Ann. Acad. Rom. Sci.

Ser. Math. Appl.

ISSN 2066-6594

Vol. 18, No. 2/2026

EUCLIDEAN-LAGRANGE AND

CANTOR-LAGRANGE QUARTIC

POLYNOMIALS AND ASSOCIATED CUBIC

CURVES^∗

Mircea Crasmareanu^†

Dedicated to Professor Gheorghe Moro¸sanu on the occasion of his 75th

anniversary

Communicated by S. Trean¸t˘a

DOI

10.56082/annalsarscimath.2026.2.103

Abstract

The purpose of this paper is to introduce and examine two classes of

quartic real polynomials P having the same Euclidean norm as their

Lagrange resolvent, respectively, having the square of the Euclidean

norm equal to the height of the Lagrange resolvent. The reduced form

of the polynomial P is provided, which eliminates the cubic term. Find-

ing such polynomials with integer coeﬃcients is always of interest. The

Lagrange resolvent associates a cubic curve with each of these polyno-

mials as a cubic polynomial. It highlights the situations in which these

cubic curves are elliptic curves.

Keywords: quartic real polynomial, Lagrange resolvent, Euclidean-

Lagrange polynomial, cubic curve, Cantor-Lagrange polynomial.

MSC: 12D05, 51N35, 65H04.

^∗Accepted for publication on November 30, 2025

^†mcrasm@uaic.ro, Faculty of Mathematics, Alexandru Ioan Cuza University, Iasi,

700506, Romania

103

Euclidean quartic polynomials and associated cubic curves

104

1 Introduction

Five techniques to solve a quartic equation are identiﬁed by the names

Descartes-Euler-Cardano, Ferrari-Lagrange, Neumark, Christianson-Brown,

and Yacoub-Fraidenraich-Brown. In this study, we concentrate on the lat-

ter, which relates a speciﬁc monic (and reduced) quartic polynomial P to

its Lagrange resolvent LR(P) represented as a monic cubic polynomial

This note begins by treating both polynomials P and LR(P) as vectors

in R⁴and R³, respectively, and consequently connecting their Euclidean

norms k · k. It pertains to a speciﬁc category of P, speciﬁcally those for

which kP_dk = kLR(P_d)k, and this collection is the focus of our investiga-

tion referred to as Euclidean-Lagrange polynomials. Here P_drepresents the

depressed version of the original quartic polynomial P.

The key condition for the isometry of the Lagrange map P → LR(P) is

formulated as a quadratic equation regarding the coeﬃcients (p, q, r) of P.

Since this equation resembles a Pythagorean theorem, we solve it entirely

using two real parameters α = 0, β = 0. The speciﬁc case α = β is fully

addressed. Another Lagrange-Euclidean polynomial is obtained as ﬁxed

point of the Lagrange map above expressed only in terms of coeﬃcients.

We also link a cubic curve LC(P) : y²= LR(P)(x) and concentrate on

situations where this curve is an elliptic curve. Indeed, we found all of them

on the renowned database https://www.lmfdb.org/ where many speciﬁcs

can be accessed.

In the ﬁnal section, we present a new category of polynomials P, speciﬁ-

2

cally those satisfying kP_dk = h(LR(P_d)), where h denotes the Cantor height

of the cubic polynomial LR(P_d). Three instances of these polynomials, re-

ferred to as Cantor-Lagrange, are examined.

2 Euclidean-Lagrange quartic polynomials

Fix a natural number n ∈ N^∗. The setting of this work is provided by the

n-dimensional real linear space of monic polynomials of grade n:

R^m_n^onic[x] := {P(x) = xⁿ+ a₁xⁿ⁻¹+ ... + a_n; a₁, ..., a_n∈ R}.

It is an Euclidean space with respect to the usual scalar product of Rⁿ

and hence the square of the induced norm is the sum of the square of the

coeﬃcients:

2

kPk := a²₁+ ... + a²_n.

(1)

M. Crasmareanu

105

Remark 1. It is well-known that in order to prove that the set of real

algebraic numbers is countable Cantor deﬁnes the height of above P as the

positive (real) number:

h(P) := n + |a₀= 1| + |a₁| + ... + |a_n|.

It follows the inequality:

2

(h(P) − n − 1)²≤ nkPk

xⁿ⁺¹−1

with equality only for the polynomial Φ_n(x) =

= xⁿ+xⁿ⁻¹+...+x+1.

x−1

Our study focuses on a ﬁxed quartic polynomial P ∈ R^m₄^onic[x]:

P(x) := x⁴+ Ax³+ Bx²+ Cx + D

A

4

which with a Cardano-type transformation x = y −

have the reduced form

(here the subscript d means depressed):

P_d(y) := y⁴+ py²+ qy + r.

(2)

The Lagrange resolvent of P is a cubic polynomial LR(P) constructed

with the coeﬃcients of P_d; for more details see [1, p. 322] while for an

universal method to solve the quartic equation see [5]:

LR(P)(u) := u³− pu²− 4ru + (4pr − q²) = 0.

(3)

Inspired by [2] we introduce:

Deﬁnition 1. i) P (or equivalently P_d) is an Euclidean-Lagrange polyno-

mial if it preserves the Euclidean norm with respect to Lagrange transfor-

mation DR:

kP_dk = kLR(P)k.

(4)

Hence, the restriction of LR to the set of Euclidean-Lagrange polynomials

is an isometry.

ii) The Lagrange-cubic curve associated to (arbitrary) P_dis:

LC(P_d) : y²= LR(P)(x).

We have immediately:

(5)

Euclidean quartic polynomials and associated cubic curves

106

Proposition 1. i) The only Euclidean-Lagrange polynomials with r = 0 are

P_d⁰(y) = y⁴and the two 1-parameter families P^±(y) = y⁴+ py²± y, p ∈ R.

d

ii) Let P with r = 0. Then P is an Euclidean-Lagrange polynomial if and

only if there exists the non-zero real numbers α ≤ β such that:

√

2αβ

15

q = α²+ β²> 0,

r =

,

p =

[(α²+ β²)²+ β²− α²].

(6)

√

8αβ

15

Proof. The equality (4) reads as a Pythagorean relation:

q²= 15r²+ (4pr − q²)²

(7)

hence, we use the well-known parametrization of Pythagorean triples ( [4]);

there exists α ≤ β such that:

√

15r = 2αβ,

4pr − q²= β²− α²,

q = α²+ β²

(8)

and the claimed relations follow immediately.

Example 1. The Lagrange-cubic curves associated to the polynomials from

i) are:

LC(P⁰) : y²= x³,

LC(P^±) : y²= x³− 1 = (x − 1)(x²+ x + 1)

(9)

which are the semicubical parabola and the elliptic curve given by

https://www.lmfdb.org/EllipticCurve/Q/144/a/3. This elliptic curve has

complex multiplication.

Example 2. We choose α = β = 0 in (8) obtaining:

s





2

s

!

√

15

2

15

2

P_d^α(y) = y⁴+ α²

y + 2y +

= y⁴+ α²

y +

,



√

2

15

(10)

!

√

15

8

481

60

2

LC(P^α) : y²= x³− α²

x −

x

, kP_d^αk = kLR(P^α)k =

α².

√

2

15

(11)

In order to obtain a cubic curve with integral coeﬃcients in the right-hand-

q

√

5!

2

side we choose α²= 2 15 =

and therefore:

ꢀ

ꢁ

√

LC P^±

: y²= x³− 15x²− 16x = x(x + 1)(x − 6).

(12)

2

15

M. Crasmareanu

107

This elliptic curve is https://www.lmfdb.org/EllipticCurve/Q/27864/c/1 and

its Weierstrass expression is:

ꢀ

ꢁ

√

2

15

LC P^±

: y²= X³− 75X − 266.

(13)

√

ꢂ

ꢃ

√

2

15

The cubic polynomial LC P^±

having real distinct roots is a strictly

hyperbolic polynomial according to a well-known terminology, see [2] or [3].

3 The ﬁxed points of the Lagrange map

A second way to ﬁnd Euclidean-Lagrange polynomials is to ﬁnd the ﬁxed

points of the Lagrange map:

LR : (p, q, r) ∈ R³→ (−p, −4r, 4pr − q²) ∈ R³.

(14)

Also, the Lagrange map is a surjective map on its image {(A, B, C) ∈

R³; AB − C ≥ 0} with:

ꢀ

ꢁ

√

B

LR⁻¹(A, B, C) = −A, ± AB − C, −

.

(15)

4

A direct computation yields:

Proposition 2. The Lagrange map (1) has only two ﬁxed points: (0, 0, 0)

ꢄ

ꢅ

1

and 0, ¹₄, −₁₆

.

This result gives a new Euclidean-Lagrange polynomial:

√

y

P_d^fixed(y) = y⁴+ −

4

1

17

u

1

,

kP_d^fixedk =

,

LR(P^fixed)(u) = u³+

−

16

4

16

(16)

and the only real (hence strictly positive) root of LR(P^fixed) is:

p

√

3

9 + 129

1

q

u =

−

' 0.21193.

(17)

√

2 × 6^2/3

3

6(9 + 129)

The Euclidean-Lagrange polynomials P_d^fixedhas only two real roots, one

strictly negative and one strictly positive and the associated cubic curve is:

x

1

LC(P^fixed) : y²= x³+

−

.

(18)

4

16

Euclidean quartic polynomials and associated cubic curves

108

Recall that for a real number α the α-it homothetical transformation of the

Weierstrass cubical curve Γ : y²= x³+ Px + Q is the cubical curve:

H_α(Γ) : y²= x³+ (α²P)x + (α³Q).

Hence, choosing α = 4 for (5) we obtain the elliptical curve:

H₄(LC(P^fixed)) : y²= x³+ 4x − 4

(19)

(20)

which is exactly the curve https://www.lmfdb.org/EllipticCurve/Q/688/a/1.

It is worth to point out that, excepting P_d(y) = y⁴, all previous obtained

polynomials have distinct roots. Indeed P_d(y) = y⁴is the only Euclidean-

Lagrangian polynomial with multiple roots since the equality P_d(y) = (y²+

ay + b)²means: a = 0 = q, p = 2b, r = b². From the characterization (7)

we have also p = r = 0.

4 Cantor-Lagrange quartic polynomials

We introduce now a new class of quartic polynomial in the same setting

above but now considering the Cantor height.

Deﬁnition 2. i) P (or equivalently P_d) is a Cantor-Lagrange polynomial

if:

2

kP_dk = h(LR(P))

(21)

which means:

p²+ q²+ r²= 4 + |p| + 4|r| + |4pr − q²|.

(22)

We analyze some interesting cases according to the pair (p, r) of coeﬃ-

cients.

Case I) Suppose that r = 0; it_√follows the characterization p²= 4 + |p|

and we have two solutions p_±

=

^17±1. Hence we have two 1-parameters

2

families P_d^(I,±,q)(y) = y⁴+ p_±y²+ qy.

Case II) p = 0 implies the characterization r²= 4+4|r| with the solutions

√

r_±= ±2( 2 + 1). In conclusion, we have again the 1-parameter families of

Cantor-Lagrange polynomials P_d^(II,±,q)(y) = y⁴+ qy + r_±.

Case III) 4pr < q², p > 0 and r > 0 give the following quadratic equation,

representing a hyperbola in the plane (p, r):

H : p²+ 4pr + r²− p − 4r − 4 = 0.

(23)

M. Crasmareanu

109

The eccentricity of H is e = 2 and H contains an inﬁnite sequence of lattice

points e.g. (p, r) = (1, 2). So, an example of this case is:

ꢆ

P_d(y) = y⁴+ y²+ 3y + 2,

LR(P)(u) = u³− u²− 8u − 1,

(24)

2

kP_dk = h(LR(P)) = 14.

References

[1] D.A. Cox, Galois Theory, Wiley, 2004.

[2] M. Crasmareanu, The diagonalization map as submersion, the cubic

equation as immersion and Euclidean polynomials, Mediterr. J. Math.

19 (2022), 65.

[3] M. Crasmareanu, Hyperbolic and weak Euclidean polynomials from

Wronskian and Leibniz maps, Axioms 13 (2024), 104.

[4] M. Crasmareanu, The Farey sum of Pythagorean and Eisenstein triples,

Math. Sci. Appl. E-Notes 12 (2024), 28-36.

[5] S.L. Shmakov, A universal method of solving quartic equations, Int. J.

Pure Appl. Math. 71 (2011), 251-259.