Hyperbolic Primality Test and Catalan–Mersenne Number Conjecture

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Hyperbolic Primality Test and Catalan–Mersenne Number Conjecture

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Youngik Lee^*

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Submitted:

10 May 2024

Posted:

14 May 2024

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Abstract

We develop a primality test for Mersenne numbers and investigate the primality of the sequence of Catalan-Mersenne numbers. We use induction as a proof to show that the 5th Catalan-Mersenne number is a prime number, thus demonstrating the infinitude of Mersenne primes.

Keywords:

Subject: Computer Science and Mathematics - Algebra and Number Theory

1. Introduction

Many primality tests have been developed, such as the Lucas–Lehmer primality [1] test, Proth’s test [2], etc., which are useful prime test to specific use cases. In this note, we develop a primality test specified for Mersenne numbers. One of the interesting prime numbers is the Mersenne prime,

M_{p}

, which has the form

2^{p} - 1

Lemma 1

(Hyperbolic Primality Test). If there exists a non-trivial x satisfying:

2^{p - 1} + x \equiv 0 mod (2 x + 1)

(1)

Then

2^{p} - 1

is prime. Here, the trivial solutions are

x = 0

and

2^{p - 1} - 1

Proof.

When p is prime, for an arbitrary Mersenne number

M_{p}

M_{p} = 2^{p} - 1 = (\sum_{i = 1}^{k} a_{i} 2^{i} + 1) (\sum_{i = 1}^{q} b_{i} 2^{i} + 1)

(2)

And when

M_{p}

is prime, we call it a Mersenne prime. [3] When all

a_{i}

(or

b_{i}

) are zero, then

M_{p}

is prime.

\begin{matrix} 2^{p} - 1 & = & 2 (2^{p - 1} - 1) + 1 \\ = & (2 x + 1) (2 y + 1) \\ = & 2 (2 x y + x + y) + 1 \end{matrix}

(3)

Therefore we can summarize the result as:

2^{p - 1} - 1 = 2 x y + x + y

(4)

This can be rewritten as:

2^{p - 1} = x y + (x + 1) (y + 1)

(5)

which represents a hyperbola. So, if there exists any integer lattice point on the hyperbola,

M_{p}

is not prime. Thus, determining the primality of

M_{p}

is equivalent to checking for lattice points on the hyperbola.

In modulo forms, we can rewrite it as:

2^{p - 1} + x \equiv 0 mod (2 x + 1)

(6)

For the computational test, the range of x is:

1 \leq x < ⌊\frac{\sqrt{2^{p} - 1} + 1}{2}⌋ = ⌊2^{\frac{p}{2} - 1}⌋

(7)

□

2. Theorems and Lemmas

One interesting problem involves the primality test for Catalan-Mersenne numbers. The Catalan-Mersenne number is a recursively defined sequence [4]:

c_{0} = 2, c_{n + 1} = 2^{c_{n}} - 1 = M_{c_{n}}

(8)

Therefore, the Catalan-Mersenne numbers are a subset of the double Mersenne numbers, which are numbers of the form:

M_{M_{n}} = 2^{2^{n} - 1} - 1

(9)

And we can choose a specific sequence where:

a_{1} = 2^{2} - 1, a_{n + 1} = 2^{a_{n}} - 1

(10)

a_{n} = M_{a_{n - 1}}

is prime, then by Lemma 1, there is no non-trivial x, other than

x = 0

A_{n - 1} - 1

, that satisfies:

2^{a_{n - 1} - 1} + x = A_{n - 1} + x \equiv 0 mod (2 x + 1)

(11)

Then in the case of

a_{n + 1} = M_{a_{n}}

, let us try to find a non-trivial x that satisfies:

2^{2^{a_{n - 1}} - 2} + x \equiv 0 mod (2 x + 1)

(12)

This equation can be rewritten as:

4^{2^{a_{n - 1} - 1} - 1} + x = 4^{A_{n - 1} - 1} + x \equiv 0 mod (2 x + 1)

(13)

However, for some

k_{n}

, we can rewrite

4^{A_{n - 1}} - 1 = A_{n - 1}^{k_{n}}

k_{n} = \frac{A_{n - 1} - 1}{{log}_{4} A_{n - 1}}, \prod_{i = 2}^{n} k_{i} = A_{n - 1} - 1

(14)

Thus, all

k_{n}

are integers, and Equation (15) holds for the primality test of

a_{n}

k_{n} = \frac{A_{n - 1} - 1}{A_{n - 2} - 1} = \sum_{i = 0}^{k_{n - 1} - 1} 4^{(A_{n - 2}) i}

(15)

Equation (13) becomes as:

A_{n - 1}^{k_{n}} + x \equiv 0 mod (2 x + 1)

(16)

Theorem 1.

When

CM

is the set of all Catalan-Mersenne numbers, if

a_{n} \in CM

is a double Mersenne prime, then

M_{a_{n}}

is also a prime number.

Proof.

A double Mersenne prime, denoted as X, is a prime number that satisfies

X = M_{M_{p}}

, where

M_{p}

is also prime. According to Equation (11),

A_{n} = 2^{a_{n} - 1}

(1): When $n = 2$ , we can perform the hyperbolic primality test:

$A_{1} + x \equiv 0 mod (2 x + 1)$

which yields:

$4 + x \equiv 0 mod (2 x + 1)$

There are no non-trivial solutions for x.

Thus, $a_{2}$ is a double Mersenne prime.
(2): When n = 3, we can perform the hyperbolic primality test:

$A_{2} + x \equiv 0 mod (2 x + 1)$

which yields:

$64 + x \equiv 0 mod (2 x + 1)$

There are no non-trivial solutions for x.

Thus, $a_{3}$ is a double Mersenne prime.

For arbitrary x, we can express:

A_{n - 1} = O (2 x + 1) + a x + b

(17)

where

a = 0

or 1.

Then let us use induction for the general case of

n \geq 3

. Let us assume that

a_{n}

(defined in Equation (10)) is a double Mersenne prime. According to Lemma 1 and Equation (17), there is no non-trivial x that satisfies:

A_{n - 1} \equiv a x + b \equiv x + 1 mod (2 x + 1)

(18)

This implies that

(a, b) \neq (1, 1)

because

x \neq 0

Next, let us show that

a_{n + 1}

is prime.

A_{n - 1}^{k} = O (2 x + 1) + {(a x + b)}^{k}

(19)

We can use Equation (16) for the primality test of

a_{n + 1}

, with the result of Equation (18) to check whether there is a non-trivial x that satisfies:

A_{n - 1}^{k} \equiv {(a x + b)}^{k} \equiv x + 1 mod (2 x + 1)

(20)

To make Equation (20) hold, b should be 1. However, Equation (18) shows that

(a, b) \neq (1, 1)

, so the available value of a is 0.

A_{n - 1}^{k} \equiv 1 \equiv x + 1 mod (2 x + 1)

(21)

So, we have:

0 \equiv x mod (2 x + 1)

(22)

This implies that x should be 0, but this is not allowed. Hence, there is no non-trivial x that satisfies Equation (20). Therefore,

a_{n + 1}

is prime. Consequently, if

a_{n - 1}

is a double Mersenne prime, then

a_{n}

is prime. □

Theorem 2.

There are infinitely many Mersenne primes.

Proof.

Theorem 1 shows that if there is one double Mersenne prime, it generates an infinite series of Mersenne primes. Since

M_{7}

is a double Mersenne prime, there are infinitely many Mersenne primes. □

3. Application

In this section, we investigate the generation of Catalan-Mersenne-like sequences using different initial values and examine the applicability of the recursive hyperbolic primality test.

3.1. $a_{1} = 3$ Case

For

a_{1} = 3

Catalan-Mersenne number sequence, we can calculate

k_{2}

k_{5}

, such as

k_{2} = \frac{4^{2 - 1} - 1}{2 - 1} = 3, k_{3} = \frac{4^{4 - 1} - 1}{4 - 1} = 21

k_{4} = \frac{4^{63} - 1}{4^{3} - 1} = 1350326852860866918505454791395905601

k_{5} = \frac{A_{4} - 1}{{log}_{4} A_{4}} = \frac{4^{4^{63} - 1} - 1}{4^{63} - 1} = \sum_{i = 0}^{k_{4} - 1} 4^{63 i}

Until

a_{4}

it is showed that it is a prime by Lucas–Lehmer primality test. Based on Theorem. 1,

a_{5}

is also prime.

a_{5} = 2^{2^{127} - 1} - 1 = M_{2^{127} - 1}

(23)

which has

\approx 5.12 \times 10^{37}

digits.

3.2. $a_{1} = 2^{11} - 1$ Case

For the

a_{1} = 2^{11} - 1

Catalan-Mersene number sequence,

k_{2} = \frac{2^{2^{11} - 2} - 1}{2^{10} - 1} = (n o n i n t e g e r)

And

a_{1}

is also not prime. So there is no infinite numbers of primes in this sequence.

3.3. $a_{1} = 2^{13} - 1$ Case

For the

a_{1} = 2^{13} - 1

Catalan-Mersene number sequence,

k_{2} = \frac{2^{2^{13} - 2} - 1}{2^{12} - 1} = (n o n i n t e g e r)

And

a_{1}

is a prime, but

a_{2}

is not a prime. So there is no infinite numbers of primes in this sequence.

4. Conclusions and Outlook

In this research, we introduced a hyperbolic primality test specified for Mersenne numbers and demonstrated that

a_{5}

is prime. Another intriguing observation is that there is an infinite number of primes that can be generated from the

a_{1} = 3

Catalan-Mersenne sequence. However, it’s noteworthy that other infinite prime sequences with different initial values are not readily apparent.

For the next step, we can investigate the general relations between lattice points on the hyperbola and the initial values of Catalan-Mersenne numbers.

References

John H. Jaroma, “Note on the Lucas–Lehmer Test", Irish Mathematical Society (2004), https://www.irishmathsoc.org/bull54/M5402.pdf.
OEIS, “Proth primes", https://oeis.org/A080076.
OEIS, “Mersenne primes", https://oeis.org/A000668.
OEIS, “Catalan-Mersenne numbers", https://oeis.org/A007013.

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Hyperbolic Primality Test and Catalan–Mersenne Number Conjecture

Abstract

1. Introduction

2. Theorems and Lemmas

3. Application

3.1. a 1 = 3 Case

3.2. a 1 = 2 11 − 1 Case

3.3. a 1 = 2 13 − 1 Case

4. Conclusions and Outlook

References

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3.1. $a_{1} = 3$ Case

3.2. $a_{1} = 2^{11} - 1$ Case

3.3. $a_{1} = 2^{13} - 1$ Case