Comptes Rendus Mathématique

Tho Nguyen Xuan

Comptes Rendus
Mathématique
Tho Nguyen Xuan
Note on a symmetric Diophantine equation
Volume 363 (2025), p. 985-988
Online since: 5 September 2025
https://doi.org/10.5802/crmath.785
This article is licensed under the
Creative Commons Attribution 4.0 International License.
http://creativecommons.org/licenses/by/4.0/
C EN T R E
MER S ENN E
The Comptes Rendus. Mathématique are a member of the
Mersenne Center for open scientific publishing
www.centre-mersenne.org — e-ISSN : 1778-3569
Comptes Rendus. Mathématique
2025, Vol. 363, p. 985-988
https://doi.org/10.5802/crmath.785
Research article / Article de recherche
Number theory / Théorie des nombres
Note on a symmetric Diophantine equation
Note sur une équation diophantienne symétrique
Tho Nguyen Xuan a
a Faculty of Mathematics and Informatics, Hanoi University of Science and Technology, Hanoi, Vietnam
E-mail: tho.nguyenxuan1@hust.edu.vn
Abstract. Using an elementary argument, we show that for all rational numbers α such that neither α nor 3α
is a rational square, the equation
x4 −4αx2 −4αy2 + y4 =−6α2
has no rational solutions. This answers Hindes’ two questions and generalizes his theorem (Theorem 1.1)
in “Rational points on certain families of symmetric equations”, Int. J. Number Theory 11 (2015), no. 6, pp.
1821–1838.
Résumé. En utilisant un argument élémentaire, nous montrons que pour tous les nombres rationnels α tels
que ni α ni 3α n’est un carré rationnel, l’équation
x4 −4αx2 −4αy2 + y4 =−6α2
n’a pas de solutions rationnelles. Ceci répond aux deux questions de Hindes et généralise son théorème
(Theorem 1.1) dans “Rational points on certain families of symmetric equations”, Int. J. Number Theory 11
(2015), no. 6, pp. 1821–1838.
Keywords. Arithmetic geometry, Diophantine equations, rational points.
Mots-clés. Géométrie arithmétique, équations diophantiennes, points rationnels.
2020 Mathematics Subject Classification. 14G05, 14G12.
Funding. The author is supported by the Vietnam National Foundation for Science and Technology Devel-
opment (NAFOSTED) (grant number 101.04-2023.21) and by Vietnam Institute for Advanced Study in Math-
ematics (VIASM) from April 2025 to May 2025. The author gratefully acknowledges the Institute for their
support and funding.
Manuscript received 4 May 2025, revised and accepted 15 July 2025.
1. Introduction
Let a,b ∈Q. Consider the symmetric quartic equation
x4 +ax2 +ay2 + y4 = b (1)
with rational numbers x, y . When a = 0, equation (1) reduces to the Fermat quartic
x4 + y4 = b. (2)
Dem’janenko [6] studied rational solutions of (2) using the elliptic curve y2 = x(x2 + b). Sil-
verman [10] extended Dem’janenko’s result to number fields. Hindes [8] further extended
ISSN (electronic): 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/
986 Tho Nguyen Xuan
Dem’janenko’s method to study rational solutions of a family of equations (1) with a ̸= 0. For
each pair of rational numbers (a,b) with b(a2 +4b)(a2 +2b) ̸= 0, let us define the curves{
F = F(a,b) : x4 +ax2 +ay2 + y4 = b,
E = E(a,b) : y2 = x
(
x2 −4ax − (16b +4a2)). (3)
There are two maps φ1,φ2 : F(a,b) → E(a,b) given by{
φ1(x, y) =
(−4x2, x(8y2 +4a)),
φ2(x, y) =
(−4y2, y(8x2 +4a)).
For each squarefree positive integer α, let F (α) = F(α·a,α2·b) and E (α) = E(α·a,α2·b). Hindes [8,
Theorem 1.1] proved the following theorem.
Theorem 1. Let F and E be the symmetric quartic and elliptic curves defined in (3) corresponding
to (a,b) = (−4,−6). Then the following statements hold.
(1) If p is a prime number such that p ≡ 1 mod 24, then F (p) is everywhere locally solvable.
(2) The global root number W (E (p)) =−1 for all positive, odd primes.
(3) If α> 7 ·1074 is square-free and rank(E (α)(Q))≤ 1, then F (α)(Q) =∅.
(4) If p ̸≡ ±1 mod 16, then rank(E (p)) ≤ 2.
(5) Assuming the parity conjecture, if p > 3 ·1074 and p ≡ 25 mod 48, then F (p)(Q) = ∅, and
F (p) breaks the Hasse principle.
Hindes posed the following two questions about Theorem 1.
(1) Is part (3) of Theorem 1 still true when α< 7 ·1074 and α ̸= 3?
(2) Is F (577)(Q) empty?
In this paper, we answer these two questions by proving the following theorem.
Theorem 2. Let α be a rational number such that α and 3α are not rational squares. Then the
equation
x4 −4αx2 −4αy2 + y4 =−6α2 (4)
has no rational solutions.
Theorem 2 provides a significant generalization of parts (3) and (5) of Theorem 1. The
conditions α > 7 ·1074 and rank(Eα(Q)) ≤ 1 in part (3) are not necessary. The assumption of the
parity conjecture and the condition p > 3 ·1074 in part (5) are not required. In contrast to Hindes’
approach, our proof of Theorem 1 is short and elementary. For other studies on the rational
solutions to (2), see Bremner and Morton [3], Serre [9], Flynn and Wetherell [7], Cohen [5], and
Bremner and Tho [4].
Combining part (1) of Theorem 1 with Theorem 2, we obtain the following corollary.
Corollary 3. Let p be a prime number such that p ≡ 1 mod 24. Then:
(1) F (p) is everywhere locally soluble;
(2) F (p)(Q) =;.
In particular, F (p) is a counterexample to the Hasse principle.
2. Proof of Theorem 2
Proof. Assume that (x, y) is a rational solution to (4). Equation (4) can be written in the form
(x2 −2α)2 + (y2 −2α)2 = 2α2. (5)
Tho Nguyen Xuan 987
Since the equation X 2 +Y 2 = 2 has a parameterization
X = −t
2 −2t −1
t 2 +1 , Y =
t 2 −2t −1
t 2 +1 ,
it follows from (5) that there exists a rational number t such that
x2 −2α = (−t
2 −2t −1)α
t 2 +1 ,
y2 −2α = (t
2 −2t −1)α
t 2 +1 .
Hence 
x2 = (t
2 −2t +3)α
t 2 +1 ,
y2 = (3t
2 −2t +1)α
t 2 +1 .
(6)
Let v = x y/α. Then we obtain
v2 = (t 2 −2t +3)(3t 2 −2t +1). (7)
Let C be the quartic curve defined by (7). Since C has a rational point (1,2), it is an elliptic curve.
MAGMA [2] shows that a Weierstrass model of C is
y2 = x3 −x2 −2x, (8)
which has the Mordell–Weil group isomorphic to (Z/2Z)2. Note that C with model (8) is labeled
as 96A in Table I in the classical book of Birch and Kuyk [1]. The table also shows that C has rank 0
with the Mordell–Weil group isomorphic to (Z/2Z)2.
Since equation (7) has four rational solutions (t , v) = (±1,±2), they are all rational solutions
of (7). It follows from (6) that if neither α nor 3α is a rational square, then (4) admits no rational
solutions, completing the proof of Theorem (2). □
Acknowledgments
The author is indebted to the anonymous referee for pointing out the error in the previous version
of the paper and sincerely appreciates the suggestion, which helped correct the error and improve
the presentation of the paper.
Declaration of interests
The author does not work for, advise, own shares in, or receive funds from any organization
that could benefit from this article, and has declared no affiliations other than their research
organizations.
References
[1] B. J. Birch and W. Kuyk (eds.), Modular functions of one variable. IV. Proceedings of the
International Summer School on Modular Functions of One Variable and Arithmetical
Applications, RUCA, University of Antwerp, Antwerp, July 17–August 3, 1972, Lecture Notes
in Mathematics, Springer, 1975, iv+151 pages.
[2] W. Bosma, J. Cannon and C. Playoust, “The Magma algebra system. I. The user language”,
J. Symb. Comput. 24 (1997), no. 3-4, pp. 235–265.
[3] A. Bremner and P. Morton, “A new characterization of the integer 5906”, Manuscr. Math. 44
(1983), no. 1-3, pp. 187–229.
988 Tho Nguyen Xuan
[4] A. Bremner and N. X. Tho, “On the Diophantine equation x4 + y4 = c”, Acta Arith. 204
(2022), no. 2, pp. 141–150.
[5] H. Cohen, Number theory. Vol. I. Tools and Diophantine equations, Graduate Texts in
Mathematics, Springer, 2007, xxiv+650 pages.
[6] V. A. Dem’janenko, “The indeterminate equations x6 + y6 = az2x6 + y6 = az3, x4 + y4 =
az4”, in American Mathematical Society Translations, Ser. 2, Vol. 119 (L. J. Leifman, ed.),
American Mathematical Society Translations, Series 2, vol. 119, American Mathematical
Society, 1983, pp. 27–34.
[7] E. V. Flynn and J. L. Wetherell, “Covering collections and a challenge problem of Serre”, Acta
Arith. 98 (2001), no. 2, pp. 197–205.
[8] W. Hindes, “Rational points on certain families of symmetric equations”, Int. J. Number
Theory 11 (2015), no. 6, pp. 1821–1838.
[9] J.-P. Serre, Lectures on the Mordell-Weil theorem (M. Brown and M. Waldschmidt, eds.),
Aspects of Mathematics, Vieweg & Sohn, 1989, x+218 pages.
[10] J. H. Silverman, “Rational points on certain families of curves of genus at least 2”, Proc.
Lond. Math. Soc. (3) 55 (1987), no. 3, pp. 465–481.


Note on a symmetric Diophantine equation

Benoît Daniel

Simon Labrunie

Victor Nistor

Uniform regularity and weight estimates for the Poisson problems when rounding off the conical points

Yiming Zhu

Comptes Rendus
Mathématique
Yiming Zhu
A remark on the effective nonvanishing of Calabi–Yau varieties
Volume 363 (2025), p. 1059-1063
Online since: 29 September 2025
https://doi.org/10.5802/crmath.735
This article is licensed under the
Creative Commons Attribution 4.0 International License.
http://creativecommons.org/licenses/by/4.0/
C EN T R E
MER S ENN E
The Comptes Rendus. Mathématique are a member of the
Mersenne Center for open scientific publishing
www.centre-mersenne.org — e-ISSN : 1778-3569
Comptes Rendus. Mathématique
2025, Vol. 363, p. 1059-1063
https://doi.org/10.5802/crmath.735
Research article / Article de recherche
Algebraic geometry / Géométrie algébrique
A remark on the effective nonvanishing of
Calabi–Yau varieties
Une remarque sur la non-annulation effective des
variétés de Calabi–Yau
Yiming Zhu a
a Department of Mathematics, Southern University of Science and Technology, 1088 Xueyuan Rd,
Shenzhen 518055, China
E-mail: 12131245@mail.sustech.edu.cn
Abstract. For a Calabi–Yau variety X , we prove that there exists a positive integer m, depending on two natural
invariants of the fiber of its Albanese map, such that the pluricanonical system |mKX | is non-empty.
Résumé. Pour une variété de Calabi–Yau X , nous démontrons qu’il existe un entier strictement positif m, dé-
pendant de deux invariants naturels de la fibre de son application d’Albanese, tel que le système pluricano-
nique |mKX | soit non vide.
2020 Mathematics Subject Classification. 14E05.
Funding. A grant from SUSTech supports the author.
Manuscript received 15 September 2023, revised 14 May 2024 and 22 February 2025, accepted 18 February 2025.
We work with varieties defined over C.
1. Introduction
In their work [2], Birkar and Zhang proved that for any smooth projective variety X of dimension d
with nonnegative Kodaira dimension, there exists a universal constant m that depends on d and
two natural invariants of the general fiber of the Iitaka fibration of X such that the pluricanonical
system |mKX | defines the Iitaka fibration.
In this note, we investigate the case of Kodaira dimension zero. We consider minimal models
of Kodaira dimension zero, i.e., projective varieties with terminal singularities and numerically
trivial canonical divisor (also called Q-Calabi–Yau varieties).
Let X be a Q-Calabi–Yau variety. By [6, Theorem 8.2], OX (mKX ) ∼= OX for some m ∈ N, and
the global index IX is defined as the minimum of such m. The canonical cover of X is defined
as the quasi-étale cover X̃ := Spec⊕IX −1i=0 OX (−i KX ) → X . By [6, Theorem 8.3], the Albanese map
f : X → A of X is an étale fiber bundle. Let F be the fiber of f . Since F has terminal singularities
and KF = KX
∣∣
F is numerically trivial, F is also a Q-Calabi–Yau variety. Let F̂ be a smooth model
of the canonical cover F̃ of F , and let βF̂ := dim H dimF (F̂ ,C) be the middle Betti number of F̂ .
Set m(IF ,βF̂ ) = IF · lcm
{
m ∈ N ∣∣ ϕ(m) ≤ βF̂ }, where ϕ is Euler’s totient function. We show the
following.
ISSN (electronic): 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/
1060 Yiming Zhu
Theorem 1. The pluricanonical system
∣∣m(IF ,βF̂ )KX ∣∣ is nonempty.
Note that m(IF ,βF̂ ) is solely dependent on F . We have the following.
Corollary 2. Let X be a smooth projective variety of Kodaira dimension zero and q(X ) =
dim H 1(X ,OX ) be its irregularity.
(1) If q(X ) = dim(X )−1, then |12KX | is non-empty.
(2) If q(X ) = dim(X )−2, then |mKX | is non-empty, where m = 2 · lcm
{
m ∈ N ∣∣ ϕ(m) ≤ 22} =
26 ·33 ·52 ·7 ·11 ·13 ·17 ·19 ·23.
The existence of such a universal bound for Q-Calabi–Yau 3-folds was initially shown by
Kawamata [7, Theorem 3.2]. Using Bogomolov–Beauville decomposition, Beauville showed that
for a Calabi–Yau 3-fold X , |BKX | is nonempty, where B = lcm
{
m ∈N ∣∣ϕ(m) ≤ 20} = 25 ·33 ·52 ·7 ·
11 ·13 ·17 ·19 [1, Proposition 8]. Morrison showed that the same number works in the terminal
singular case [9, Corollary]. The proofs of Kawamata and Morrison both rely on analyzing the
terminal 3-fold singularities.
2. Preliminaries
2.1. Boundedness of pluricanonical representation
Let X be a projective variety with terminal singularities. The pluricanonical representation of the
birational automorphism group of X is denoted as ρm : Bir(X ) → Aut H 0(mKX ). We adopt the
convention ρm(g ) := (g−1)∗ so that ρm(g ) ·ρm(h) = ρm(g ◦h).
Theorem 3 ([11, Theorem 14.10] and [4, Corollary 3.10]). For a Q-Calabi–Yau variety X , let
d = dim X and m = IX its canonical index, then
ϕ
(|Imρm |)≤ dim H d (X̂ ,C),
where ϕ is Euler’s totient function and X̂ is a smooth model of the canonical cover X̃ of X .
Proof. By [4, Corollary 3.10], Imρm is a finite subgroup of Aut H 0
(
X ,OX (mKX )
) = C∗, hence is
cyclic. Choose g ∈ Bir(X ) and 0 ̸= ω ∈ H 0(X ,OX (mKX )). We have g−1∗ω = αω and α is a root
of unity. On the canonical cover π : X̃ → X , we have π∗ω = ω̃m , where ω̃ is a holomorphic d-
form on X̃ , and g induces a g̃ ∈ Bir(X̃ ) such that g̃−1∗π∗ω = π∗g−1∗ω. Hence g̃−1∗ω̃ = α1/mω̃.
Since X̃ has terminal singularities, ω̃ can be viewed as a holomorphic d-form ω̂ on X̂ , and g̃
induces a ĝ ∈ Bir(X̂ ) such that ĝ−1∗ω̂ = α1/mω̂. By the same argument of [11, first paragraph
of the proof of Proposition 14.4], we have
[
Q(α1/m) : Q
] ≤ dim H d (X̂ ,Q), and thus ϕ(|Imρm |) ≤
dim H d (X̂ ,Q). □
2.2. Fiber bundles
Let F be a projective variety, and let f : X → Y be an F -fiber bundle defined by a cocycle
φi j : Yi j → Aut(F ), where {Yi } is an open covering of Y with Yi j := Yi ∩ Y j . Assume that X
is terminal and Y is smooth. Take m to be a positive integer such that mKX is Cartier, then
f∗OX (mKX /Y ) is a rank h0(mKF ) := dim H 0
(
F,OF (mKF )
)
vector bundle. The following Lemma
describes this bundle in terms of φi j .
Lemma 4. The vector bundle f∗OX (mKX /Y ) is defined by the cocycleρm◦φi j , whereρm : Aut(F ) →
Aut H 0
(
F,OF (mKF )
)
is the pluricanonical representation of the automorphism group of F .
Yiming Zhu 1061
Proof. See also [11, p. 186] for a proof. We have isomorphisms αi : Xi := f −1(Yi ) ∼= Yi ×F over Yi ,
where αi ◦α−1j = (id,φi j ). By taking a set of basis {ωk }k=1,...,h0(mKF ) of H 0
(
F,OF (mKF )
)
, we can
write each ρm(φi j ) as a matrix: ρm(φi j )ωk = φ∗j iωk =
∑
l ωl (Φi j )l k , where (Φi j )lk is a matrix
indexed by lk. Let s be a local section of f (mKX /Y ). Under the isomorphisms αi , s can be
identified on each Yi with si = ∑k ti k ⊗ωk satisfying s j = (id,φi j )∗si . Where ti k ∈ H 0(Yi ,OYi ).
Consequently, we find that
∑
k t j k ⊗ωk =
∑
l ,k ti k ⊗ωl (Φ j i )lk and thus ti k =
∑
l (Φi j )kl t j l . □
We say an F -fiber bundle f : X → Y is locally constant if there exists a representation
γ : π1(Y ) → Aut(F ), such that X ∼= Ỹ ×F /π1(Y ) where Ỹ is the universal cover of Y , and π1(Y )
acts on Ỹ ×F diagonally. Viewing Ỹ → Y as a principal π1(Y )-bundle defined by a cocycle ui j ,
then the fiber bundle X → Y is defined by the cocycleγ(ui j ), and the vector bundle f∗OX (mKX /Y )
is defined by the cocycle ρm ◦γ(ui j ) by Lemma 4. In other words, f∗OX (mKX /Y ) is a local system
defined by the representation ρm ◦γ.
2.3. Structure of Albanese maps
A morphism between two normal varieties will be called a fibration if it is surjective and has
connected fibers. Let X be a projective variety with terminal singularities. The irregularity of X is
defined as q(X ) = dim H 1(X ,OX ). Take Y to be any smooth model of X . Since X is terminal, it has
rational singularities. Hence q(X ) = q(Y ). Let Y → A be the Albanese map of Y , and the Albanese
map of X is defined as the rational map X 99K A which is a morphism by [10, Proposition 2.3].
If κ(X ) = 0, then κ(Y ) = 0. Then Y → A is a fibration by [5, Theorem 1]. Hence X → A is also a
fibration.
Now, let X be aQ-Calabi–Yau variety. Kawamata showed that its Albanese map f : X → A is an
étale fiber bundle [6, Theorem 8.3]. Cao and later Wang showed more in the smooth and singular
case respectively.
Theorem 5 ([3, Theorem 1.2] and [12, Theorem B] ). The fiber bundle f : X → A is locally
constant.
Cao and Wang in fact proved this theorem under a much weaker hypothesis: −KX and
−(KX +∆) (for klt pairs (X ,∆)) is nef instead of numerically trivial. Since the fiber bundle f : X → A
is locally constant, it is defined by a representation γ : π1(A) → Aut(F ). By Lemma 2.2, we have
the following.
Lemma 6. The vector bundle f∗OX (mKX ) is a local system defined by the representation ρm ◦γ.
3. Proof of Theorem 1 and Corollary 2
Proof of Theorem 1. Let f : X → A be the Albanese map of X . It is a locally constant fiber bundle
by Theorem 5. Let F be the fiber of f . Let iX and iF be the Cartier index of X and F respectively.
We claim that iX = iF . Since iX KX is Cartier, iX KF = iX KX
∣∣
F is Cartier. Hence iF
∣∣ iX . On the other
hand, let {Ai } be an open covering of A such that Xi := f −1 Ai ∼= Ai ×F . Since iF KX
∣∣
Xi
is Cartier
for each Xi , we have iF KX is Cartier. Thus iX
∣∣ iF , and iX = iF is proved.
The Cartier index iF divides the global index IF , hence OX (IF KX ) is a line bundle. By Theo-
rem 3 and Lemma 6, the line bundle f∗OX (IF KX ) is a local system defined by the representation
ρIF ◦γ : π1(A) → Aut H 0
(
F,OF (IF KF )
)
, and the kernel K of this map is of finite index such that
ϕ
([
π1(A) : K
])≤βF̂ . This induces a finite étale cover q : A1 → A with ϕ(deg q) ≤βF̂ , and the pull-
back of f∗OX (IF KX ) through this cover is trivial. Let X1 = A1 ×A X be the fiber product, and let
f1 : X1 → X , p : X1 → X be the induced morphisms as in the following commutative diagram.
1062 Yiming Zhu
X1 X
A1 A
f1
p
f
q
Therefore we have:
q∗ f∗OX (IF KX ) =OA1 ⊗H 0
(
F,OF (IF KF )
)
. (1)
On X1, we have
H 0
(
X1,OX (IF KX1 )
)= H 0(X1, p∗OX (IF KX ))
= H 0(A1, f1∗p∗OX (IF KX ))
= H 0(A1, q∗ f∗OX (IF KX ))
= H 0(A1,OA1 )⊗H 0
(
F,OF (IF KF )
)
=C.
The first equality holds as f : X1 → X is finite étale and thus KX1 = p∗KX . The third equality
follows from flat base change. The fourth equality is a consequence of (1). Thus, IF KX1 ∼ 0.
Since X1 → X is étale with the same degree as q , p∗OX1 is locally free of rank m := deg q . By the
projection formula, we have ∧m p∗OX1 = ∧m p∗OX1 (IF KX1 ) = ∧m p∗
(
p∗OX (IF KX )
) = ∧m p∗OX1 ⊗
OX (mIF KX ). Since ∧m p∗OX1 is a line bundle, we conclude that mIF KX ∼ 0. □
Proof Corollary 2. Let f : X → A be the Albanese map of X . By running a KX -MMP over A, we
have a minimal model X ′ of X over A which is actually minimal over C by [8, second paragraph
of the proof of Theorem 4.2]. By Theorem 5, X ′ → A is a locally constant fiber bundle, hence the
singular locus of X ′ is of dim ≥ dim A. But X ′ has at most terminal singularities whose singular
locus is of codim ≥ 3, hence the fiber F ′ of X ′ → A and X ′ are both smooth. We may assume
from the beginning that X is already minimal, i.e., that X is a smoothQ-Calabi–Yau variety. In the
following, let F be the fiber of the Albanese map of X and m := m(IF ,βF̂ ).
If dimF = 1, then F is an elliptic curve, m = 12.
If dimF = 2:
• if F is a K3 surface, m = lcm{m ∈N ∣∣ϕ(m) ≤ 22};
• if F is an Enriques surface, its canonical cover is a K3 surface, m = 2 · lcm{m ∈N ∣∣ϕ(m) ≤
22
}
;
• if F is an Abelian surface, m = lcm{m ∈N ∣∣ϕ(m) ≤ 6};
• if F is a bielliptic surface, its canonical cover is an Abelian surface, m = 2,3,4 or 6·lcm{m ∈
N
∣∣ϕ(m) ≤ 6}.
Then we conclude by Theorem 1. □
Acknowledgments
The author thanks his advisor Zhan Li, and his classmate Xingying Li for helpful discussions. The
author also thanks the reviewers for their comments to improve the article.
Declaration of interests
The author does not work for, advise, own shares in, or receive funds from any organization
that could benefit from this article, and has declared no affiliations other than their research
organizations.
Yiming Zhu 1063
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[12] J. Wang, Positivity of direct images and projective varieties with nonnegative curvature, PhD
thesis, Institut Polytechnique de Paris (France), 2020. Online at https://theses.hal.science/
tel-02982921.


A remark on the effective nonvanishing of Calabi–Yau varieties

Badr Elmansouri

Reflected BSDEs with default time and irregular obstacles

Shalom Eliahou

Jean-Louis Verger-Gaugry

Comptes Rendus
Mathématique
Shalom Eliahou and Jean-Louis Verger-Gaugry
The number system in rational base 3/2 and the 3x +1 problem
Volume 363 (2025), p. 329-336
Online since: 20 May 2025
https://doi.org/10.5802/crmath.662
This article is licensed under the
Creative Commons Attribution 4.0 International License.
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C EN T R E
MER S ENN E
The Comptes Rendus. Mathématique are a member of the
Mersenne Center for open scientific publishing
www.centre-mersenne.org — e-ISSN : 1778-3569
Comptes Rendus. Mathématique
2025, Vol. 363, p. 329-336
https://doi.org/10.5802/crmath.662
Research article / Article de recherche
Combinatorics, Number theory / Combinatoire, Théorie des nombres
The number system in rational base 3/2 and
the 3x +1 problem
Le système de numération en base rationnelle 3/2 et le
problème 3x +1
Shalom Eliahou ,a,b and Jean-Louis Verger-Gaugry ,c
a Univ. Littoral Côte d’Opale, UR 2597 – LMPA – Laboratoire de Mathématiques Pures et Appliquées
Joseph Liouville, F-62100 Calais, France
b CNRS FR 2037, France
c Université-Grenoble-Alpes, Université Savoie Mont Blanc, F-73000 Chambéry, France
E-mails: eliahou@univ-littoral.fr, verger-gaugry.jean-louis@unsechat-math.fr
Abstract. The representation of numbers in rational base p/q was introduced in 2008 by Akiyama, Frougny
& Sakarovitch, with a special focus on the case p/q = 3/2. Unnoticed since then, natural questions related to
representations in that specific base turn out to intimately involve the Collatz 3x +1 function. Our purpose
in this note is to expose these links and motivate further research into them.
Résumé. La représentation des nombres dans la base rationnelle p/q a été introduite en 2008 par Akiyama,
Frougny & Sakarovitch, avec un accent particulier sur le cas p/q = 3/2. Des questions naturelles liées aux
représentations dans cette base spécifique, passées inaperçues jusqu’ici, s’avèrent impliquer intimement la
fonction de Collatz 3x + 1. Le but de cette note est d’exposer ces liens et de motiver des recherches plus
approfondies sur ceux-ci.
Keywords. Collatz conjecture, iteration, numeration, odometer.
Mots-clés. Conjecture de Collatz, itération, numération, odomètre.
2020 Mathematics Subject Classification. 11B83, 68R15.
Manuscript received 24 April 2024, revised 22 May 2024 and 21 June 2024, accepted 29 June 2024.
1. Introduction
The representation of numbers in rational base p/q with p > q coprime positive integers was
introduced in 2008 in [2], with a focus on the case p/q = 3/2 to study Mahler’s Z -number problem
and the Josephus problem.
Given n ∈N, its representation in rational base 3/2 is the unique expression of the form
n =
k∑
i=0
ai
2
(
3
2
)i
ISSN (electronic): 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/
330 Shalom Eliahou and Jean-Louis Verger-Gaugry
with digits ai ∈ {0,1,2} for all i ≥ 0 and ak ̸= 0 if n ≥ 1. We then denote the corresponding word in
{0,1,2}∗ by
〈n〉 = ak ak−1 · · ·a0.
We set 〈0〉 = ϵ, the empty word, and we identify any word 0w ∈ {0,1,2}∗ with w .
Denoting L = {〈n〉 | n ∈N} ⊂ {0,1,2}∗, here are two natural questions about L :
(1) If w ∈L , which extensions among w0, w1, w2 still belong to L ?
(2) Given k ≥ 1, what is the lexicographically largest word of length k in L ? Equivalently,
what is the largest n ∈N such that 〈n〉 is of length k?
Apparently unnoticed so far, answers to these questions turn out to intimately involve the Collatz
3x +1 function, namely T : N→N defined by
T (n) =
{
n/2 if n ≡ 0 mod 2,
(3n +1)/2 if n ≡ 1 mod 2. (1)
The purpose of this note is to expose these links and motivate further research into them. Its
contents are as follows. In Section 2, we recall basic properties of the numeration system in
rational base 3/2, in particular the recursive construction of the word 〈n〉 in Section 2.1 and the
odometer allowing to construct 〈n + 1〉 from 〈n〉 in Section 2.2. We also propose a variant of
the Collatz conjecture in Section 2.3. In Sections 3.1 and 3.2, we address the above questions (1)
and (2), respectively. Section 4 concludes the note with two questions and two conjectures closely
related to the Collatz conjecture and possibly of the same order of difficulty.
2. Basics
Following [2], every n ∈N has a unique expression of the form
n = 1
2
(
ak (3/2)
k +ak−1(3/2)k−1 +·· ·+a0
)
with ai ∈ {0,1,2} for all i and ak ̸= 0 if n ≥ 1. Accordingly, we associate to n the word
〈n〉 = ak ak−1 · · ·a0 ∈ {0,1,2}∗,
called the representation of n in rational base 3/2. We set 〈0〉 = ϵ, the empty word, and we identify
any word 0w ∈ {0,1,2}∗ with w .
By [5, Lemma 4.1], the length of the word 〈n〉 is given by |〈n〉| = log3/2(n)+O(1).
Definition 1. We say that a word w ∈ {0,1,2}∗ is admissible if w = 〈n〉 for some n ∈N. We denote
by L ⊂ {0,1,2}∗ the language of admissible words, i.e.
L = {〈n〉 | n ∈N}.
As observed in [2], the language L is prefix-closed, i.e. stable under taking prefixes. That is,
every prefix w ′ of an admissible word w = w ′w ′′ is itself admissible.
Conversely, given an admissible word w ∈ L , which extensions among w0, w1, w2 remain
admissible? The answer involves the Collatz 3x +1 function. See Proposition 7.
2.1. Construction of 〈n〉
Given n ∈ N, a recursive construction of 〈n〉 ∈ {0,1,2}∗ may be described as follows. Set 〈0〉 = ϵ,
the empty word.
Shalom Eliahou and Jean-Louis Verger-Gaugry 331
Proposition 2. Let n0 ∈N\ {0}. Denote 〈n0〉 = ak · · ·a0 with ai ∈ {0,1,2} for all i . Then:{
a0 ≡ 2n0 mod 3;
〈n0〉 = 〈n1〉a0, where n1 = (2n0 −a0)/3.
Moreover, we have a0 = 2n0 −3n1 and a0 ≡ n1 mod 2.
Proof. As 〈n0〉 = ak · · ·a0, we have
n0 = 1
2
(
ak (3/2)
k +·· ·+a1(3/2)+a0
)
=
(
1
2
(
ak (3/2)
k−1 +·· ·+a1
))3
2
+ a0
2
.
Hence a0 ≡ 2n0 mod 3, and setting n1 = (2n0 − a0)/3, we have 〈n1〉 = ak · · ·a1. The claims
follow. □
2.2. The odometer
The odometer constructs the word 〈n+1〉 from the word 〈n〉. It is described as a finite automaton
in [2]. Here is an equivalent description.
Let σ = (2 1 0) be the cyclic permutation 2 7→ 1 7→ 0 7→ 2. For a word u ∈ {0,1,2}∗, let uσ ∈
{0,1,2}∗ denote the word obtained by applyingσ to each letter of u. For instance, 12012σ = 01201,
then identified to 1201.
Proposition 3. Let n ∈N. If 〈n〉 = u0v ∈ {0,1,2}∗ with v ∈ {1,2}∗, then 〈n +1〉 = u2vσ.
Proof. See [2, Section 3.2.3, pp. 66–67]. □
The rule applies even if u or v is the empty word. In particular, if 〈n〉 = v ∈ {1,2}∗, then
identifying v with 0v , we get 〈n + 1〉 = 2vσ. Table 1, listing 〈n〉 for 1 ≤ n ≤ 27, was constructed
by hand using this description of the odometer. The interested reader will effortlessly extend it
much farther.
Table 1. The words 〈n〉 for 1 ≤ n ≤ 27.
n 〈n〉
1 2
2 21
3 210
4 212
5 2101
6 2120
7 2122
8 21011
9 21200
n 〈n〉
10 21202
11 21221
12 210110
13 210112
14 212001
15 212020
16 212022
17 212211
18 2101100
n 〈n〉
19 2101102
20 2101121
21 2120010
22 2120012
23 2120201
24 2120220
25 2120222
26 2122111
27 21011000
2.3. The Collatz conjecture
Dating back to the 1930’s, this conjecture deals with iterates T (k) of the Collatz function T : N→N
defined in (1).
Conjecture 4. Let n0 ∈N, n0 ≥ 1. Then there exists k ≥ 1 such that T (k)(n0) = 1.
332 Shalom Eliahou and Jean-Louis Verger-Gaugry
Here is an equivalent version, as shown below.
Conjecture 5. Let n0 ∈N, n0 ≥ 1. Then the proportion of even numbers in the sequence of iterates
T (k)(n0) tends to 1/2 as k tends to infinity.
Proposition 6. Conjectures 4 and 5 are equivalent.
Proof. If Conjecture 4 holds, then at some point the sequence of iterates T (k)(n0) collapses to
1,2,1,2,1,2, . . . , whence Conjecture 5.
Conversely, assume Conjecture 5 holds. Given an integer n ≥ 1, letΩ(n) = (n,T (n),T (2)(n), . . .)
denote the trajectory of n under iteration of T . Moreover, given k ≥ 1, let
Ωk (n) =
(
n,T (n),T (2)(n), . . . ,T (k−1)(n)
)
,
and let k1 denote the number of odd integers in Ωk (n). Assume for a contradiction that Conjec-
ture 4 fails. Then there is an integer m ≥ 10 such that minΩ(m) = m. (Note that minΩ(n) = 1 for
n ≤ 9.) We claim that for all k ≥ 1, we have
T (k)(m) ≤ (3.1)
k1
2k
m. (2)
The proof is straightforward, by induction on k. Indeed, (2) holds for k = 1, since m is odd
by minimality in Ω(m) and T (m) = (3m + 1)/2 = (3 + 1/m)m/2 ≤ (3.1)m/2 since m ≥ 10. Let
now k ≥ 1 and assume that T (k)(m) ≤ (3.1)k1 /2k m. Denote n = T (k)(m). If n is even, then
T (k+1)(m) = T (n) = n/2 ≤ (3.1)k1 /2k+1m, as desired. If n is odd, then
T (k+1)(m) = T (n) = (3+1/n)n/2 ≤ (3+1/m)n/2 ≤ (3.1)n/2
since n ≥ m ≥ 10. Hence T (k+1)(m) ≤ (3.1)k1+1/2k+1m, as desired. The claim is proved.
Now (3.1)k1 /2k ≥ 1 if and only if k1/k ≥ ln(2)/ln(3.1), and ln(2)/ln(3.1) > 0.6. By Conjecture 5,
for k large enough, the proportion k1/k will fall below 0.6. Hence T (k)(m) ≤ (3.1)k1 /2k m < m,
contradicting the minimality of m. This completes the proof of the proposition. □
At the time of writing, Conjecture 4 has been verified up to n0 ≤ 268. See [3]. With this bound,
the results of [4] and its Table 2 on page 54 imply that, besides the trivial cycle {1,2} under T , no
other cycle, if any, can have length less than 114,208,327,604. See [1] for a general reference on
the conjecture, and [6] for recent strong results about it.
3. Links with the Collatz function
We now address the two questions about L = {〈n〉 | n ∈ N} asked in the Introduction, and show
that in both cases, the answer naturally involves the Collatz function T .
3.1. Extending admissible words
Here is the first question. Given w ∈L , which extensions among w0, w1, w2 still belong to L ?
Proposition 7. Let w ∈L , say w = 〈n〉 for some n ∈N. Then:{
w1 ∈L ⇐⇒ n ≡ 1 mod 2, and then w1 = 〈(3n +1)/2〉 = 〈T (n)〉;
w0 ∈L ⇐⇒ n ≡ 0 mod 2 ⇐⇒ w2 ∈L , and then w0 = 〈3n/2〉 and w2 = 〈(3n +2)/2〉.
Proof. If 〈n〉1 ∈L , Proposition 2 implies 1 ≡ n mod 2 and w1 = 〈(3n+1)/2〉. Similarly, if 〈n〉0 ∈L
then 0 ≡ n mod 2 and w0 = 〈3n/2〉. Finally, if 〈n〉2 ∈ L then 2 ≡ n mod 2 and w2 = 〈(3n +2)/2〉.
The stated equivalences follow. □
Corollary 8. Let n ∈N be odd. Then 〈T (n)〉 = 〈n〉1.
Shalom Eliahou and Jean-Louis Verger-Gaugry 333
Proof. Immediate from the above proposition. □
Remark 9. The map n 7→ n/2 for n even is straightforward in the base 2 representation of n:
just remove the last digit, 0. Symmetrically, Corollary 8 shows that the map n 7→ (3n + 1)/2 for
n odd is straightforward in the base 3/2 representation of n: just add 1 as rightmost digit. By
any chance, could it happen that considering the map n 7→ T (n) simultaneously in base 2 and in
rational base 3/2 may lead to new insights on the Collatz conjecture?
3.2. Saturated words
Given a number system B over a finite set of digits, we say that a positive integer n ≥ 1 is saturated
if its representation 〈n〉B in base B is of length strictly less than that of 〈m〉B for all m ≥ n +1.
Accordingly, we say that 〈n〉B is a saturated word. For instance in base 10, the saturated integers
and corresponding words are the n = 10ℓ−1 = 9 · · ·9.
The digits 1,2 of the saturated words in rational base 3/2 turn out to be unpredictable in some
sense and closely related to the 3x + 1 problem. Recall that |w | denotes the length of the finite
word w .
Notation 10. Given k ≥ 1, we denote by sat(k) the largest positive integer whose representation
in base 3/2 is of length k. That is,
sat(k) = max{n ∈N, |〈n〉| = k}.
Proposition 11. For all k ≥ 1, 〈sat(k)〉 is the unique admissible word of length k with digits in
{1,2}.
Proof. By the odometer as described in Proposition 3, if 〈n〉 is of length k and contains a 0 digit,
then 〈n+1〉 is of the same length k. This shows that the inequality |〈n+1〉| > |〈n〉| happens exactly
when 〈n〉 contains no 0 digit except possibly as an irrelevant prefix. □
For instance, a look at 〈n〉 for 1 ≤ n ≤ 27 in Table 1 yields sat(k) and the corresponding word
for k ≤ 7 in Table 2. Observe in Table 2 that 〈sat(k −1)〉 is a prefix of 〈sat(k)〉. Indeed, this is true
for all k ≥ 1.
Proposition 12. For all k ≥ 2, we have 〈sat(k)〉 = 〈sat(k − 1)〉a0 where a0 ∈ {1,2} and a0 ≡
sat(k −1) mod 2.
Proof. Since L is prefix-closed, it follows from Proposition 11 that the length k − 1 prefix of
〈sat(k)〉 equals 〈sat(k − 1)〉. Thus 〈sat(k)〉 = 〈sat(k − 1)〉a0 where a0 ∈ {1,2}, and a0 ≡ sat(k −
1) mod 2 by Proposition 2. □
Table 2. The first seven saturated numbers and words.
k sat(k) 〈sat(k)〉
1 1 2
2 2 21
3 4 212
4 7 2122
5 11 21221
6 17 212211
7 26 2122111
334 Shalom Eliahou and Jean-Louis Verger-Gaugry
3.3. The function U
It turns out that the sequence
(
sat(k)
)
k≥1 coincides with the sequence of iterates U
(k)(0) of a
certain function U closely related to the Collatz function T . See Proposition 16 below.
Notation 13. Let U : N→N be defined as
U (n) =
{
(3n +2)/2 = T (n)+n +1 if n ≡ 0 mod 2,
(3n +1)/2 = T (n) if n ≡ 1 mod 2.
Remark 14. We have U (n) ≡ T (n)+n +1 mod 2 for all n ∈N.
Starting from any n0 ∈ N, let us consider the sequence of iterates U (k)(n0) for k ≥ 1. The
following proposition shows that, for the sequence of corresponding words 〈U (k)(n0)〉 in rational
base 3/2, each term is a prefix of the next one and the added digit is either 1 or 2.
Proposition 15. Let n ∈N. Then:
〈U (n)〉 =
{
〈n〉1 if n ≡ 1 mod 2,
〈n〉2 if n ≡ 0 mod 2.
Proof. Straightforward from Proposition 7. □
A particularly interesting sequence of iterates U (k)(n0) is for n0 = 0.
Proposition 16. For all k ≥ 1, we have sat(k) =U (k)(0).
Proof. For k = 1, we have sat(1) = 2 and U (1) = T (1) = 2. For k ≥ 2, we only need show
sat(k) =U (sat(k −1)). Indeed, it follows from Proposition 12 that
sat(k) = 3sat(k −1)+a0
2
where a0 ∈ {1,2} and a0 ≡ sat(k − 1) mod 2. Hence sat(k) = U
(
sat(k − 1)) by definition of U , as
desired. □
Consequently, the growth factor of the sequence sat(k) tends to 3/2.
4. Open questions
We conclude this note with two questions and two conjectures.
Question 17. Is the Collatz conjecture equivalent to some behavior of the iteration of the func-
tion U ? For instance, starting with any n0 ∈ N, is it true that the proportion of 1’s in 〈U (k)(n0)〉
tends to 1/2 as k goes to infinity, as supported by numerical evidence? Note that by Proposition 15,
the number of digits “0” remains constant along the trajectory of n0.
More specifically, for n0 = 0, we have U (k)(0) = sat(k) for all k ≥ 1 by Proposition 16. Having
computed sat(k) up to k = 220 = 1,048,576, we find that the word w = 〈sat(220)〉, of length k = 220,
counts exactly 525,067 digits “1” and 523,509 digits “2”. Moreover, for all 1 ≤ ℓ < 17, the word w
contains all 2ℓ words in {1,2}ℓ as subwords. See Table 3 for the word 〈sat(200)〉, with exactly 100
digits “1”. This numerical evidence leads us to the following conjecture.
Conjecture 18. The proportion of digits “1” in the word 〈sat(k)〉 tends to 1/2 as k tends to infinity.
Shalom Eliahou and Jean-Louis Verger-Gaugry 335
Is Conjecture 18 equivalent to the Collatz conjecture? The question arises from the close
relationship between the functions T and U , Remark 14 and the analogy with Conjecture 5. At
any rate, even if not equivalent to Conjecture 5, Conjecture 18 might well be of the same order of
difficulty.
In fact, the numerical evidence below Question 17 leads to an even stronger conjecture. Since
〈sat(k −1)〉 is a prefix of 〈sat(k)〉 for all k ≥ 2, we may define the limit
〈sat(∞)〉 = lim
k→∞
〈sat(k)〉,
an infinite word over {1,2}. Informally, recall that an infinite word w over a finite alphabet B of
cardinality |B| = b is said to be normal if for all ℓ ≥ 1, all bℓ words in Bℓ occur as subwords in
w[n], the length n prefix of w , with frequency tending to 1/bℓ as n →∞. Our stronger conjecture
runs as follows.
Conjecture 19. The word 〈sat(∞)〉 is normal.
Table 3. The word 〈sat(200)〉 = 〈U 200(0)〉.
2122111221211221212112212112121111122121111111212222211221212212121
1211222212212222221211111221221212211211222211122111222122221211212
122122212122212121211111111212122222111112211121212222122111112112
Our second question is about fully understanding the Collatz map T in rational base 3/2.
Question 20. For n odd, we have seen that 〈T (n)〉 = 〈n〉1. For n even, how does one determine
〈T (n)〉 from 〈n〉? Is any answer susceptible to lead to new insights on the Collatz conjecture?
Note that determining the parity of n ∈ N from the digits ai of the word 〈n〉 = ak · · ·a0 ∈ L is
more cumbersome than in base 2 or 10. Indeed, the following criterion is straightforward but not
quite practical for calculations by hand.
Proposition 21. Let n ∈N\ {0} and 〈n〉 = ak · · ·a0 ∈L . Then
n ≡ 0 mod 2 ⇐⇒
k∑
i=0
2k−i 3i ai ≡ 0 mod 2k+2.
Proof. We have n = 12
(∑k
i=0(3/2)
i ai
) = 2−(k+1)(∑ki=0 2k−i 3i ai ). Hence n ≡ 0 mod 2 ⇐⇒ 2k+1n ≡
0 mod 2k+2 ⇐⇒ ∑ki=0 2k−i 3i ai ≡ 0 mod 2k+2. □
Acknowledgments
We thank the anonymous referee for their careful reading and constructive comments, in partic-
ular for insisting to formally state Proposition 6 and carefully prove it, as opposed to just a remark
with informal arguments as in the original version of this article.
Declaration of interests
The authors do not work for, advise, own shares in, or receive funds from any organization
that could benefit from this article, and have declared no affiliations other than their research
organizations.
336 Shalom Eliahou and Jean-Louis Verger-Gaugry
References
[1] J. C. Lagarias (ed.), The ultimate challenge: the 3x + 1 problem, American Mathematical
Society, 2010, pp. xiv+344.
[2] S. Akiyama, C. Frougny and J. Sakarovitch, “Powers of rationals modulo 1 and rational base
number systems”, Isr. J. Math. 168 (2008), pp. 53–91.
[3] D. Barina, “Convergence verification of the Collatz problem”, J. Supercomput. 77 (2021),
pp. 2681–2688.
[4] S. Eliahou, “The 3x + 1 problem: new lower bounds on nontrivial cycle lengths”, Discrete
Math. 118 (1993), no. 1-3, pp. 45–56.
[5] J. F. Morgenbesser, W. Steiner and J. M. Thuswaldner, “Patterns in rational base number
systems”, J. Fourier Anal. Appl. 19 (2013), no. 2, pp. 225–250.
[6] T. Tao, “Almost all orbits of the Collatz map attain almost bounded values”, Forum Math. Pi
10 (2022), article no. E12 (56 pages).


The number system in rational base $3/2$ and the $3x+1$ problem

Yong Hu

Yisheng Tian

Trivialité des groupes de Whitehead réduits avec applications à l’approximation faible et l’approximation forte

Xavier Roulleau

Comptes Rendus
Mathématique
Xavier Roulleau
The cuspidal cubic and line arrangements with only triple points
Volume 363 (2025), p. 593-598
Online since: 5 June 2025
https://doi.org/10.5802/crmath.741
This article is licensed under the
Creative Commons Attribution 4.0 International License.
http://creativecommons.org/licenses/by/4.0/
C EN T R E
MER S ENN E
The Comptes Rendus. Mathématique are a member of the
Mersenne Center for open scientific publishing
www.centre-mersenne.org — e-ISSN : 1778-3569
Comptes Rendus. Mathématique
2025, Vol. 363, p. 593-598
https://doi.org/10.5802/crmath.741
Research article / Article de recherche
Algebraic geometry / Géométrie algébrique
The cuspidal cubic and line arrangements
with only triple points
La cubique cuspidale et les arrangements de droites avec
seulement des points triples
Xavier Roulleau a
a Université d’Angers, CNRS, LAREMA, SFR MATHSTIC, F-49000 Angers, France
E-mail: xavier.roulleau@univ-angers.fr
Abstract. We describe a new infinite family of line arrangements in the projective plane with only triple point
singularities, and we recover previously known examples.
Résumé. Nous décrivons une nouvelle famille infinie d’arrangements de droites dans le plan projectif ne
présentant que des singularités triples, et nous retrouvons des exemples déjà connus.
2020 Mathematics Subject Classification. 05B35, 14N20, 52C30.
Manuscript received 30 December 2024, revised 11 March 2025, accepted 14 March 2025.
1. Introduction
Line arrangements, i.e. finite unions of lines in the projective plane, are studied in various
domains: algebra, topology, combinatorics... A double point (respectively triple point) on a
line arrangement is a point where exactly 2 (respectively 3) lines of the arrangement meet. The
celebrated Sylvester–Gallai Theorem asserts that any line arrangement over the real field has at
least one double point, or is the trivial example: the union of lines passing through the same
point.
Hirzebruch [2] then proved that over the complex field, a line arrangement is either the trivial
example or possesses double or triple points. To date, there exists a unique non-trivial example
of a complex line arrangement with only triple points. Even before that result, the construction
of line arrangements with many or exclusively triple points had attracted attention, see e.g. [3] for
a historical account.
In this paper, we present a new construction of line arrangements over finite fields that
exhibit only triple points. Line arrangements with only triple points are also interesting from
a combinatorial perspective, as they are linked to Steiner triple systems. A Steiner triple system
is a set T of 3-element subsets of a finite set, such that any 2-element subset is contained in a
unique 3-element subset t ∈ T .
Before presenting our results, let us introduce some notations and recall some properties of
line arrangements. If L = ℓ1 +·· ·+ℓn is a line arrangement labeled by {1, . . . ,n}, the data M(L )
ISSN (electronic): 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/
594 Xavier Roulleau
of the triples {i , j ,k} ⊂ {1, . . . ,n} such that the lines ℓi , ℓ j and ℓk meet at a common point is called
the matroid associated with L ; it encodes the combinatorics of L . If L has only triple points,
then M(L ) is a Steiner triple system.
If γ is a projective transformation of the plane, one may define γL = γℓ1 + ·· · + γℓn . The
line arrangements L , γL have the same combinatorics, meaning that M(L ) = M(γL ). Given a
matroid M , a labelled line arrangement L such that M(L ) = M is said to be a realization of M .
Given a field K , there exists a scheme parametrizing the realizations over K , and the quotient
R(M)/K of these realizations modulo the action of PGL3(K ) is called the moduli space of M . A
line arrangement over a field K is said to be rigid if R(M)/K is zero-dimensional.
In the present paper, we describe a new infinite family of line arrangements in the plane with
only triple points, and study some of their associated moduli spaces:
Theorem 1. For q = 3n , n ≥ 1, there exists a line arrangement Lq of q lines over Fq with q(q−1)6
triple points and no other singularities.
Let Mq = M(Lq ) be the matroid associated to Lq . For n ≥ 3, the matroid Mq has no realizations
over any field of characteristic ̸= 3.
The moduli space R(M27)/F3 is 3-dimensional with a 2-dimensional immersed component of
multiplicity 2.
One obtains a model of the reduced scheme of R(M27)/F3 as an open sub-scheme of a quadric
in P4.
Our first example of a line arrangement Lq was obtained as the dual of the q points defined
over Fq on the smooth part of the cuspidal cubic y2z = x3. However the generic realization of M27
is such that the 27 points of its dual are not contained in any plane cubic.
The previously known line arrangements with only triple points are:
(a) The trivial example: three lines meeting at one point.
(b) Over a field containing a primitive third root of unity (in particular over any algebraically
closed field of characteristic ̸= 3): the Ceva(3) line arrangement with 9 lines, given by
(x3 − y3)(x3 − z3)(z3 − y3) = 0.
It has 12 triple points and it is the dual of the Hesse configuration, i.e. of the 9 flex points
of a smooth cubic. In characteristic 3, there is also a line arrangement of 9 lines having
the same associated matroid as Ceva(3). This is the 9 lines in P2(F3) not containing a
given point p in P2(F3). These line arrangements are rigid.
(c) The Fano plane: the 7 lines in P2(F2). It is also a rigid line arrangement.
(d) The sequence of line arrangements C ′2n−1: for n ≥ 4, the line arrangement C ′2n−1 consists
of 2n −1 lines defined over F2. It is constructed in [3] and obtained by taking the dual of a
generic projection to P2 of the 2n −1 points Pn−1(F2) contained in Pn−1.
(e) A rigid example over F11: a rigid line arrangement with 19 lines and 57 triple points,
constructed in [3].
It is an open question [5] whether or not (a) and (b) are the unique examples of complex line
arrangements with only triple points.
For any n ≥ 2, we also obtain line arrangements C2n−1 over F2 with 2n −1 lines and only triple
points. These are constructed by using the dual of the non-zero points in A1(F2n ) in the smooth
part of the cuspidal cubic curve (which is isomorphic toA1).
For n ≥ 4, we obtain that our line arrangement C2n−1 and the line arrangement C ′2n−1 of
example (d) define isomorphic matroids. Thus C2n−1 is another construction of C ′2n−1.
Our construction of Cn generalizes a bit the one of C ′n since it holds for n = 2 and 3. The line
arrangement C3 is the trivial example, and C7 is the Fano plane (example (c)). Regarding the
line arrangement Ln : for n = 3, it is the trivial example, and L9 corresponds to example (b). We
Xavier Roulleau 595
checked that the 19 points in the dual of example (e) are not contained in a cubic curve. This
example is therefore the only one known that does not appear to be related to cubic curves.
2. Line arrangements using the cuspidal cubic
2.1. The cuspidal cubic
Let C ,→P2 be the cuspidal cubic over a field K , with a flex point O defined over K . Let C∗ be the
complement of the singular point. As for a smooth cubic curve, one can define a composition
law (denoted by +) on C∗ by chords and tangents, so that O is the neutral element.
Theorem 2 ([4, Chapter III, Proposition 2.5]). The composition law + makes C∗(K ) an abelian
group isomorphic to (K ,+). Three points a,b,c ∈ C∗(K ) (identified with elements of K ) are on a
line if and only if
a +b + c = 0 in K .
In particular the tangent to a point a in C∗ cuts C∗ at a residual point b such that 2a+b = 0. As
a concrete example, one may consider the curve C∗ :
{
x3 − z y2 = 0} ,→P2 with flex point (0 : 1 : 0)
and the parametrization t → (t : 1 : t 3).
Now, suppose that K is a finite field of order #K = q = pn , where p is the characteristic of K .
From Theorem 2, the lines containing three distinct points of K correspond bijectively to the sets
of the form
Sa,b =
{
a,b,−(a +b)}
that have cardinality 3. One has |Sa,b | = 3 if and only if
a ̸= b and a ̸= −(a +b) and b ̸= −(a +b)
which is equivalent to condition
a ̸= b and b ̸= −2a and a ̸= −2b. (1)
2.2. Characteristic 3
2.2.1. General construction
Suppose that Char(K ) = 3. Let a,b ∈ K and Sa,b =
{
a,b,−(a + b)}. Since −2 = 1 in K , the
conditions in (1) for #Sa,b = 3 reduce to a ̸= b. Therefore the number of lines containing 3
points is 13
(q
2
) = q(q−1)6 , and through each point in K , there are 12 (q − 1) such lines. This leads
to a configuration (
q 1
2 (q−1)
,
(
q(q−1)
6
)
3
)
of points and lines. Let Lq denote the dual of the q points of K .
Theorem 3. The line arrangement Lq of q lines has
q(q−1)
6 triple points and no other singularities.
Proof. The dual of the lines containing three points of K are triple points of the line arrangement
Lq . Thus Lq has
q(q−1)
6 triple points. A line arrangement with q lines can have at most
1
3
q(q−1)
2
triple points, and if that bound is reached there are no other singularities. □
We note that every point a on C∗ is a flex point: for any a one has 3a = 0. Geometrically, the
intersection of the tangent line at a with the cubic curve is three times the point a.
596 Xavier Roulleau
Remark 4. As a group, or Fp -vector space, Fq is isomorphic to (Fp )n , where q = pn . Since
the relation a +b + c = 0 is preserved under such an isomorphism, the group GLn(Fp ) induces
symmetries on the matroid Mq associated to Lq . For p = 3, since for any element t ∈ Fq , we have
(a + t )+ (b + t )+ (c + t ) = a +b + c, the relation a +b + c = 0 is preserved by translation. Thus, the
automorphism group of M3n contains the general affine group of An/F3 , which is the semi-direct
product of (Z/3Z)n with GLn(F3).
2.2.2. Examples in characteristic 3
• The line arrangement L3 consists of 3 lines passing through the same point.
• The line arrangement L9 consists of 9 lines with 12 double points. The matroid associated
to L9 is the same as the matroid associated to the Ceva(3) line arrangement, after appropriately
labeling the lines.
• Let M27 be the matroid associated with the line arrangement L27. Using [1] and OSCAR
software, one computes that the moduli space R(M27) over F3 of M27 is 3-dimensional with
a non-reduced immersed component Z of dimension 2 and multiplicity 2. A model for this
moduli space is an open sub-scheme of a rank 3 quadric inA4. The equations for this model, as
well as the associated 27 normal vectors of the line arrangements, are provided in the ancillary
file of the arXiv version of this paper.
Using the generic point of R(M27), one can check using MAGMA software that the 27 points
of the dual of the generic realization of M27 are not contained on a cuspidal cubic. However,
for the generic element of Z , (which is such that Zred is a sub-scheme of R(M27)), the 27 points
dual to a realization of M27 lie on a cuspidal cubic curve.
We now state the following result:
Proposition 5. For any q = 3n with n ≥ 3, the matroid Mq has no realizations over any field of
characteristic ̸= 3.
Proof. Using [1] and OSCAR software, one computes that the matroid M27 has no realizations
over any field of characteristic ̸= 3. The matroid Mq , q = 3n , can also be described as triples
{a,b,c} in (F3)n such that a +b + c = 0. For n ≥ 3, the group (F3)3 is a sub-group of (F3)n , and a
realization of Mq gives a realization of M27 by taking the labels indexed by that sub-group. This
forces the realization to be over a field of characteristic 3. □
2.3. Characteristic 2
2.3.1. General construction
Suppose that Char(K ) = 2 and K = Fq , where q = 2n for n > 1. Conditions in (1) for Sa,b ={
a,b,−(a +b)} to be of order 3 are then a ̸= b and a ̸= 0, b ̸= 0. Geometrically, since for any a ∈ K ,
one has 2a = 0, the tangent line trough a contains 0, and the point a with multiplicity 2, so it
contains no other points of the cuspidal cubic. This implies that for any pair a ̸= b in K ∗, the line
passing through a and b intersects the cubic at a third point c =−(a+b) ̸∈ {0, a,b}. Through each
point a ̸= 0 of K there are 12 (q−2) lines containing a and another point of K ∗. There are
(q−1)(q−2)
6
such lines, and each line contains 3 points of K ∗. This results in a(
(q −1) 1
2 (q−2)
,
(
(q−1)(q−2)
6
)
3
)
configuration of points and lines. Let Cq−1 be the dual of the q −1 points in K ∗. We obtain:
Theorem 6. The line arrangement Cq−1 of q − 1 lines has (q−1)(q−2)6 triple points and no other
singularities.
Xavier Roulleau 597
2.3.2. Examples in characteristic 2
• The line arrangement C22−1 is 3 lines passing through the same point.
• The line arrangement C23−1 is the Fano plane.
• Let N2n−1 be the matroid associated to C2n−1; this is the set of triples {a,b, a +b} ⊂ F2n with
a ̸= b, a ̸= 0, b ̸= 0. Using [1], one computes that the moduli space R(N15)/F2 is an open
subscheme in P3
/F2
, in particular it is 3-dimensional. We checked that the generic element C in
R(N15) has the property that there is a cuspidal cubic containing the 15 points that are dual to
the 15 lines of C .
For n > 3, in [3], are constructed some line arrangements C ′2n−1 (over the field F23n−1 ) with
only triple points and with the same number q −1 = 2n −1 of lines as C2n−1. The 2n −1 points
dual to C ′2n−1 are the image of the 2
n−1 points ofPn−1(F2) inPn−1 by a generic projection toP2.
In fact, one can check that the line arrangements Cq−1 and C ′q−1 define isomorphic ma-
troids as follows. As a group, the field K = Fq (for q = 2n) is the group (F2)n and the set (F2)n \{0}
is contained inPn−1(F2). Two distinct points a, b of (F2)n \{0} generate a line inPn−1 which con-
tains the points a, b, c = a +b (so that a +b +c = 0), and no other points of Pn−1(F2). As shown
in [3], a generic projection f : Pn 99K P2 induces a bijection between the set Pn−1(F2) and its
image, and f preserves the alignments. Specifically, for a,b,c ∈Pn−1(F2), one has a+b+c = 0 in
(F2)n if and only if f (a), f (b) and f (c) are on a line inP2. Thus a bijective morphism η of groups
between K and (F2)n defines an isomorphism between the matroids M(Cq ) and M(C ′q ), i.e.
sends the triples {a,b,c} of N = M(Cq ) bijectively to the triples of N ′ = M(C ′q ). If C ′q = (ℓa)a∈Fn2
is a realization of N ′, then (ℓη(a))a∈K is a realization of N .
The construction of C ′15 in [3] explains why the moduli space R(N15) is an open sub-scheme
of P3
/F2
, since the line arrangement C ′15 is obtained by the projection to P
2 from a generic point
in P3 of the 15 points of P3(F2) in P3.
Remark 7. For n ≥ 3, the projections Pn 99K P2 are parametrized by the Grassmannian G(n −
2,n + 1). For n > 4, it would be of interest to determine whether, as in the case n = 3, an open
sub-scheme of the Grassmannian also corresponds to the moduli spaces of the matroid N2n−1.
For n = 4, we have verified that the moduli space R(N31) is 6-dimensional and rational, as G(2,5).
Remark 8. The line arrangements C2n−1 are defined over F2n . This addresses the question posed
before Theorem 3.4 in [3] regarding the identification of a smaller field over which C ′q can be
defined.
2.4. Characteristic ̸= 2,3
Suppose that Char(K ) = p > 3. Then for b =−2a and b =− 12 a, Sa,b has order 2. Thus, the number
of lines containing 3 points and passing through a ∈C∗ is 12 (q −3). Furthermore, the number of
lines containing 3 distinct points of K is
1
6 q(q −3)
and the number of lines containing exactly two points is q . This leads to a(
q 1
2 (q−3)
,
(
q(q−3)
6
)
3
)
configuration of points and lines. The dual is a line arrangement of q lines and q(q−3)6 triple
points. It has
q(q−1)
2 −3
q(q−3)
6 = q
double points (corresponding to the q 2-rich lines in the dual space).
598 Xavier Roulleau
Declaration of interests
The author does not work for, advise, own shares in, or receive funds from any organization
that could benefit from this article, and has declared no affiliations other than their research
organizations.
References
[1] D. Corey, L. Kühne and B. Schröter, “Matroids in OSCAR”. To appear as a chapter in the book
The Computer Algebra System OSCAR.
[2] F. E. P. Hirzebruch, “Arrangements of lines and algebraic surfaces”, in Arithmetic and geom-
etry, Vol. II , Progress in Mathematics, Birkhäuser, 1983, pp. 113–140.
[3] L. Kühne, T. Szemberg and H. Tutaj-Gasińska, “Line arrangements with many triple points”,
Rend. Circ. Mat. Palermo (2) 73 (2024), no. 7, pp. 2501–2512.
[4] J. H. Silverman, The arithmetic of elliptic curves, Second edition, Graduate Texts in Mathe-
matics, Springer, 2009, pp. xx+513.
[5] G. Urzúa, “Some open questions about line arrangements in the projective plane”, Proyec-
ciones 41 (2022), no. 2, pp. 517–536.


The cuspidal cubic and line arrangements with only triple points

Mark M. Malamud

Vladimir V. Marchenko

On zero eigenvalue of invariant Schrödinger operators with point interactions at vertices of some regular polyhedra

Hussein Mourtada

Bernd Schober

Teissier singularities

Ricardo García López

Claude Sabbah

Comptes Rendus
Mathématique
Ricardo García López and Claude Sabbah
Corrigendum to “Hodge–Lyubeznik numbers”
Volume 363 (2025), p. 843-844
Online since: 15 July 2025
https://doi.org/10.5802/crmath.773
This article is licensed under the
Creative Commons Attribution 4.0 International License.
http://creativecommons.org/licenses/by/4.0/
C EN T R E
MER S ENN E
The Comptes Rendus. Mathématique are a member of the
Mersenne Center for open scientific publishing
www.centre-mersenne.org — e-ISSN : 1778-3569
Comptes Rendus. Mathématique
2025, Vol. 363, p. 843-844
https://doi.org/10.5802/crmath.773
Erratum / Erratum
Algebraic geometry, Singularity theory / Géométrie algébrique, Théorie des
singularités
Corrigendum to “Hodge–Lyubeznik numbers”
Corrigendum à « Nombres de Hodge–Lyubeznik »
Ricardo García López a and Claude Sabbah b
a Departament de Matemàtiques i Informàtica, Universitat de Barcelona, Gran Vía 585,
08007 Barcelona, Spain
b CMLS, CNRS, École polytechnique, Institut Polytechnique de Paris, 91128 Palaiseau cedex, France
E-mails: ricardogarcia@ub.edu, claude.sabbah@polytechnique.edu
Abstract. A Tate twist is missing in the exact sequence of Example (1), leading to incorrect formulas for some
Hodge–Lyubeznik numbers, as noticed by Bradley Dirks. We provide the correct formulas.
Résumé. Un twist de Tate est manquant dans la suite exacte de l’Exemple (1), induisant des formules
incorrectes pour certains nombres de Hodge–Lyubeznik, comme remarqué par Bradley Dirks. Nous donnons
les formules correctes.
Keywords. Hodge numbers, Lyubeznik numbers, mixed Hodge modules.
Mots-clés. Nombres de Hodge, nombres de Lyubeznik, modules de Hodge mixtes.
2020 Mathematics Subject Classification. 13D45, 14B15, 32S35.
Funding. The first author received partial support from grant PID2022-137283NB-C22 funded by
MCIN/AEI/10.13039/501100011033.
Manuscript received 4 June 2025, accepted 21 June 2025.
The exact sequence of [1, Example (1)] should read
H s−3(M) L−→ H s−1(M)(1) −→ H s{0}(X )(1) −→ H s−2(M)
L−→ H s (M)(1).
The Tate twist in the middle term was mistakenly omitted, leading to incorrect formulas for
the Hodge–Lyubeznik numbers computed by means of it. The consequences of this error were
noticed by Bradley Dirks. With the exact same reasoning, the formulas for the Hodge–Lyubeznik
numbers are:
• In Example (1):
(1) λp,q0,s (OX ,0) = h−p,−q
(
H s−1(M)prim
)
for 2 ≤ s ≤ d −1;
(2) λp,qr,d (OX ,0) = hp+d−1,q+d−1
(
H d+r−2(M)coprim
)
for 2 ≤ r ≤ d ;
(3) all other λp,qr,s (OX ,0) vanish.
• In Example (2):
λ
−p,−q
0,4 (Ai ) = hp,q
(
H 3(Mi )prim
)= hp,q (H 3(Mi ))−hp−1,q−1(H 1(Mi )).
In particular, if N1 = X3(70,16,16,14,7,6) we obtain, according to [2],
λ−1,−20,4 (A1) = h1,2
(
H 3(N1)
)= 3365330286081,
ISSN (electronic): 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/
844 Ricardo García López and Claude Sabbah
while if N2 = X3(56,49,8,6,5,4,4),
λ−1,−20,4 (A2) = h1,2
(
H 3(N2)
)= 3343868254721.
• In Remark (1) on page 220,
hp,qK (−n) = hp,q (H d{0}(X ))= hp,q (H d−1(M)prim),
and K (−n) is pure of weight d − 1. It follows that Q(n) is pure of weight 1 − d and
hp,qQ(n) = h−p,−q (H d−1(M)prim).
We take this opportunity to correct a typo: on page 218, two lines after (6), one should read “its
restriction to U isQHU [d ](n)”.
Acknowledgments
We thank Bradley Dirks for noticing erroneous formulas and the typo.
Declaration of interests
The authors do not work for, advise, own shares in, or receive funds from any organization
that could benefit from this article, and have declared no affiliations other than their research
organizations.
References
[1] R. García López and C. Sabbah, “Hodge–Lyubeznik numbers”, C. R. Math. 363 (2025),
pp. 213–221.
[2] J. Wang, Z. Yu and Y. Wang, “Examples of diffeomorphic complete intersections with differ-
ent Hodge numbers”, Topology Appl. 339 (2023), article no. 108690 (6 pages).


Corrigendum to “Hodge–Lyubeznik numbers”

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