MODULAR CUBIC SEQUENCE PATTERN AND ITS APPLICATION TO
THE DIOPHANTINE EQUATION $X^3 + Y^3 + Z^3 = K$

A PREPRINT

Eduardo J. Acuña T.

eacuna1@uc.edu.ve

Samuel J. Flores

sflores4@uc.edu.ve

Paul F. Marrero. R.

paulqed0@protonmail.com

September 10, 2022

ABSTRACT

We propose a new sequence-based algorithm to track and find solutions to the Diophantine equation $X^3 + Y^3 + Z^3 = k$, for a fixed integer $k > 0$.

# 1 Introduction

The following Diophantine equation, known as the sum of the 3 cubes, represents one of the most complex problems one can attempt to solve, in 2019 Booker. A [1] proposes an algorithm to find the number 33, later the solution for the number 42 will be discovered, but there are still numbers to be discovered.

However, although there are the following numbers in the list, it does not answer the question of how many numbers are solutions to K? for this reason by studying the patterns generated with Primordial Algebra (2021)[5] it was possible to determine the sequence in which all the cubed numbers are found once classified by the module (9).

## 1.1 Notation

* Let $n$ be any integer number

* Let $\Phi^Z$ be any of the class of any $\mathbb{Z}$, we replace "Z" to denote the value of the class.

* Let $Z_{\phi}$ be any Primal Number, which is the representative value of the class and can be operated with.

# 2 Set of cubes congruent to primordial numbers

A primordial number 2021 [5] is the representative element of a primordial class, these elements form a finite group, which when operated can generate members of the same group, the sub-group of the primordial numbers are the set of integers modulo 9, whose remainder is congruent to that primordial number, this denotes a direct relationship of behavior of a primordial number and its integer component.

Let the set $C=\{\pm 1_{\phi}, \pm 8_{\phi}, \pm 9_{\phi}\}$ where $C \subseteq \mathbb{Z}$, be the set formed by the only 6 values that a primordial number raised to the cube can take.
# 3 Sequential Functions $3n + m$

## 3.1 Definition 0

Let $f$ be the sequential function such that $f(n) = (3n + m)^3$ with $n, m$ integers and $m \in \{\pm 1, \pm 2, \pm 3\}$ from where:

$$f(n) \equiv c \pmod{9}, \text{ for any } c \in \mathbb{C}.$$

## 3.2 Definition 1: specific cases

Let $f : \mathbb{Z} \to \mathbb{Z}$

$$f(n) = \begin{cases} m^3, & \text{if } n = 0, \\ \frac{(3n + m)^3}{1 \le m \le 3} & \text{if } n > 0, \\ \frac{(3n + m)^3}{-3 \le m \le -1} & \text{if } n < 0. \end{cases}$$

## 3.3 Definition 2

For a $n > 0$ we have that $f : \mathbb{N} \to \mathbb{N}$. Therefore the sequential functions $3n + m$ have the form:

$$f(n) = (3n + 1)^3 \equiv 1 \pmod{9}$$

$$f(n) = (3n + 2)^3 \equiv 8 \pmod{9}$$

$$f(n) = (3n + 3)^3 \equiv 9 \pmod{9}$$

In the cases of functions $3n + m$ where $n < 0$, shall be characterized by $f : \mathbb{Z}^- \to \mathbb{Z}^-$ and:

$$f(n) = (3n - 1)^3 \equiv -1 \pmod{9}$$

$$f(n) = (3n - 2)^3 \equiv -8 \pmod{9}$$

$$f(n) = (3n - 3)^3 \equiv -9 \pmod{9}$$

# 4 Algorithm Modular Cubic Sequences (A.M.C.S.)

From the definition 0. 3.1, we have that $f : \mathbb{Z} \to \mathbb{Z}$ from where:

$$f(n) = (3n + m)^3.$$

**Input:** $k > 0$.

**Output:** There exists a solution $(x, y, z)$ for $x^3 + y^3 + z^3 = k$ or the message "there is no solution because you entered a $k \equiv 4,5 \pmod{9}$" if there is not.

**Step 1:** Calculate $k \pmod{9}$. And write "k belongs to the class of congruence ($k \pmod{9}$)$^\phi$"

**Step 2:** Let $\{X\} = 3n + a$; $\{Y\} = 3n + b$; $\{Z\} = 3n + d$, and $|\{X\}| \le |\{Y\}| \le |\{Z\}| \wedge \{Z\} > 0$. Continuously take $n = 0$. Then apply $n$ in $(\{X\}, \{Y\}, \{Z\})$ such that $\{X\} = 3(0) + a = a$; $\{Y\} = 3(0) + b = b$; $\{Z\} = 3(0) + d = d$ with integers $a, b, d \in \{\pm 1, \pm 2, \pm 3\}$.

**Step 3:** Raise to the third power to $(\{X\}, \{Y\}, \{Z\})$ to get $(\{X^3\}, \{Y^3\}, \{Z^3\}) = (a^3, b^3, d^3)$. Then $a^3, b^3, d^3 \in \mathbb{C}$.

**Step 4:** Call to Step 1 and apply the SAM algorithm congruent combinations for a given $k_\phi$ class so that: $a^3 + b^3 + d^3 \equiv k \pmod{9}$, with $d^3 > 0$. So now select a valid congruent combination $k_\phi$ and write "There exist a congruent solution for the Diophantine equation".
**Step 5:** Select the triad $(a^3, b^3, d^3)$ resulting from Step 4 and apply cube root on such triad i.e.,
$$ (\sqrt[3]{a^3}, \sqrt[3]{b^3}, \sqrt[3]{d^3}) = (a, b, d). \text{ Then do } \{X\} = a \wedge \{Y\} = b \wedge \{Z\} = d. $$

**Step 6:** Add $3n$ to each element of $(\{X\}, \{Y\}, \{Z\})$ to get $\{X\} = 3n+a; \{Y\} = 3n+b; \{Z\} = 3n+d$, with $(a, b, d)$ selected from (Step 4).

**Step 7:**

**Case 1.** If $a > 0$, then apply any integer $n > 0$ in $\{X\}$ and develop the sequence as follows
$$ \{X\} = x_1, x_2, x_3, \dots, x_{n-1}, x_n. \text{ Else you have an } a < 0, \text{ now apply all integer } n < 0 \text{ in } \{X\} \text{ to get:} $$
$$ \{X\} = x_1, x_2, x_3, \dots, x_{n-1}, x_n. $$

**Case 2.** If $b > 0$, then apply any integer $n > 0$ in $\{Y\}$ and develop the sequence as follows
$$ \{Y\} = y_1, y_2, y_3, \dots, y_{n-1}, y_n. \text{ Else you have an } b < 0, \text{ now apply all integer } n < 0 \text{ in } \{Y\} \text{ to get:} $$
$$ \{Y\} = y_1, y_2, y_3, \dots, y_{n-1}, y_n. $$

**Case 3.** If $d > 0$, then apply any integer $n > 0$ in $\{Z\}$ and develop the sequence as follows
$$ \{Z\} = z_1, z_2, z_3, \dots, z_{n-1}, z_n. $$

**Step 8:** Raise to the third power to $(\{X\}, \{Y\}, \{Z\})$ such that:
$$ \{X\}^3 = x_1^3, x_2^3, x_3^3, \dots, x_{n-1}^3, x_n^3; \{Y\}^3 = y_1^3, y_2^3, y_3^3, \dots, y_{n-1}^3, y_n^3; \{Z\}^3 = z_1^3, z_2^3, z_3^3, \dots, z_{n-1}^3, z_n^3. $$

Of the above successions **If** we found what for values $(x_i^3, y_i^3, z_i^3)$ with an integer $i$ such that $1 \le i \le n$ we have that $x_i^3 + y_i^3 + z_i^3 \equiv k \pmod 9$, **then** write "This is a congruent solution to $k$ modulo 9 for the Diophantine equation". **Else** write "nonexistence".

**Step 9:** If when $(\{X\}^3, \{Y\}^3, \{Z\}^3)$ is evaluated, we found values $(x_i^3, y_i^3, z_i^3)$ such that $x_i^3 + y_i^3 + z_i^3 = k$, **Then** write "There is a fixed solution for the Diophantine equation".

# 5 Apply the AMSC

Let $k = 42$ and start applying **AMSC**.

**Input:** $k = 42$. "There exists a solution $(x, y, z)$ for $x^3 + y^3 + z^3 = 42$."

1) $42 \pmod 9 = 6$ "42 belongs to the congruence class $6\phi$".

2) Let $\{X\} = 3n + a; \{Y\} = 3n + b; \{Z\} = 3n + d$. Now we take $n = 0$, so $\{X\} = a; \{Y\} = b; \{Z\} = d$, with $a, b, d \in \{\pm 1, \pm 2, \pm 3\}$.

3) Let us raise to the third power to $(\{X\}, \{Y\}, \{Z\})$. $\{X^3\} = a; \{Y^3\} = b; \{Z^3\} = d$, where each element $a^3, b^3, d^3 \in \mathbb{C}$.

4) Let us call step 4. Then by the congruent combinations of the SAM algorithm, we have that for $a^3 + b^3 + d^3 \equiv 42 \pmod{6\phi}$ (mod 9) the following combinations are available:

$$ 8\phi + (-1\phi) + (-1\phi) \equiv 6\phi \pmod 9, \quad (1) $$

$$ (-1\phi) + (8\phi) + (8\phi) \equiv 6\phi \pmod 9, \quad (2) $$

$$ (8\phi) + (8\phi) + (8\phi) \equiv 6\phi \pmod 9, \quad (3) $$

We chose the combination (2), ergo $a^3 = -1$; $b^3 = 8$; $d^3 = 8$. Triad $(-1, 8, 8)$.
5) Let us take the Last (-1, 8, 8) and apply the cubic root to each member of the triad to get:
($\sqrt[3]{-1}$, $\sqrt[3]{8}$, $\sqrt[3]{8}$) = (-1, 2, 2). It then remains for us that {$X$} = -1 $\land$ {$Y$} = 2 $\land$ {$Z$} = 2.

6) Adding 3n to each element of ({$X$}, {$Y$}, {$Z$}) we got that: {$X$} = 3n - 1; {$Y$} = 3n + 2; {$Z$} = 3n + 2.

7) Case 1. Since $a = -1$, we apply all integer $n < 0$ over {$X$} and develop the sequence as follows:

$$ \{X\} = -4, -7, -10, -13, \dots, -80538738812075974, \dots, -x_{n-1}, -x_n. $$

Case 2. Since $b = 2$, we apply all integer $n > 0$ over {$Y$} and develop the sequence as follows:

$$ \{Y\} = 5, 8, 11, 14, \dots, 80435758145817515, \dots, y_{n-1}, y_n. $$

Case 3. Similarly to case 2 we have that:

$$ \{Z\} = 5, 8, 11, 14, \dots, 12602123297335631, \dots, z_{n-1}, z_n. $$

8) Now raising to the third power to ({$X$}, {$Y$}, {$Z$}) we will have:

$$ \{X\}^3 = -64, -343, -1000, -2197, \dots, (-80538738812075974)^3, \dots, (-x_{n-1})^3, (-x_n)^3. $$

$$ \{Y\}^3 = 125, 512, 1331, 2744, \dots, (80435758145817515)^3, \dots, (y_{n-1})^3, (y_n)^3. $$

$$ \{Z\}^3 = 125, 512, 1331, 2744, \dots, (12602123297335631)^3, \dots, (z_{n-1})^3, (z_n)^3. $$

We found that if for example $x_2^3 = -343$; $y_3^3 = 1331$; $z_4^3 = 2744$, then: $x_2^3 + y_3^3 + z_4^3 = 3732$, and how $3732 \equiv 6 \pmod 9$, "This is a congruent solution to 6 modulo 9 for the Diophantine equation."

9) Finally if we add the following terms of the sequences:

$$ X_i^3 = (-80538738812075974)^3 $$

$$ Y_i^3 = (80435758145817515)^3 $$

$$ Z_i^3 = (12602123297335631)^3. $$

we can verify that $X_i^3 + Y_i^3 + Z_i^3 = 42$. Ergo "there is a fixed solution for this Diophantine equation."

## 5.1 Relation to the S.A.M. algorithm.

The AMSC algorithm and the SAM existence algorithm (2021) [3] are complementary, since with SAM for a given K we can know the primal class paths of that number, and with AMSC we can run the sequence based on the paths that SAM gives us.

## 5.2 Example

Let's take the number 33 from Booker [1].

**Input:** K=33

**Step 1:** SAM Evaluating 33 for SAM, we find that 33 is a class 6 number.

$$ 33 \equiv \Phi^6 = 6_{\phi} $$

**Step 2:** SAM will give us the following combinations of paths for the sum of the 3 cubes with class 6 results:

$$ a) (8_{\phi}) + (-1_{\phi}) + (-1_{\phi}) $$

$$ b) (-1_{\phi}) + (8_{\phi}) + (8_{\phi}) $$

$$ c) (8_{\phi}) + (8_{\phi}) + (8_{\phi}) $$

**Step 3:** AMCS Turning roads into sequences.

$$ a.s) (3 * n_x + 2)^3 + (3 * -n_y - 1)^3 + (3 * -n_z - 1)^3 $$

$$ b.s) (3 * -n_x - 1)^3 + (3 * n_y + 2)^3 + (3 * n_z + 2)^3 $$

$$ c.s) (3 * n_x + 2)^3 + (3 * n_y + 2)^3 + (3 * n_z + 2)^3 $$
As we know 33 has only one known solution so far, SAM indicates that the path to that solution is $(8_\phi) + (-1_\phi) + (-1_\phi)$, which means that the sequence is the $(3 * n_x + 2)^3 + (3 * -n_y - 1)^3 + (3 * -n_z - 1)^3$.

$$ (8_\phi) + (-1_\phi) + (-1_\phi) = (3 * n_x + 2)^3 + (3 * -n_y - 1)^3 + (3 * -n_z - 1)^3 = 33 $$

**Output:** 33 is a valid solution:

$$ 33 = (3 * 2955376325095842 + 2)^3 + (3 * -912037156269013 - 1)^3 + (3 * -2926135147620746 - 1)^3 $$

# 6 The n-core solutions.

For all integer $s \neq 0$, such that $s = 3n + m$, with $n \in \mathbb{Z}$, $m \in \{\pm 1, \pm 2, \pm 3\}$ and $s^3 \equiv m^3 \pmod 9$, its n-core solution will be:

$$ n = \frac{s-m}{3}. $$

If $n = 0$, then $s = m$.

## 6.1 Remark.

The usefulness of this method lies in the possibility that it offers us to verify the behavior, congruence and result of the sequence $3n + m$ when raised to the cube when we previously know some of the values of the triad $(x, y, z)$ that give the entire k solution for the Diophantine equation $x^3 + y^3 + z^3 = k$.

# 7 Unique results of the algorithm.

Using the AMCS Algorithm we have found a result for $k = 8$ not reported before, and although the case $x^3 + y^3 + z^3 = 8$ can be considered trivial because classical results such as $(0)^3 + (0)^3 + (2)^3 = 8$ are known, and any other result where $x = n \land y = -n \land z = 2$, with $n \in \mathbb{Z}$ can satisfy the Diophantine equation, we have found a different type of solution $(x, y, z)$ such that $x \neq y \neq z \neq 0$.

## Result.

Let $x = -12$, $y = -16$, $z = 18$ we have that: $x^3 + y^3 + z^3 = (-12)^3 + (-16)^3 + (18)^3 = 8$.

This new different solutions $(x, y, z)$ for each k integers are in agreement with the Heath-Brown conjecture about the existence of infinite solutions for all $k \neq 4, 5 \pmod 9$ (1992) [2].

# 8 Some Remarks.

1. The AMCS algorithm can find all the data reported by Huisman S. (2017). [4].

2. The sequence, $(3n + m)$ without being cubed, generates the whole set $\mathbb{Z} - \{0\}$.

3. From the previous point, since the matrix $E_9^\phi$ is composed of all integers less zero, $\mathbb{Z} - \{0\}$, we can confirm that the sequence generates all the elements of the matrix $E_9^\phi$.

4. Moving in the sequence $(3n + m)$ is the same as adding 1 to n as it is to adding 3 to m.

$$ (3(n + 1) + m)^3 = (3n + (m + 3))^3 $$

7. If we wanted to have the possibility of get integers such that $k < 0$ we would have to remove the restriction on the strict positivity of $\{Z\}$, i.e., we would have to define in step 2 normally that $|X| \leq |Y| \leq |Z|$ without $\{Z\} > 0$.

6. Also in step 7, for Case 3, we should add: Else you have an $d < 0$ in $\{Z\}$ to get: $\{Z\} = z_1, z_2, z_3, \dots, z_{n-1}, z_n$, in order to have the possibility of get an integer $k < 0$
5 Although the sequence itself is not a solution to the problem, it is an approximation to it, since any valid $(x^3, y^3, z^3)$ triad for a given $k$ will belong to the sequence.

Bibliography

[1] A. R. Booker. Cracking the problem with 33. *Research in Number Theory*, 5(3):1–6, 2019.

[2] A. R. Booker and A. V. Sutherland. On a question of mordell. *Proceedings of the National Academy of Sciences*, 118(11):e2022377118, 2021.

[3] S. Flores, E. Acuña, and P. Marrero. Existential refinement on the search of integer solutions for the diophantine equation $x^3 + y^3 + z^3 = n$, 2021. URL https://arxiv.org/abs/2103.17037.

[4] S. G. Huisman. Newer sums of three cubes, 2016. URL https://arxiv.org/abs/1604.07746.

[5] E. J. A. Tarazona, S. Flores, and P. F. Marrero. Matrize_φ^9. clasificación de números enteros. *algebra primordial*. *Revista MATUA* ISSN: 2389-7422, 8(1):10-45, 2021.