Puzzle—Solving the <i>n</i>-Fractions Puzzle as a Constraint Programming Problem

Malapert, Arnaud; Provillard, Julien

doi:10.1287/ited.2017.0193

Cited by 3 publications

(7 citation statements)

References 2 publications

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“…Euler's factorisation method is suited to semi-primes whose construction are prime factors that are said to be Pythagorean [21,22] and can further be improved upon by considering the parity of the squares. The limitations of this method pertain only to semi-prime constructs that are Pythagorean primes (p = 1 mod 4).…”

Section: Theory and Proposed Methodsmentioning

confidence: 99%

A Fast Factorisation of Semi-Primes Using Sum of Squares

Overmars¹,

Venkatraman²

2019

MCA

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For several centuries, prime factorisation of large numbers has drawn much attention due its practical applications and the associated challenges. In computing applications, encryption algorithms such as the Rivest–Shamir–Adleman (RSA) cryptosystems are widely used for information security, where the keys (public and private) of the encryption code are represented using large prime factors. Since prime factorisation of large numbers is extremely hard, RSA cryptosystems take advantage of this property to ensure information security. A semi-prime being, a product of two prime numbers, has wide applications in RSA algorithms and pseudo number generators. In this paper, we consider a semi-prime number whose construction consists of primes, N = p 1 p 2 , being Pythagorean and having a representation on the Cartesian plane such that, p = x 2 + y 2 . We prove that the product of two such primes can be represented as the sum of four squares, and further, that the sums of two squares can be derived. For such a semi-prime, if the original construction is unknown and the sum of four squares is known, by Euler’s factorisation the original construction p 1 p 2 can be found. By considering the parity of each of the squares, we propose a new method of factorisation of semi-primes. Our factorisation method provides a faster alternative to Euler’s method by exploiting the relationship between the four squares. The correctness of the new factorisation method is established with mathematical proofs and its practical value is demonstrated by generating RSA-768 efficiently.

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Section: Theory and Proposed Methodsmentioning

confidence: 99%

A Fast Factorisation of Semi-Primes Using Sum of Squares

Overmars¹,

Venkatraman²

2019

MCA

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“…(4, 6,10,15) = 15, 2(5, 2, 3) ⇒ (5, 2), (2, 3) ⇒ 2 2 + 5 2 2 2 + 3 2 = (29)(13) = 377 (7) This requires the sums of four squares to be found until the correct sum of four squares is factored. This is not viable for the factorization of large semi-primes.…”

Section: Case Example 1 Let Us Considermentioning

confidence: 99%

“…The current perception is that there appears to be only a limited understanding of their underlying structure, and several mathematicians are constantly trying to uncover the mysteries behind these prime numbers. There is still much to be carried out, and areas of further interest are channeled towards a better understanding of the structure of primes for arriving at faster prime number generating algorithms and faster solutions to the prime factorization problem [4][5][6][7]. There is a need for generating more robust primes that are less susceptible to known factorization methods.…”

Section: Introductionmentioning

confidence: 99%

New Semi-Prime Factorization and Application in Large RSA Key Attacks

Overmars¹,

Venkatraman²

2021

JCP

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Semi-prime factorization is an increasingly important number theoretic problem, since it is computationally intractable. Further, this property has been applied in public-key cryptography, such as the Rivest–Shamir–Adleman (RSA) encryption systems for secure digital communications. Hence, alternate approaches to solve the semi-prime factorization problem are proposed. Recently, Pythagorean tuples to factor semi-primes have been explored to consider Fermat’s Christmas theorem, with the two squares having opposite parity. This paper is motivated by the property that the integer separating these two squares being odd reduces the search for semi-prime factorization by half. In this paper, we prove that if a Pythagorean quadruple is known and one of its squares represents a Pythagorean triple, then the semi-prime is factorized. The problem of semi-prime factorization is reduced to the problem of finding only one such sum of three squares to factorize a semi-prime. We modify the Lebesgue identity as the sum of four squares to obtain four sums of three squares. These are then expressed as four Pythagorean quadruples. The Brahmagupta–Fibonacci identity reduces these four Pythagorean quadruples to two Pythagorean triples. The greatest common divisors of the sides contained therein are the factors of the semi-prime. We then prove that to factor a semi-prime, it is sufficient that only one of these Pythagorean quadruples be known. We provide the algorithm of our proposed semi-prime factorization method, highlighting its complexity and comparative advantage of the solution space with Fermat’s method. Our algorithm has the advantage when the factors of a semi-prime are congruent to 1 modulus 4. Illustrations of our method for real-world applications, such as factorization of the 768-bit number RSA-768, are established. Further, the computational viabilities, despite the mathematical constraints and the unexplored properties, are suggested as opportunities for future research.

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“…Since each fraction is at least 1/99, this family of problems has solutions for at most n ≤ 99. Malapert and Provillard prove in a recent paper [2] that the puzzle has no solution for n ≥ 45. Two models are described in the literature (see [2]) to solve the n-fractions puzzle.…”

Section: Introductionmentioning

confidence: 99%

“…Malapert and Provillard prove in a recent paper [2] that the puzzle has no solution for n ≥ 45. Two models are described in the literature (see [2]) to solve the n-fractions puzzle. The division model handles Equation (1) with floating point arithmetic.…”

Section: Introductionmentioning

confidence: 99%