) = -G(s, 1, 1/X)/X and R(1, X) ≤ 0,
ν L = µ L Res s=1 (ζ L (s)) ≤ 1 8 log 2 d L and ν L ≤ (log d L + 4µ Q -2) 2 /8 for d L > 145.
Take d 2 = p a prime integer. Then

I(p, X) ≤ 1 2 1 - 1 p × log X + 2µ Q -1 + log p p -1 2 -µ Q - p log 2 p (p -1) 2 + log p 2X (2 log X -4µ Q + 2 -log p) + R(p, X), where R(p, X) = 1 2πi (s)=α (p s-1 -1) 1 (s -1) 2 - 1 s 2 + µ Q 1 s -1 + 1 s X -s Z Q (s) ds ≤ Z 2 (X/p, 1) + µ Q (Z 1 (X/p, 1) + Z 1 (X/p, 0)) p ≤ p(M 2 (1) + µ Q (M 1 (1) + M 1 (0))) 12X 2 ≤ p 5X 2
, by Remark 4.1. Hence, we get Theorem 5.3 (second bound of Corollary 1.5). Let p ≥ 2 be a prime integer. Let L be a real quadratic number field of discriminant d L . Assume that p divides d L . Assume that d L ≥ 4p, which is the case for p = 2, or that p ≥ 337. Then Finally, for p = 2, we have

ν L = µ L Res s=1 (ζ L (s)) ≤ 1 8 1 - 1 p × log d L + 4µ Q -2 + 2 log p p -1 2 . For p = 2 we have 4µ Q -2 + 2 log p p-1 = 2(1 + γ -log(2π)) = -0.52132 . . . . Proof. Recall that we take X = √ f ≥ √ p, since p divides f > 1. Con- sider the function f (X) := - 1 2 1 - 1 p × µ Q + p log 2 p (p -1) 2 + log p 2X (2 log X -4µ Q + 2 -log p) + p 5X 2 . Then f (X) = - log p 2X 2 (2 log X -4µ Q -log p) - 2p 5X 3 is negative for X ≥ √ p.
R(2, X) ≤ µ Q + log 2 3 M 1 (0) 24X 2 + M 2 (0) 12X 2 + M 2 (1) 48X 2 ≤ 3 20X 2
and Res s=0 (H(s, 2, X)) + R(2, X) ≤ -1 4X 2 X log 2 X -3 5 is negative for

X = √ d L ≥ √ 5.

  ))/2 = 0.02309 . . . .Whereas we have the positivity of λK , µ K , ν K and explicit bounds (see [Lou01, Theorem 1]) (3) Res s=1 (ζ K (s)) ≤ e log d K 2(m -1) m-1 and ν K ≤ e log d K 2m m ,

  as in the proof of Theorem 4.2. Hence, we obtain (compare with Theorem 4.2):Theorem 5.2 (First bound of Corollary 1.5). Let L be a real quadratic number field of discriminant d L . Then

p+21 21 log 2 p+21 21 2 .

 2 Finally, f ( √ p) ≤ 0 for p ≥ 337 and f ( √ 4p) ≤ 0 for p ≥ 2.5.2. Proof of the third and fourth bounds in Corollary 1.5Theorem 5.4. Let p ≥ 2 be a prime integer. Let κ p ≤ 5.14084 be as in (39) below. Let L be a real quadratic number field of discriminantd L ≥ 4 Assume that χ L (p) = -1. Then ν L = µ L Res s=1 (ζ L (s)) ≤ 1 8 × p -1 p + 1 × log d L +4µ Q -2+ 4p log p p 2 -1 2 +4κ p .If p = 2, this bound holds true for d L ≥ 5. Since 4κ 2 = 19.52507 . . . and 4µ Q -2 + 4p log p p 2 -1 = -0.059224 . . . , we get the third and fourth bounds in Corollary 1.5.Proof. As in Section 3.1, we apply (25), (29), (30), (31) and Lemma 3.6 with d 1 = p , a = 1/2 and d 2 = 1. We obtain 1 + 1 p ν L ≤ I(√ d L ),whereI(X) = Res s=1 (G(s, p, X)) + Res s=0 (G(s, p, X)) + R(p, X), with G(s, p, X) := µ Q + log p p + 1 Π 1 (s, 1/2) + Π 2 (s, 1) X s-1 Z Q (s), Π k (s, a) := a 1-s (s -1) k 1 -1 p s + a s s k 1 -1 -s, 1/2) + Π 2 (1 -s, 1) X -s Z Q (s) ds.Now,Res s=1 (G(s, p, X)) 1) 2 log 2 p (39)+ log(16π) -γ p + 1 log p + 6 -µ Q + 2µ Q log 2 satisfies κ p ≤ κ 7 =5.14083 . . . and κ 2 = 4.88126 . . . . (Notice that 2µ Q -1 + 2p(log p)/(p 2 -1) = A + C = c 2 (p, 1) and κ p = c 3 (p, 1), with the notation of Section 3.1.) Moreover,Res s=0 (G(s, p, X)) = --log π p + 1 log p + µ Q -2µ Q log 2 -µ 2 Q satisfies κ p ≥ κ 2 = 0.62924 . . . . Noticing that Π 1 (1 -s, a) = 2 -s s k (p s-1 -1) -2 s-1 (s -1) k 1 -1 -s, a) = 1 s 2 (1 -p s-1 ) +

  Theorem 4.4. Let p ≥ 2 be a given prime. Let χ be a primitive even Dirichlet character of conductor f ≥ 5 not divisible by p. Assume that either 2 ≤ p ≤ 61 or f ≥

		p	1)	≤	5p 12X 2 ,
	by Remark 4.1.
	p 6 log p	2 . Then

  then X > 1, as v/log v ≥ e for v > 1, and since X → X log 2 X increases with X > 1, we have

	and
	Res s=0 (H(s, p, X)) + R(p, X) ≤ -	log 2 X 4X	+	p + 21 84X 2
							= -	1 4X 2 X log 2 X -	p + 21 21
	is negative for X ≥ 4	p+21 21 log 2 p+21 21	. Indeed, if u > 1 and X ≥ 4u log 2 u =	√ log √ u	u	2
	X log 2 X ≥	4u log 2 u	log 2	4u log 2 u	= u 1 + 4	log u 1/4 2 log u 1/4 log u
							1 (s -1) k 1 -	2 p	+	1 p s
	and using (38) and Remarks 4.1, we have
	R(p, X) ≤	1 p	µ Q +	log p p + 1	Z 1 (2X/p, 0) + Z 2 (X, 0)
		+ 1 -	2 p	Z 2 (X, 1) + Z 2 (pX, 1)
	≤ µ Q +	log 3 4	pM 1 (0) 48X 2 +	M 2 (0) 12X 2 +	(p -1) 2 M 2 (1) 12p 2 X 2
	≤	p 84X 2 +	M 2 (0) + M 2 (1) 12X 2

2

≥ 1,

as v/(2 log v) ≥ e/2 ≥ 1 for v > 1.

Abstract (will appear on the journal's web site only) Various bounds on the absolute values of L-functions of number fields at s = 1 and on residues at s = 1 of Dedekind zeta functions of a number field L are known. Also, better bounds depending on the splitting behavior of given prime ideals of L of small norms are known. These bounds involve a term ν L in the series expansion at s = 1 of the Dedekind zeta function of L. We explain why one should expect to have bounds on ν L which also depend on the splitting behavior in L of given prime integers. We explicitly do that for L a real quadratic number field. We deduce very good upper bounds on the residue at s = 1 of the Dedekind zeta function of a non-normal totally real cubic number field K, bounds depending on the splitting behavior of the prime p = 2 in K.


Stéphane R Louboutin
email: stephane.louboutin@univ-amu.fr

On Ennola's conjecture on non-Galois cubic number fields with exceptional units

Keywords: Mathematics Subject Classification, Primary, 11R16, 11R27, Key words and phrases, Units, Exceptional units, Cubic number fields, Unit index, Thue's Lemma, The abc conjecture

Let ε be a non-Galois totally real cubic special unit, i.e. a unit such that ε -1 is also a unit. Then ε and ε -1 are multiplicatively independent and the unit index j ε of the groups of units generated by -1, ε and ε -1 in the group of units of the ring of algebraic integers of Q(ε) is finite. It is known that {ε, ε -1} is a system of fundamental units of the cubic order Z[ε]. V. Ennola conjectured that {ε, ε -1} is always a system of fundamental units of the maximal order of Q(ε), i.e. that j ε is always equal to 1. Fix an algebraic closure of Q. We prove that for any given prime p there are only finitely many cases for which p divides j ε . We explain how this result makes Ennola's conjecture very reasonable for its possible exceptions would be few and far between. Our proof is conditional: we conjecture that the degrees of some explicit rational fractions that clearly are Laurent polynomials are always negative and given by conjectured explicit formulas. These degrees being easy to compute by using any formal language for algebraic computation, we checked enough of them to obtain that for any given prime p ≤ 1875 there are only finitely many cases for which p divides j ε . We also prove that under the assumption of the ABC conjecture there are only finitely many exceptions to Ennola's conjecture. 0 2010

Introduction

Let ε > 1 with ε = (3 + √ 5)/2 be a real quadratic algebraic unit. It is not that difficult to prove that ε is a fundamental unit of the order Z[ ], i.e. that the group of units of this order is generated by -1 and ε. (If ε = (3 + √ 5)/2, then ε = (ε -1) 2 and ε -1 = (1 + √ 5)/2 > 1 is the fundamental unit of the order Z[ ].) However, since Q(ε) = Q(ε n ) for any n ≥ 2, we cannot expect a real quadratic algebraic unit to always be a fundamental unit of the maximal order of the real quadratic field it generates. In contrast to this situation, V. Ennola's conjectured that if ε is a non-Galois totally real cubic exceptional algebraic unit, i.e. a unit ε for which ε-1 is also a unit, then not only is {ε, ε -1} a system of fundamental unit of the totally real cubic order Z[ε] (see [START_REF] Thomas | Fundamental units for orders in certain cubic number fields[END_REF]Proposition (3.6)]), but it always is a system of fundamental units of the maximal order of the non-Galois totally real cubic field Q(ε). The aim of this paper is to prove a weak form of Ennola's conjecture which we now expound.

Fix an algebraic closure of Q. A totally real cubic algebraic unit ε is called an exceptional unit if ε-1 is also a unit. In that situation, ε and ε-1 are multiplicatively independent (see Theorem 22 below) and we let

j ε = (U ε : -1, ε, ε -1 ) < ∞
denote the finite index of the group of units -1, ε, ε -1 generated by -1, ε and ε -1 in the group U ε of units of the ring of algebraic integers of the number field Q(ε). If ε is exceptional, then any η ∈ V ε is also exceptional and (i) Q(η) = Q(ε), (ii) -1, η, η -1 = -1, ε, ε -1 and (iii) j η = j ε for η ∈ V ε := {ε, 1/ε, 1 -ε, 1/(1 -ε), (ε -1)/ε, ε/(ε -1)}.

Proposition 1. (i). Let ε l denote any of the three distinct real roots of the Qirreducible polynomial P l (X) := X 3 + (l -1)X 2 -lX -1 ∈ Z[X]. Then ε is a totally real cubic exceptional unit with Q(ε) not normal if and only if ε is one of the three real conjugates of some η ∈ V ε l for some l ≥ 3. (ii). Let θ m denote any of the three distinct real roots of the

Q-irreducible polynomial Q m (X) := X 3 +mX 2 -(m+3)X +1 ∈ Z[X].
Then ε is a totally real cubic exceptional unit with Q(ε) normal if and only if ε is one of the three real conjugates of some η ∈ V θm for some m ≥ -1 or some η ∈ V -θ -1 .

Proof. A cubic unit ε is exceptional if and only if Irr

(ε, Q, X) = X 3 -aX 2 + bX -c ∈ Z[X], with c = N Q(ε)/Q (ε) ∈ {±1} and -(1-a+b-c) = -Irr(ε, Q, 1) = N Q(ε)/Q (ε-1) ∈ {±1}
. It amounts to asking c = 1 and b-a ∈ {±1}, or c = -1 and b-a ∈ {-1, -3}, i.e. to asking that Irr(ε, Q, X) = X 3 -aX 2 + (a -1)X -1 = P -a+1 (X) = Irr(ε -a+1 , Q, X),

Irr(ε, Q, X) = X 3 -aX 2 + (a + 1)X -1 = -X 3 P -a (1/X) = Irr(1/ε -a , Q, X), Irr(ε, Q, X) = X 3 -aX 2 + (a -1)X + 1 = -P a-2 (1 -X) = Irr(1 -ε a-2 , Q, X)
in which cases ε is one of the three real conjugates of some η ∈ V ε l , or Irr(ε, Q, X) = X 3 -aX 2 + (a -3)X + 1 = X 3 Q a-3 (1/X) = Irr(1/θ a-3 , Q, X), in which case ε is one of the three real conjugates of some η ∈ V θm .

Since P -l-3 (X) = -(X -1) 3 P l (X/(X -1)), we have Irr(ε -l-3 , Q, X) = Irr(ε l /(ε l -1), Q, X), and we may assume that l ≥ -1. Since the discriminant (l 2 + 3l -1) 2 -32 of P l (X) is positive if and only if l ≤ -5 or l ≥ 2, we may moreover assume that l ≥ 2. In that case, this discriminant is a square if and only if l = 2, in which case

P 2 (X) = -Q -1 (-X). Since Q -m-3 (X) = X 3 Q m (1/X), we have Irr(θ -m-3 , Q, X) = Irr(1/θ m , Q, X)
and we may assume that m ≥ -1. Finally, the discriminant (m 2 + 3m + 9) 2 of Q m (X) is a square.

Hence, as in [START_REF] Ennola | Cubic number fields with exceptional units[END_REF], [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF] and [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF], consider the non-Galois totally real cubic number fields Q(ε l ), where

Irr(ε l , Q) = X 3 + (l -1)X 2 -lX -1 (l ≥ 3)
is of positive and non-square discriminant

D l = (l 2 + 3l -1) 2 -32.
Then ε l is an exceptional unit, ε l and ε l -1 are multiplicativily independent and the set {ε l , ε l -1} is a system of fundamental units of the totally real cubic order Z[ε l ], by [START_REF] Thomas | Fundamental units for orders in certain cubic number fields[END_REF]Proposition (3.6)]. Moreover, Z[ε l ] is equal to the ring of algebraic integers of Q(ε l ) whenever D l is square-free. By the usual conjecture on square-free values of polynomials, see for example [START_REF] Granville | ABC means we can count squarefrees[END_REF]Theorem 1], this should happen infinitely often with positive probability

ρ = p 1 - ω(p 2 ) p 2 = 6 7 p≡±1 (mod 8) 1 - ω(p) p 2 = 0.839 • • • .
Here, ω(p 2 ) and ω(p) are the number of solutions mod p 2 , respectively mod p, to the congruence F (l) := (l 2 + 3l -1) 2 -32 ≡ 0 (mod p 2 ), respectively F (l) ≡ 0 (mod p). Hence ω(p 2 ) = 0 implies p ≡ ±1 (mod 8), ω(7 2 ) = 7 = 1 = ω(7) and ω(p 2 ) = ω(p) ∈ {0, 2, 4} for p > 7, as F (l) is of discriminant -2 14 7 3 (Hensel's Lemma).

In [START_REF] Ennola | Cubic number fields with exceptional units[END_REF], V. Ennola conjectured that {ε l , ε l -1} is always a fundamental pair of units for the maximal order of Q(ε l ). He checked numerically that this conjecture holds true for 3 ≤ l ≤ 500 and supported it by proving that the unit index

j l := (U l : -1, ε l , ε l -1 )
of the groups of units generated by -1, ε l and ε l -1 in the group of units U l of the ring of algebraic integers of Q(ε l ) is always coprime to 2, 3 and 5. In [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF], we added a lot more credit to Ennola's conjecture by proving (i) that gcd(j l , 19!) = 1 for l ≥ 3 and (ii) that j l = 1 for 3 ≤ l ≤ 5 • 10 7 (not by using softwares for numerical computation with number fields like Pari GP, but by using Proposition 4 below). In [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF] we introduced new tools that enabled us to prove that for l effectively large enough, we have gcd(j l , N !) = 1 with N = 97. However, proving that j l = 1 for l effectively large enough seems intractable at the moment.

The aim of this paper is to prove a weak form of Ennola's conjecture:

Theorem 2. Assume that Conjectures 12 and 20 below hold true. Then for any given prime p ≥ 3 there are only finitely many l ≥ 3 for which p divides the unit index j l . Hence, for any given integer N ≥ 2 we have gcd(j l , N !) = 1 for l ≥ l N effectively large enough.

Conjectures 12 and 20 below assert that the explicit identities of Tables 1 and2 in the ring of the Laurent polynomials 1 and3, where for

Q[T, T -1 ] := n∈Z q n T n ; q n ∈ Q and q n =
G(T ) = n∈Z q n T n ∈ Q[T, T -1 ] we set deg G(T ) := max{n ∈ Z; q n = 0}.
This should not be that difficult to prove (see Conjecture 14). Maybe it could even be proved by using a formal language for algebraic computation. It is just that at the moment we have no brighter idea than to use the trinomial expansion formula

(x + y + z) n = i+j+k=n n! i!j!k! x i y j z k to expand each term R a,b (T ) u R -a,-b (T ) v (or S a,b (T ) u S a,b (1/T ) v
) and then collect all the terms in G a,b (T ) of a given degree to check that those of non-negative degree cancel out and that it remains only few terms of negative degree. Nobody would want to read in extenso such a proof. Moreover, whereas it would enable us to check our formulae in Table 1, it would not explain why we end up with such simple results for the G a,b (T )'s.

At least, for a given p is is rather easy to check that Conjectures 12 and 20 hold true. The hardest one to check is Conjecture 20 because the number of non-zero coefficients of G a,b,p (X, Y ) increases rapidly with p and the computation of G a,b,p (X, Y ) in the ring Q[T, T -1 ] becomes slow and requires a lot of memory (see the proof of Theorem 21 for details). We stopped our computation at p ≤ 1875 and obtain: Theorem 3. For any given odd prime p ≤ 1875 there are only finitely many l ≥ 3 for which p divides the unit index j l . 2 Ennola's conjecture follows from the ABC conjecture

Theorem 4 below makes Ennola's conjecture asserting that j l = 1 for l ≥ 3 a very reasonable conjecture, for its possible exceptions should be few and far between. We also used it in [Lou17, Proposition 5] with p 0 = 7 to prove that j l = 1 for 3 ≤ l ≤ 5•10 7 .

We will finally use it with p 0 = 5 to prove Theorem 6, which asserts that Ennola's conjecture holds true for l large enough under the assumption of the ABC conjecture which asserts that rad(ABC) :

= p|ABC p C 1-
for any triples (A, B, C) of coprime positive integers satisfying A + B = C.

To begin with, let ε be a totally real cubic exceptional unit. Let D ε be the discriminant of Irrr(ε, Q, X) and I ε be the index of Z[ε] in the ring of algebraic integers of Q(ε). Hence,

D ε = I 2 ε d Q(ε)
. Using [Cus] we obtained in [Lou17, proof of Theorem 2] that

j ε ≤ Reg(ε, ε -1) 1 16 log 2 (d Q(ε) /4) = Reg(ε, ε -1) 1 16 log 2 (D ε /4I 2 ε )
.

(1)

Now, (i) the ABC conjecture yields lower bounds on d Q(ε) and upper bounds on I ε for the parametrized families of exceptional units given in Proposition 1 and (ii) asymptotics for the regulators Reg(ε, ε -1) are easy to obtain. Hence the ABC conjecture yields conditional upper bounds on j ε . If we know beforehand that the primes p less than or equal to this conditional upper bound do not divide j ε , at least for the parameter large enough, then we get that j ε = 1, except possibly for finitely many cases.

The totally real non-normal cubic case

Theorem 4. Assume that p j l for 2 ≤ p < p 0 and l ≥ 3 (e.g. take p 0 = 5 by Proposition 8 or p 0 = 23 by [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF]), respectively for 2 ≤ p < p 0 and l large enough (e.g. take p 0 = 101 by [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF] or p 0 = 1877 by Theorem 3). Write the discriminant of

Irr(Q, ε l , X) as D l = (l 2 + 3l -1) 2 -32 = a l b 2 l , with a l > 1 square-free and 1 ≤ b l ≤ l 2 . If b l ≤ l 2-2/ √ p 0 /
2 then j l = 1 and {ε l , ε l -1} is a system of fundamental units of the ring of algebraic integers of Q(ε l ) for l ≥ 3, respectively for l large enough.

Proof. Since D l = I 2 ε l d Q(ε l ) = a l b 2 l and a l is square-free, it follows that I ε l divides b l . Hence, I ε l ≤ b l . Now, j l = 1 if I ε l ≤ l 2-2/ √ p 0 /2, by [Lou17, Theorem 2] (notice that [Enn2] gives j l = 1 only if I ε l ≤ l/3). Lemma 5. Assume that the ABC conjecture holds true. Let g(X) ∈ Z[X] be of non-zero discriminant. Fix > 0. Then b l |l| 1+ , where g(l) = a l b 2 l with a l square-free. Proof. First, a l |l| d /b 2 l , where d = deg g(X)
. By [Gr, Corollary 1], under the assumption of ABC conjecture, we have

a l b l ≥ p|g(l) p |l| d-1-. Hence, |l| d /b l |l| d-1-
and the desired first result follows.

Theorem 6. Assume that the ABC conjecture holds true. Fix an algebraic closure of Q. Then, j l = 1 for l large enough. Hence, by Proposition 1, if ε is a non-normal totally real cubic exceptional unit then {ε, ε -1} is a system of fundamental units of the cubic number field Q(ε), except possibly for finitely many cases.

Proof. Now, take g(X) = (X 2 + 3X -1) 2 -32 and < 1 -2/ √ 5. We have 1 + < 2-2/ √ 5. Therefore, for l large enough we have b l ≤ l 2-2/

The totally real normal cubic case

According to Proposition 1, consider by [Was, Proposition 1 andits proof]. Fix > 0 By [START_REF] Granville | ABC means we can count squarefrees[END_REF]Corollary 1], under the assumption of ABC conjecture, we have

Irr(θ m , Q, X) = X 3 + mX 2 -(m + 3)X + 1 (m ≥ -1), of discriminant D m = f 2 m with f m = m 2 + 3m + 9. Write f m = b m c 3 m , with b m cube- free. Then d Q(θm) ≥ rad(b m ) 2 , with rad(b m ) := p|bm p,
m 1-/6 rad(f m ) = rad(b m c m ) ≤ rad(b m )rad(c m ) rad(b m ) m 2 /rad(b m ) 1/3 and rad(b m ) m (1-)/2 , by using rad(b m )rad(c m ) 3 ≤ b m c 3 m = f m m 2 . Hence, d Q(θm) ≥ rad(b m ) 2 m 1-. Now, as in [Lou17, proof of Theorem 2] it is readily seen that Reg(θ m , θ m -1) ≤ log m + 2 m 2 for m ≥ 1.
Hence, by ( 1) we obtain j θm < p 0 = 17 for m large enough. Now, by [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF]Lemmas 12,13], we know that if p < 17 divides j θm for some m ≥ -1, then p ∈ {3, 7, 13}. By [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF]Theorems 17,18] we know that neither 3 nor 7 divides j θm for m > 5. Finally, by [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF]Remark 20] we know that 13 does not divide j θm for m large enough. Hence we have:

Theorem 7. Assume that the ABC conjecture holds true. Fix an algebraic closure of Q. If ε is a normal totally real cubic exceptional unit then {ε, ε -1} is a system of fundamental units of the cubic number field Q(ε), except possibly for finitely many cases.

Notice that contrary to the non-normal cubic case where no example of j ε l > 1 are known, here 7 values of m for which j θm > 1 are known: m 3 5 12 54 66 1259 2389 j θm 3 7 13 19 13 97 31

3 How to prove that p does not divide j l

To begin with, recall (see [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF]Proposition 10] or [Lou20, Proposition 3]):

Proposition 8. Assume that l ≥ 3. The prime numbers 2 and 3 never divide the unit index j l and an odd prime number p ≥ 5 divides j l if and only if one of the p -4 units

ε k l (ε l -1) is a pth power in Q(ε l ), where 2 ≤ k ≤ p -3.
Corollary 9. Let p ≥ 5 a prime. Let

E p := {(a n , b n ); 1 ≤ n ≤ p -4} ⊆ Z =0 × Z ≥1
be a set of pairs of coprime rational integers with p b such that

Z/pZ \ {0, 1, p -2, p -1} = {a n /b n ; 1 ≤ n ≤ p -4}.
Then, p divides j l if and only if one of the p -4 units

ε a,b = (-1) a+b ε a l (ε l -1) b is a pth power in Q(ε l ), where (a, b) ∈ E p .
Proof. The unit ε k l (ε l -1) is a pth power if and only so is any ε kb l (ε l -1) b for b not divisible by p, hence if and only if so is ε a l (ε l -1) b for b not divisible by p and any a verifying a ≡ kb (mod p). Since -1 = (-1) p is a pth power, the desired result follows.

According to the following result known as Thue's Lemma, we can find a set E p such that (a, b) ∈ E p implies max(|a|, |b|) ≤ √ p:

Lemma 10. (See also [LM]). For any k not divisible by a prime p ≥ 3, there exist a and b = 0 such that a ≡ kb (mod p) and 0 < max(|a|, |b|) ≤ √ p.

Proof. The set F p := {(x, y) ∈ Z 2 ; 0 ≤ x, y ≤ g := √ p } contains (g + 1) 2 > p elements. Thus, 2 of the (g + 1) 2 elements x -yk's are equal mod p when (x, y) range over F p , say x -yk ≡ x -y k (mod p) with (x, y) = (x , y ). Then x -x ≡ (y -y )k (mod p), with |x -x | ≤ g < p and |y -y | ≤ g < p. Hence, b = y -y = 0 and the desired result follows.

In the present paper we will need to prove that we can find a set E p such that (a, b) ∈ E p implies a 2 + ab + b 2 ≤ p (Proposition 16). The number of points (a, b) ∈ Z 2 satisfying max(|a|, |b|) ≤ √ p is asymptotic to 4p. The number of those satisfying a 2 + ab + b 2 ≤ p is asymptotic to 2π √ 3 p, where 2π √ 3 < 4. Hence, Proposition 16 below is stronger than Thue's Lemma.

We now explain how we prove that a given unit

ε a,b = (-1) a+b ε a l (ε l -1) b is not a pth power in Q(ε l ) for p ≥ B a,b and l ≥ l a,b,p effectively large enough. Notice that the rational fractions R a,b (T ), G a,b (T ), R a,b,m (T ) and G a,b,m (T ) that crop up in this paper are in the ring Q[T, T -1 ] of Laurent polynomials. Proposition 11. Let a, b ∈ Z not both equal to 0 be given. Set s = max(a + b, -a, -b) ≥ 0, t = max(-a -b, a, b) ≥ 0 and R a,b (T ) := T -a + (-1) a+b T -b + T a+b ∈ Q(T ). Suppose there exists 0 = F a,b (X, Y ) = u,v f u,v X u Y v ∈ Z[X, Y ] such that G a,b (T ) := F a,b (R a,b (T ), R -a,-b (T )) = F a,b (R a,b (T ), R a,b (1/T )) ∈ Z[T, T -1 ] is of negative degree. Set M a,b = max{us + vt; f u,v = 0} and N a,b = -deg G a,b (T ) ≥ 1.
Then for any given odd prime

p ≥ B a,b := M a,b + N a,b + 1 the unit ε a l (ε l -1) b is not a pth power in Q(ε l ) for l ≥ l a,b,p := w p a,b
effectively large enough, for some effective w a,b > 1 not depending on p.

Proof. If α, α and α are the three conjugates of a totally real cubic algebraic number α and m ≥ 3 is an odd integer, we set

S m (α) = α 1/m + α 1/m + α 1/m ∈ R.
(2)

Hence, S m (α) ∈ Z if α in a mth power in Q(α). Set ε = (-1) a+b ε a l (ε l -1) b . Let p ≥ 3 be given. Set w := l 1/p > 1. Letting θ stand for an effective real number such that 0 ≤ θ ≤ 1, not necessarily the same at different places, the three conjugates of ε l satisfy

ε l = -l(1 -θl -1 ) ε l -1 = -l (1 + θl -1 ) ε l = -l -1 (1 -θl -1 ) ε l -1 = -(1 + θl -1 ) ε l = 1 + θl -1 ε l -1 = l -1 (1 -2θl -1 ) (evaluate Irr(ε l , Q) = X 3 + (l -1)X 2 -lX -1 at -l and -l + 1, at -l -1 and -l -1 + l -2
and at 1 and 1 + l -1 to check sign changes and do the same for Irr(

ε l -1, Q) = X 3 + (l + 2)X 2 + (l + 1)X -1).
Hence, with the notation in (2), as l goes to infinity we have

S p (ε) = R a,b (w) + O(w s-p ) and S p (1/ε) = R -a,-b (w) + O(w t-p ).
For P (X, Y ) = X u Y v with u, v ≥ 0 we thus have

P (S p (ε), S p (1/ε)) = P (R a,b (w), R -a,-b (w)) + O(w us+vt-p ).
Hence,

F a,b (S p (ε), S p (1/ε)) = G a,b (w) + O(w M a,b -p ) = q N a,b w -N a,b + O(w -N a,b -1 ) + O(w M a,b -p ),
where q N a,b = 0. Now assume that p ≥ M a,b + N a,b + 1. We obtain

0 < |F a,b (S p (ε), S p (1/ε))| < 1 if w ≥ w a,b > 1 is effectively large enough, i.e. if l = w p ≥ w p a,b = l p is effectively large enough. Since S p (ε) ∈ Z and S p (1/ε) ∈ Z and hence F a,b (S p (ε), S p (1/ε)) ∈ Z whenever ε = η p is a pth power in Q(ε l ), we get the desired result. 4 Conjectural suitable polynomials F a,b (X, Y )
In [START_REF] Ennola | Cubic number fields with exceptional units[END_REF] and [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF] the authors laboriously constructed some such suitable polynomials 1] with a guess for an explicit formula for such F a,b (X, Y )'s. Finally, after the acceptance for publication of [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF], we realized that when we computed the B a,b 's defined in Proposition 11 for these F a,b (X, Y )'s in Table 1, we almost always got B a,b = a 2 + ab + b 2 + 1. We have also just now realized that with the notation of [Lou20, Lemma 8] we have V d (X, Y ) = -P d (-Y, X)!! Hence, we have now Conjecture 12 which is much clearer and more complete than [Lou20, Conjecture 7] and from which we will deduce Theorem 18. Conjecture 12 can be checked easily on a given pair (a, b) by using any formal language for computation (we checked it on a MacBook Air laptop computer using Maple in 3120 seconds for max(|a|, b) ≤ 50):

Conjecture 12. Let (a, b) ∈ Z =0 × Z ≥1 , with a + b = 0. For d ≥ 1, set P d (X, Y ) = d k,l≥0 0≤2k+3l≤d (-1) k-1 k + l k d -k -2l k + l X k Y d-2k-3l d -k -2l ∈ Z[X, Y ].
(with the convention 0! = 1). Let F a,b (X, Y ) ∈ Z[X, Y ] be as in Table 1. Keep the notation of Proposition 11. Then, M a,b and

G a,b (T ) := F a,b (R a,b (T ), R -a,-b (T )) = F a,b (R a,b (T ), R a,b (1/T )) ∈ Z[T, T -1 ]
are as in Table 1. In particular, G a,b (T ) is of negative degree and

B a,b = a 2 + ab + b 2 + 1, except if a = -2b with b ≥ 1 odd, in which case B a,b = B -2b,b = 6b 2 + 1. The fact that P d (X, Y ) is in Z[X, Y ] follows from the identity k + 2l d -k -2l k + l k d -k -2l k + l = k -1 + l l + 2 k + l -1 k × d -k -2l -1 k + l -1 .
Notice that with the notation of Proposition 11, we have s = deg R a,b (T ) and t = deg R -a,-b (T ), except in the case that a = -2b with b ≥ 1 odd, in which case s

= 2b = deg R a,b (T ) but t = b = -2b = deg R -a,-b (T ). Table 1: Conjecture for F a,b (X, Y ), for a = 0, b ≥ 1 and c := a + b = 0 1. a ≥ 1 odd and b ≥ 1 odd F a,b (X, Y ) = -P a (Y, X) -P b (Y, X) + P c (X, Y ), M a,b = c max(a, b), G a,b (T ) = T -a 2 + T -b 2 -T -c 2 + 2T -ab for which N a,b = min(a, b) 2 . 2. a ≤ -1 odd, b ≥ 1 odd and c > 0 F a,b (X, Y ) = P |a| (X, Y ) -P b (Y, X) + P c (X, Y ), M a,b = b max(|a|, c), G a,b (T ) = -T -a 2 + T -b 2 -T -c 2 -2T -|a|c for which N a,b = min(|a|, c) 2 . 3. a ≤ -1 odd, b ≥ 1 odd and c < 0 F a,b (X, Y ) = P |a| (X, Y ) -P b (Y, X) -P |c| (Y, X), M a,b = |a| max(b, |c|), G a,b (T ) = -T -a 2 + T -b 2 + T -c 2 + 2T -b|c| for which N a,b = min(b, |c|) 2 . Table 1 (continued) 4. a ≥ 1 odd, b ≥ 1 even F a,b (X, Y ) = -P a (-Y, -X) -P b (-Y, -X) + P c (-X, -Y ), M a,b = c max(a, b), G a,b (T ) = -T -a 2 + T -b 2 + T -c 2 + 2T -ab for which N a,b = min(a, b) 2 . 5. a ≤ -1 odd, b ≥ 1 even and c > 0 F a,b (X, Y ) = P |a| (-X, -Y ) -P b (-Y, -X) + P c (-X, -Y ), M a,b = b max(|a|, c), G a,b (T ) = T -a 2 + T -b 2 + T -c 2 + 2T -|a|c for which N a,b = min(|a|, c) 2 . 6. a ≤ -1 odd, b ≥ 1 even and c < 0 F a,b (X, Y ) = P |a| (-X, -Y ) -P b (-Y, -X) -P |c| (-Y, -X), M a,b = |a| max(b, |c|), G a,b (T ) = T -a 2 + T -b 2 -T -c 2 + 2T -b|c| for which N a,b = min(b, |c|) 2 . Table 1 (continued) 7. a ≥ 2 even, b ≥ 1 odd F a,b (X, Y ) = -P a (-Y, -X) -P b (-Y, -X) -P c (-X, -Y ), M a,b = c max(a, b), G a,b (T ) = T -a 2 + T -b 2 -T -c 2 for which N a,b = min(a, b) 2 . 8. a ≤ -2 even, b ≥ 1 odd and c > 0 F a,b (X, Y ) = P |a| (-X, -Y ) + P b (-Y, -X) + P c (-X, -Y ), M a,b = b max(|a|, c), G a,b (T ) = -T -a 2 -T -b 2 + T -c 2 for which N a,b = min(|a|, c) 2 . 9. a ≤ -2 even, b ≥ 1 odd and c < 0 F a,b (X, Y ) = P |a| (-X, -Y ) + P b (-Y, -X) + P |c| (-Y, -X), M a,b = |a| max(b, |c|), G a,b (T ) = -T -a 2 -T -b 2 + T -c 2 for which N a,b = min(b, |c|) 2 if a = -2b, 4b 2 if a = -2b. Table 1 (continued) 10. a ≥ 2 even, b ≥ 1 even F a,b (X, Y ) = -P a (Y, X) -P b (Y, X) + P c (X, Y ), M a,b = c max(a, b), G a,b (T ) = T -a 2 + T -b 2 -T -c 2 + 2T -ab for which N a,b = min(a, b) 2 . 11. a ≤ -2 even, b ≥ 1 even and c > 0 F a,b (X, Y ) = P |a| (X, Y ) -P b (Y, X) + P c (X, Y ), M a,b = b max(|a|, c), G a,b (T ) = -T -a 2 + T -b 2 -T -c 2 -2T -|a|c for which N a,b = min(|a|, c) 2 . 12. a ≤ -2 even, b ≥ 1 even and c < 0 F a,b (X, Y ) = P |a| (X, Y ) -P b (Y, X) -P |c| (Y, X), M a,b = |a| max(b, |c|), G a,b (T ) = -T -a 2 + T -b 2 + T -c 2 + 2T -b|c| for which N a,b = min(b, |c|) 2 .
Remark 13. Cases 11 and 12 in Table 1 correct the awkward last two cases in [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF]Table 1]. This awkwardness has in fact no impact on the results obtained in [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF] where one deals only with coprime integers a and b. In fact, here again we deal with pairs of coprime integers, by Corollary 9. Hence, we do not even need cases 10-11-12 of the present Table 1.

Conjecture 12 would be a consequence of the behavior of the P d (X, Y )'s:

P |d| (-R a,b (T ), -R a,b (1/T )) = -S a,b (1/T |d| ) if d = a, R a,b (1/T |d| ) if d ∈ {b, c}.
Proposition 15. Assume that Conjecture 14 holds true. Let the F a,b (X, Y )'s be as in Table 1.

Then G a,b (T ) := F a,b (R a,b (T ), R a,b (1/T )) is as Table 1.
Proof. For example, in case 1 of Table 1, we have R a,b (T ) = S a,b (T ) and taking F a,b (X, Y ) as given in Table 1 and using Conjecture 14 we do obtain

-c ) = (T -a 2 + T -ab + T ac ) + (T -ab + T -b 2 + T bc ) -(T ac + T bc + T -c 2 ) = T -a 2 + T -b 2 -T -c 2 + 2T -ab .
In the same way, as a second example, in case 9 of Table 1, taking F a,b (X, Y ) as given in Table 1 and using Conjecture 14 we do obtain

G a,b (T ) = F a,b (R a,b (T ), R a,b (1/T )) = P |a| (-R a,b (T ), -R a,b (1/T )) + P b (-R a,b (1/T ), -R a,b (T )) + P |c| (-R a,b (1/T ), -R a,b (T )) = -S a,b (1/T |a| ) + R a,b (T b ) + R a,b (T |c| ) = -S a,b (T a ) + R a,b (T b ) + R a,b (T -c ) = -(T -a 2 + T -ab + T ac ) + (T -ab -T -b 2 + T bc ) + (T ac -T bc + T -c 2 ) = -T -a 2 -T -b 2 + T -c 2 .
Cases -2-8 being treated in the same way we obtain the following Table :   Cases

b ) + S a,b (T -c ) 4 S a,b (-T a ) + S a,b (T b ) -S a,b (-T -c ) 5 -S a,b (-T a ) + S a,b (T b ) -S a,b (-T -c ) 6 -S a,b (-T a ) + S a,b (T b ) + S a,b (-T -c ) 7 S a,b (T a ) -R a,b (T b ) -R a,b (T -c ) 8 -S a,b (T a ) + R a,b (T b ) + R a,b (T -c ) 9 -S a,b (T a ) + R a,b (T b ) + R a,b (T -c )
The desired result follows.

5 Proof of Theorem 2

∈ {2, • • • , p -3} computes some (a k , b k ) ∈ Z =0 × Z ≥1 such that Q -3 (a k , b k ) = min{Q -3 (a, b); (a, b) ∈ Z =0 × Z ≥1 , p b, a ≡ kb (mod b)}.
We then computed Bound(p) = max{Q -3 (a k , b k ); 2 ≤ k ≤ p -3} for various p's. We hoped that we would always have Bound(p) ≤ p -1, in which case, by Corollary 9, Proposition 11 and Conjecture 12, we would have conditionally proved Theorem 2. We readily guessed Proposition 16 (which does not always hold true with the better bound a 2 + ab + b 2 ≤ p -1, by Corollary 17):

Proposition 16. For any k not divisible by a prime p ≥ 5, there exist a and b ≥ 1 such that a ≡ kb (mod p) and 0 < a 2 + ab + b 2 ≤ p, which implies p b.

Proof. Consider the quadratic form

Q -3 (X, Y ) = X 2 + XY + Y 2 and the lattice L k = {(a, b) ∈ Z 2 ; a ≡ kb (mod p)} = Ze 1 + Ze 2
, where e 1 = (k, 1) and e 2 = (p, 0). We want to find 0 = e ∈ L k such that Q -3 (e) ≤ p.

Consider the quadratic form Q(e) = Q(x, y) = Q -3 (kx + py, x) = Ax 2 + Bxy + Cy 2 = (k 2 + k + 1)x 2 + (2k + 1)pxy + p 2 y 2 of discriminant ∆ = B 2 -4AC = -3p 2 , where e = xe 1 + ye 2 = (kx + py, x) ∈ L k . Let e 1 ∈ L k be such that Q(e 1 ) = min{Q(e); 0 = e ∈ L k } and e 2 ∈ L k be such that Q(e 2 ) = min{Q(e); e ∈ L k \ Ze 1 }. Then {e 1 , e 2 } is a Z-basis of L k and if e = xe 1 + ye 2 = x e 1 + y e 2 ∈ L k , then Q(e) = Q(x, y) = Q (x , y ) = A x 2 + B x y + C y 2 is a quadratic form in x , y of the same discriminant ∆ = B 2 -4A C = -3p 2 = ∆ as the quadratic form Q. The key point is that Q is reduced, i.e. that 0 ≤ |B | ≤ A ≤ C . Indeed, A = Q (1, 0) = Q(e 1 ) ≤ Q(e 2 ) = Q (0, 1) = C yields A ≤ C and C = Q(e 2 ) ≤ Q(e 1 ± e 2 ) = A ± B + C yields |B | ≤ A . Hence, 3p 2 = 4B C -A 2 ≥ 4A 2 -A 2 = 3A 2 = 3Q (1, 0) 2 = 3Q(e 1
) 2 and 0 < Q(e 1 ) ≤ p. The desired result follows.

Corollary 17. Let p ≥ 5 be a prime integer.

1. Assume that p ≡ 5 (mod 6). For any k not divisible by p there exist a and b not divisible by p such that a ≡ kb (mod p) and 0 < a 2 + ab + b 2 ≤ p -1. Hence, there exists a set

E p ⊆ Z =0 × Z ≥1 as defined in Corollary 9 such that (a, b) ∈ E p implies 0 < a 2 + ab + b 2 ≤ p -1.
2. Assume that p ≡ 1 (mod 6). Write p = A 2 + AB + B 2 , with A > B ≥ 1 (in a unique way). Let k 1 and k 2 be such that A ≡ k 1 B (mod p) and B ≡ k 2 A (mod p). (i.e. k 1 mod p and k 2 mod p are the two non trivial cubic roots of unity mod p). Then k 1 mod p and k 2 mod p are the only k mod p not divisible by p for which there do not exist a and b not divisible by p such that kb ≡ a (mod p) and 0 < a 2 + ab + b 2 ≤ p -1. Hence, there exists a set E p ⊆ Z =0 × Z ≥1 with p -4 elements as defined in Corollary 9 such that (a, b) ∈ E p implies 0 < a 2 + ab + b 2 ≤ p -1 for p -6 of its elements and a 2 + ab + b 2 = p for its remaining 2 elements (a, b) ∈ {(A, B), (B, A)}.

Proof. Suppose there exists some k not divisible by p for which there do not exist a and b not divisible by p such that kb ≡ a (mod p) and a 2 + ab + b 2 ≤ p -1. By Proposition 16, there exist a and b not divisible by p such that kb ≡ a (mod p) and a 2 +ab+b 2 = p. Then b 2 (k 2 + k + 1) ≡ a 2 + ab + b 2 ≡ 0 (mod p), hence k 2 + k + 1 ≡ 0 (mod p) and the image of k in (Z/pZ) * is of order 3 in this multiplicative group of order p -1. Hence, 3 divides p -1 and p ≡ 1 (mod 6), which proves the first assertion. Conversely, suppose that p ≡ 1 (mod 6) and k is a solution of the congruence k 2 + k + 1 ≡ 0 (mod p). Then p does not divide k and if kb ≡ a (mod b) with a and b not divisible by p, then a 2 + ab + b 2 ≡ b 2 (k 2 + k + 1) ≡ 0 (mod p) and hence a 2 + ab + b 2 ≥ p (notice that for a and b not both equal to 0 we have a 2 + ab + b 2 > 0).

Proof of Theorem 2 for p ≡ 5 (mod 6)

The first point of Corollary 17 and Proposition 11 yield:

Theorem 18. Let p ≡ 5 (mod 6) be a given prime. Assume that Conjecture 12 holds true for the p -4 pairs (a, b) in a set E p as in the first point of Corollary 17. Then p does not divide the unit index j l for l effectively large enough. In particular, if p ≡ 5 (mod 6) is a prime less than 1875, then p does not divide the unit index j l for l effectively large enough.

Proof. We checked that Conjecture 12 holds true in the range 1 ≤ max(|a|, b) ≤ 50 on a MacBook Air laptop computer, using Maple. This computation took 52 minutes. Since a 2 + ab + b 2 ≤ p implies max(|a|, |b|) ≤ 4p/3, Theorem 18 holds true for all the primes p ≡ 5 (mod 6) less than 1875.

5.2 Proof of Theorem 2 for p ≡ 1 (mod 6)

Let p ≡ 1 (mod 6) be a prime. By the second point of Corollary 17 and Proposition 11, it remains to prove that a,b is not pth powers in Q(ε l ), where a and b are such that p = a 2 + ab + b 2 . For dealing with such occurrences we will use the following modification of Proposition 11 (notice that for given a and b, Proposition 11 deals with all the odd integers large enough, whereas Proposition 19 deals with only one odd integer at a time):

Proposition 19. Let a, b ∈ Z not both equal to 0 be given. Let m ≥ 3 be odd. Let the notation be as in Proposition 11. Set

R a,b,m (T ) = R a,b (T ) + b -a m T -a-m + (-1) a+b a -2b m T -b-m + b m T a+b-m .
Assume that

G a,b,m (T ) = F a,b (R a,b,m (T ), R -a,-b,m (T )) ∈ Q[T, T -1 ] is of negative degree. Set N a,b,m = -deg G a,b,m (T ) ≥ 1 and B a,b,m := (M a,b + N a,b,m + 1)/2. If B a,b,m ≤ m, then the unit ε a,b = (-1) a+b ε a l (ε l -1) b is not a mth power in Q(ε l ) for l ≥ l m effectively large enough.
Proof. The proof is similar to that of Proposition 11, except that we now use

ε l = -l(1 -θl -2 ) ε l -1 = -l (1 + l -1 -θl -2 )) ε l = -l -1 (1 -l -1 + 3θl -2 ) ε l -1 = -(1 + l -1 -θl -2 ) ε l = 1 + l -1 -2θl -2 ε l -1 = l -1 (1 -2l -1 + 4θl -2 ).
We obtain

S p (ε) = R a,b,m (w) + O(w s-2m ) and S m (1/ε) = R -a,-b,m (w) + O(w t-2m )
(see [START_REF] Louboutin | Non-Galois cubic number fields with exceptional units[END_REF]Lemma 8]). For P (X, Y ) = X u Y v with u, v ≥ 0 we thus have

P (S m (ε), S m (1/ε)) = P (R a,b,m (w), R -a,-b,m (w)) + O(w us+vt-2m ).
Hence,

F a,b (S m (ε), S m (1/ε)) = G a,b,m (w) + O(w M a,b -2m ) = q N a,b,m w -N a,b,m + O(w -N a,b,m -1 ) + O(w M a,b -2m ),
where q N a,b,m = 0. Now, assume that m ≥ (M a,b + N a,b,m + 1)/2. We obtain

0 < |F a,b (S m (ε), S m (1/ε))| < 1
if w is effectively large enough, i.e. if l = w m is effectively large enough. Since S m (ε) ∈ Z and S m (1/ε) ∈ Z and hence F a,b (S m (ε), S m (1/ε)) ∈ Z whenever ε is a mth power in Q(ε l ), we get the desired result. Notice that there is some cancelation between the 3 terms of R a,b (T ) if and only if we are in one of the 3 cases excluded in Conjecture 20.

Since an explicit formula for

G a,b,m a,b (T ) = n≤-N a,b,m a,b
q n T n would be very complicated, contrary to Table 1, in Table 2 we only give a formula for the non zero coefficient q N a,b,m a,b of the leading term.

2 , q N a,b,m a,b = - 2 N a,b,m a,b × bc if b > -2a a 2 if b < -2a. 3. a ≤ -1 odd, b ≥ 1 odd and c < 0: M a,b = |a| max(b, |c|), N a,b,m a,b = min(b, |c|) 2 , q N a,b,m a,b = 2 N a,b,m a,b × bc if a > -2b a 2 if a < -2b. Table 2 (continued) 4. a ≥ 1 odd, b ≥ 1 even: M a,b = c max(a, b), N a,b,m a,b = min(a, b) 2 , q N a,b,m a,b = 2 N a,b,m a,b × -bc if a < b a 2 if a > b. 5. a ≤ -1 odd, b ≥ 1 even and c > 0: M a,b = b max(|a|, c), N a,b,m a,b = min(|a|, c) 2 , q N a,b,m a,b = 2 N a,b,m a,b × a 2 if b > -2a bc if b < -2a. 6. a ≤ -1 odd, b ≥ 1 even and c < 0: M a,b = |a| max(b, |c|), N a,b,m a,b = min(b, |c|) 2 , q N a,b,m a,b = 2 N a,b,m a,b × -a 2 if a > -2b bc if a < -2b. Table 2 (continued) 7. a ≥ 2 even, b ≥ 1 odd: M a,b = c max(a, b), N a,b,m a,b = min(a, b) 2 , q N a,b,m a,b = 2bc N a,b,m a,b . 8. a ≤ -2 even, b ≥ 1 odd and c > 0: M a,b = b max(|a|, c), N a,b,m a,b = min(|a|, c) 2 , q N a,b,m a,b = 2bc N a,b,m a,b × -1 if b > -2a +1 if b < -2a. 9. a ≤ -2 even, b ≥ 1 odd and c < 0: M a,b = |a| max(b, |c|), N a,b,m a,b = min(b, |c|) 2 , q N a,b,m a,b = 2bc N a,b,m a,b × +1 if a > -2b -1 if a < -2b.
Notice that for 1 ≤ k ≤ 9, the value of N a,b,m a,b for the kth case in Table 2 is equal to the value of N a,b for the similar kth case in Table 1.

In fact, since we will only deal with pairs of positive coprime integers (a, b) such that m a,b is a prime greater than 3 (see Theorem 21), we need only 3 cases out of the 12 ones in Table 1, those given in Table 3.

Remarks on totally real exceptional units

Let ε be a totally real exceptional unit. If ε is quadratic, then the rank of the group of units of the number field Q(ε) is equal to 1 and ε and ε-1 are therefore multiplicatively dependent. The following Theorem 22 shows that this is the only such situation: Proof. Let ε be a totally real exceptional unit. Assume that ε and ε -1 are multiplicatively dependent, i.e. that that

ε a (ε -1) b = ±1,
with a, b ∈ Z not both equal to 0. Clearly, a = 0 and b = 0. To begin with, it is easy to see that the only exceptional totally real quadratic units are (i) ε = (1 ± √ 5)/2, in which case ε(ε -1) = 1; (ii) ε = (-1 ± √ 5)/2, in which case ε 2 (ε -1) -1 = -1; and (iii) ε = (3 ± √ 5)/2, in which case ε(ε -1) -2 = 1. Now, assume that K := Q(ε) is a totally real number field of degree d := (K : Q) ≥ 3. Since ±1 are the only complex roots of unity in the totally real number field K, we may assume that gcd(a, b) = 1. Let u, v ∈ Z be such that au + bv = 1. Then , where n ∈ Z n≥1 . Moreover, the sign is necessarily the plus sign. Indeed, otherwise we would have ξ = -η 2 in K for one of the two totally real exceptional units ξ = ε or ξ = 1 -ε, which would give the contradiction 1 = |N K/Q (ξ-1)| = N K/Q (1+η 2 ) > 1, as η 2 is totally positive. Therefore, one of the two totally real exceptional units ξ = ε or ξ = 1-ε satisfies ξ = (ξ-1) ±m for some m = 2 n ≥ 2. Consequently, ξ is an algebraic integer of degree d = (K : Q) ≥ 3 and the d ≥ 3 conjugates of ξ are real roots of either P m (x) = (x -1) m -x with m ≥ 4 even or Q m (x) = x(x -1) m -1 with m ≥ 2 even. This can never happen. Indeed, let m ≥ 4 be even. Then x → P m (x) = m(x -1) m-1 -1 is strictly increasing, and P m (x) has only one real root x m = 1 + m -1/(m-1) . Since

1 -ε = -(±1) u (ε -1) u ε v
P m (x m ) = 1 m • m 1 m-1 -1 - 1 m 1 m-1 < -1 < 0,
it follows that P m (x) has exactly 2 real roots. In the same way, let m ≥ 2 be even. Then Q m (x) = (x -1) m-1 ((m + 1)x -1) and we have the following table of sense of variation:

x -∞ 1 m+1 1 +∞ Q n (x) + 0 -0 + Q n (x) 1 m+1 m m+1 m -1 < 0 -1
It follows that Q m (x) has exactly 1 real root, which is greater than 1.

  0 for |n| large enough . always hold true. Notice that we do not need the full of these conjectures to be proved. The only part we need is the one on the expected degrees N a,b and N a,b,m a,b of the G a,b (T )'s and G a,b,m a,b 's as conjectured in Tables

  F a,b (X, Y ) ∈ Z[X, Y ] for small values of a, b. After that, we wrote an (unpublished) algorithm which for given a and b yields such a F a,b (X, Y ) ∈ Z[X, Y ]. After the computation of a lot of them for various choices of a and b and a lengthly research of some pattern in those F a,b (X, Y )'s, we came up in [Lou20, Table

Conjecture 14 .

 14 Let a, b ∈ Z be such that a = 0, b = 0 and c := a + b = 0. Set S a,b (T ) := T -a + T -b + T a+b . Let P d (X, Y ) ∈ Z[X, Y ] be as in Conjecture 12. Then P |d| (S a,b (T ), S a,b (1/T )) = -S a,b (1/T |d| ) for d ∈ {a, b, c}. Moreover, if a is even and b is odd, then with R a,b (T ) := T -a -T -b + T a+b we have

G

  a,b (T ) = F a,b (R a,b (T ), R a,b (1/T )) = -P a (R a,b (1/T ), R a,b (T )) -P b (R a,b (1/T ), R a,b (T )) + P c (R a,b (T ), R a,b (1/T )) = -P a (S a,b (1/T ), S a,b (T )) -P b (S a,b (1/T ), S a,b (T )) + P c (S a,b (T ), S a,b (1/T )) = S a,b (T a ) + S a,b (T b ) -S a,b (T

  S a,b (T a ) + S a,b (T b ) -S a,b (T -c ) 2 and 11 -S a,b (T a ) + S a,b (T b ) -S a,b (T -c ) 3 and 12 -S a,b (T a ) + S a,b (T

Conjecture 20 .

 20 Let (a, b) ∈ Z =0 × Z ≥1 be such that m a,b = a 2 + ab + b 2 is odd and m a,b ≥ 5. Assume that the pair (a, b) is not of the form (-2b, b) with b ≥ 1 odd, (b, b) with b ≥ 1 odd, or (-b/2, b) with b ≥ 2 even. Then Proposition 19 is satisfied for m = m a,b with M a,b and N a,b,m a,b as in Table 2, hence with B a,b,m a,b := (M a,b + N a,b,m a,b + 1)/2 = (a 2 + ab + b 2 + 1)/2 = (m a,b + 1)/2 ≤ m a,b .

  G a,b,m a,b (T ), for a = 0, b ≥ 1, a or b odd and c := a + b = 0 1. a ≥ 1 odd and b ≥ 1 odd: M a,b = c max(a, b), N a,b,m a,b = min(a, b) 2 , q N a,b,m a,b = 2 min(a,b) N a,b,m a,b × a if a < b c if a > b. 2. a ≤ -1 odd, b ≥ 1 odd and c > 0: M a,b = b max(|a|, c), N a,b,m a,b = min(|a|, c)

Table 3 :

 3 Conjecture for G a,b,m a,b (T ), with a ≥ 1 and b ≥ 1 and c := a + b 1. a ≥ 1 odd and b ≥ 1 odd: M a,b = c max(a, b) and N a,b,m a,b = min(a, b) 2 , 4. a ≥ 1 odd and b ≥ 1 even: M a,b = c max(a, b) and N a,b,m a,b = min(a, b) 2 , 7. a ≥ 2 even and b ≥ 1 odd: M a,b = c max(a, b) and N a,b,m a,b = min(a, b) 2 . The second point of Corollary 17, Propositions 11 and 19 yield: Theorem 21. Let p ≡ 1 (mod 6) be a given prime. Write p = A 2 + AB + B 2 , with A > B ≥ 1 (in a unique way ). Let E p be a set of p -4 pairs (a, b) as in the second point of Corollary 17. Assume that Conjecture 12 holds true for the p -6 pairs (a, b) in E p satisfying a 2 + ab + b 2 ≤ p and that Conjecture 20 holds true for the remaining 2 pairs of positive coprime integers (a, b) ∈ {(A, B), (B, A)} satisfying a 2 + ab + b 2 = p. Then p does not divide the unit index j l for l effectively large enough. In particular, if p ≡ 1 (mod 6) is a prime less than 1875, then p does not divide the unit index j l for l effectively large enough. Proof. First, we checked Conjecture 12 in the range 1 ≤ max(|a|, b) ≤ 50 on a Mac-Book Air laptop computer using Maple. This computation took 52 minutes. (To show that the time needed to check this conjecture increases fast with the bound on a and b, we also did the computation using Maple on a not that new IMac computer with a 2,66 Intel Core 2 Duo processor. for 1 ≤ max(|a|, b) ≤ 50 the computation took 78 minutes, for 1 ≤ max(|a|, b) ≤ 60 the computation took 4 hours, for 1 ≤ max(|a|, b) ≤ 70 the computation took 11 hours and for 1 ≤ max(|a|, b) ≤ 80 the computation took 28 hours). Second, we checked that Conjecture 20 as given in Table 2 holds true (i) in the range max(|a|, b) ≤ 25 (on a MacBook Air laptop computer using Maple, this computation took 8 hours and 30 minutes) and also (ii) in the range 7 ≤ p ≤ 1875 for the 2 pairs (a, b) ∈ {(A, B), (B, A)} satisfying a 2 + ab + b 2 = p (on a MacBook Air laptop computer using Maple, this computation took 14 hours).

Theorem 22 .

 22 Let ε be a totally real exceptional unit. Then ε and ε -1 are multiplicatively dependent if and only if ε is quadratic, i.e. if and only if ε ∈

  , Lemma 4.1], the totally real exceptional units 1 -ε and ε are not n-powers in Q(ε), where ±1 = n ∈ Z is odd. Hence a, b ∈ {±2 n ; n ∈ Z ≥0 }. Since gcd(a, b) = 1 we have a = ±1 or b = ±1. Since d ≥ 3 we have a = ±1 or b = ±1. Hence, either ε = ±(ε -1) ±2 n if a = ±1, or 1 -ε = ±ε ±2 n if b = ±1

  Theorem 2 will follow from Theorems 18 and 21 below. Theorem 18 is a consequence of Proposition 11, Conjecture 12 and the first point of Corollary 17. Theorem 21 is a consequence of Proposition 11 and Conjecture 12, Proposition 19, and Conjecture 20 and the second point of Corollary 17.Set Q -3 (x, y) = x 2 + xy + y 2 . After having proved Proposition 11 and formulated Conjecture 12, we wrote a program which for a given prime p and for any k

Table 2 :

 2 Conjecture for

/2 and j l = 1, by Theorem 4 applied with p 0 = 5. (Notice that since Lemma 5 does not yield b l |l|,[START_REF] Ennola | Fundamental units in a family of cubic fields[END_REF] is not good enough to prove this Theorem

6).

Acknowledgements

We thank Prof. A. Granville who on October 23rd (2019) sent us the proof of Lemma 5, Lemma which we needed to prove Theorem 6. We thank Prof. H. Lenstra who on October 25th (2019) sent us the proof of Proposition 16.