The document describes research into primality tests for specific classes of numbers. It presents 6 conjectures relating to primality tests for numbers of the form N = k*b^n - c or N = k*b^n + c, where b, k, n, and c meet certain criteria. It then attempts to prove the conjectures by relating them to Lucas sequences. The proof analyzes how the Lucas sequences behave modulo N depending on whether N is congruent to 1 or 3 (mod 8). It shows that the conjectures are true based on the behavior of the Lucas sequences. The document also presents some basic lemmas and theorems relating to Lucas sequences and defines the sequences to be used.
(Neamen)solution manual for semiconductor physics and devices 3edKadu Brito
This document contains solutions to problems from Chapter 1 of Semiconductor Physics and Devices: Basic Principles, 3rd edition. The problems calculate properties of semiconductor unit cells such as number of atoms, density, and volume percentage occupied by atoms. Lattice structures including simple cubic, body-centered cubic, face-centered cubic and diamond are considered. Properties of common semiconductors such as silicon and gallium arsenide are also calculated.
Kittel c. introduction to solid state physics 8 th edition - solution manualamnahnura
1. The document discusses crystallographic planes and directions in a cube, the Miller indices of planes with respect to primitive axes, and the spacing between dots projected onto different planes of a crystal structure.
2. Key concepts from crystallography such as Miller indices, primitive lattice vectors, reciprocal lattice vectors, and the first Brillouin zone are defined. Calculations of interplanar spacing and lattice parameters are shown for simple cubic and face-centered cubic lattices.
3. Binding energies, cohesive energies, and equilibrium properties are calculated and compared for body-centered cubic and face-centered cubic crystal structures. Approximations made in describing crystal binding using Madelung energies and pair potentials are
The document contains 18 multi-part questions from an exam for the Brazilian Naval Academy in 2015. The questions cover topics such as calculus, geometry, trigonometry, complex numbers, and systems of equations.
This document provides model solutions to questions from the JEE Advanced 2013 exam. It includes solutions to multiple choice and numerical questions across physics, chemistry and mathematics. Some key details include:
- Solutions to 20 multiple choice questions from Paper I Section I and II of the exam.
- Solutions involving concepts like circular motion, momentum, electrostatics, thermodynamics and more.
- Solutions to 4 numerical questions involving calculations related to decay of particles, fundamental frequency of a vibrating string, and more.
- Solutions to 20 multiple choice questions from Paper II Section I, II and III covering topics in chemistry, organic chemistry, coordination chemistry and biochemistry.
- Solutions to 4 numerical questions related
This document provides a summary of common derivatives and integrals encountered in calculus. It lists basic derivative rules and formulas for derivatives of polynomials, trigonometric functions, inverse trigonometric functions, exponential and logarithmic functions, and other common functions. It also provides the formulas and properties for basic integrals as well as techniques for evaluating more complicated integrals using substitution, integration by parts, trigonometric substitutions, partial fractions, and other standard methods.
The document provides solutions to problems from an IIT-JEE 2004 mathematics exam. Problem 1 asks the student to find the center and radius of a circle defined by a complex number relation. The solution shows that the center is the midpoint of points dividing the join of the constants in the ratio k:1, and gives the radius. Problem 2 asks the student to prove an inequality relating dot products of four vectors satisfying certain conditions. The solution shows that the vectors must be parallel or antiparallel.
This document provides a sample question paper for Class XII Mathematics. It consists of 3 sections - Section A has 10 one-mark questions, Section B has 12 four-mark questions, and Section C has 7 six-mark questions. The paper is for 3 hours and carries a total of 100 marks. Some questions provide internal choices. Calculators are not permitted. Sample questions include finding inverse functions, evaluating integrals, solving differential equations, and probability questions.
The document discusses expanding and factorizing algebraic expressions. It provides examples of expanding expressions using the distributive property, such as expanding (a + b)(c + d) to get ac + ad + bc + bd. It also discusses factorizing expressions by finding common factors, such as factorizing a2 + 2ab + b2 to get (a + b)2. Tips and techniques are presented for expanding, factorizing, finding common factors, and using the distributive property to manipulate algebraic expressions.
(Neamen)solution manual for semiconductor physics and devices 3edKadu Brito
This document contains solutions to problems from Chapter 1 of Semiconductor Physics and Devices: Basic Principles, 3rd edition. The problems calculate properties of semiconductor unit cells such as number of atoms, density, and volume percentage occupied by atoms. Lattice structures including simple cubic, body-centered cubic, face-centered cubic and diamond are considered. Properties of common semiconductors such as silicon and gallium arsenide are also calculated.
Kittel c. introduction to solid state physics 8 th edition - solution manualamnahnura
1. The document discusses crystallographic planes and directions in a cube, the Miller indices of planes with respect to primitive axes, and the spacing between dots projected onto different planes of a crystal structure.
2. Key concepts from crystallography such as Miller indices, primitive lattice vectors, reciprocal lattice vectors, and the first Brillouin zone are defined. Calculations of interplanar spacing and lattice parameters are shown for simple cubic and face-centered cubic lattices.
3. Binding energies, cohesive energies, and equilibrium properties are calculated and compared for body-centered cubic and face-centered cubic crystal structures. Approximations made in describing crystal binding using Madelung energies and pair potentials are
The document contains 18 multi-part questions from an exam for the Brazilian Naval Academy in 2015. The questions cover topics such as calculus, geometry, trigonometry, complex numbers, and systems of equations.
This document provides model solutions to questions from the JEE Advanced 2013 exam. It includes solutions to multiple choice and numerical questions across physics, chemistry and mathematics. Some key details include:
- Solutions to 20 multiple choice questions from Paper I Section I and II of the exam.
- Solutions involving concepts like circular motion, momentum, electrostatics, thermodynamics and more.
- Solutions to 4 numerical questions involving calculations related to decay of particles, fundamental frequency of a vibrating string, and more.
- Solutions to 20 multiple choice questions from Paper II Section I, II and III covering topics in chemistry, organic chemistry, coordination chemistry and biochemistry.
- Solutions to 4 numerical questions related
This document provides a summary of common derivatives and integrals encountered in calculus. It lists basic derivative rules and formulas for derivatives of polynomials, trigonometric functions, inverse trigonometric functions, exponential and logarithmic functions, and other common functions. It also provides the formulas and properties for basic integrals as well as techniques for evaluating more complicated integrals using substitution, integration by parts, trigonometric substitutions, partial fractions, and other standard methods.
The document provides solutions to problems from an IIT-JEE 2004 mathematics exam. Problem 1 asks the student to find the center and radius of a circle defined by a complex number relation. The solution shows that the center is the midpoint of points dividing the join of the constants in the ratio k:1, and gives the radius. Problem 2 asks the student to prove an inequality relating dot products of four vectors satisfying certain conditions. The solution shows that the vectors must be parallel or antiparallel.
This document provides a sample question paper for Class XII Mathematics. It consists of 3 sections - Section A has 10 one-mark questions, Section B has 12 four-mark questions, and Section C has 7 six-mark questions. The paper is for 3 hours and carries a total of 100 marks. Some questions provide internal choices. Calculators are not permitted. Sample questions include finding inverse functions, evaluating integrals, solving differential equations, and probability questions.
The document discusses expanding and factorizing algebraic expressions. It provides examples of expanding expressions using the distributive property, such as expanding (a + b)(c + d) to get ac + ad + bc + bd. It also discusses factorizing expressions by finding common factors, such as factorizing a2 + 2ab + b2 to get (a + b)2. Tips and techniques are presented for expanding, factorizing, finding common factors, and using the distributive property to manipulate algebraic expressions.
(Www.entrance exam.net)-sail placement sample paper 5SAMEER NAIK
This document provides a 75 question multiple choice test paper with questions ranging in topics from algebra, trigonometry, geometry, and calculus. The test is allotted 90 minutes and covers concepts such as functions, equations, properties of circles, ellipses, parabolas, and hyperbolas, trigonometric identities, and geometric shapes. Several questions involve finding lengths, areas, angles between lines, points of intersection, tangents, and properties of conic sections.
This document contains a 75 question test paper covering topics in mathematics including algebra, trigonometry, coordinate geometry, and calculus. The test has 90 minutes allotted and covers topics such as functions, quadratic equations, trigonometric identities, binomial coefficients, and geometric concepts like angles, triangles, and coordinate planes.
Solving the energy problem of helium final reportJamesMa54
The document discusses solving the ground state energy of a helium atom. It involves computing the Hamiltonian and overlap matrices (H and S) of the system by representing the wavefunction as a linear combination of basis functions. Computing the entries of H and S requires evaluating triple integrals over the internal coordinates of the atom. The main work is to derive a general closed form for these integrals. This involves repeatedly using integration by parts to reduce the exponents in the integrands, yielding sums of terms that can be directly evaluated or fed into computational software for further analysis. Solving these integrals is the crucial step to enable determining the ground state energy by solving the eigenvalue problem Hc = λSc.
The Solovay-Kitaev Theorem guarantees that for any single-qubit gate U and precision ε > 0, it is possible to approximate U to within ε using Θ(logc(1/ε)) gates from a fixed finite universal set of quantum gates. The proof involves first using a "shrinking lemma" to show that any gate in an ε-net can be approximated to within Cε using a constant number of applications of gates from the universal set. This is then iterated to construct an approximation of the desired gate U to arbitrary precision using a number of gates that scales as the logarithm of 1/ε.
Antwoorden fourier and laplace transforms, manual solutionsHildemar Alvarez
The document provides solutions to selected exercises from chapter 1 of a textbook. It first shows the steps to express the sum of two time-harmonic signals as another time-harmonic signal. It then defines several integrals involving trigonometric and exponential functions, and calculates their values. Finally, it summarizes properties of linear time-invariant systems and provides example responses to different input signals.
R package 'bayesImageS': a case study in Bayesian computation using Rcpp and ...Matt Moores
There are many approaches to Bayesian computation with intractable likelihoods, including the exchange algorithm, approximate Bayesian computation (ABC), thermodynamic integration, and composite likelihood. These approaches vary in accuracy as well as scalability for datasets of significant size. The Potts model is an example where such methods are required, due to its intractable normalising constant. This model is a type of Markov random field, which is commonly used for image segmentation. The dimension of its parameter space increases linearly with the number of pixels in the image, making this a challenging application for scalable Bayesian computation. My talk will introduce various algorithms in the context of the Potts model and describe their implementation in C++, using OpenMP for parallelism. I will also discuss the process of releasing this software as an open source R package on the CRAN repository.
(1) The document discusses various integration techniques including: review of integral formulas, integration by parts, trigonometric integrals involving products of sines and cosines, trigonometric substitutions, and integration of rational functions using partial fractions.
(2) Examples are provided to demonstrate each technique, such as using integration by parts to evaluate integrals of the form ∫udv, using trigonometric identities to reduce powers of trigonometric functions, and using partial fractions to break down rational functions into simpler fractions.
(3) The key techniques discussed are integration by parts, trigonometric substitutions to transform integrals involving quadratic expressions into simpler forms, and partial fractions to decompose rational functions for integration. Various examples illustrate the
It turns out that an equation x^p+y^q=z^w has a solution in Z+ if and only if at least one exponent p or q or w equals 2.DOI:10.13140/RG.2.2.15459.22567/7
This document summarizes key concepts in singular perturbation problems. It provides an example problem of determining concentration profiles in a catalyst pellet where the reaction rate is fast compared to diffusion. Singular perturbation problems require rescaling to identify dominant terms as the perturbation parameter approaches zero. Solutions in different regions must then be matched asymptotically. The example demonstrates obtaining outer and inner solutions, and matching them to solve the full problem.
Fast and efficient exact synthesis of single qubit unitaries generated by cli...JamesMa54
The document describes a presentation on an algorithm for exact synthesis of single qubit unitaries generated by Clifford and T gates. The algorithm reduces the problem of implementing a unitary to the problem of state preparation. It then uses a series of HT gates to iteratively decrease the smallest denominator exponent of the state entries until it reaches a base case that can be looked up. The algorithm runs in time linear in the initial smallest denominator exponent and provides an optimal sequence of H and T gates for implementing the input unitary exactly.
On Triplet of Positive Integers Such That the Sum of Any Two of Them is a Per...inventionjournals
In this article we discussed determination of distinct positive integers a, b, c such that a + b, a + c, b + c are perfect squares. We can determine infinitely many such triplets. There are such four tuples and from them eliminating any one number we obtain triplets with the specific property. We can also obtain infinitely many such triplets from a single triplet.
The document discusses several theorems related to twin prime conjectures:
- Theorem 1 states that a prime p can be written in the form 3k+1 or 3k-1, with k being even.
- Theorem 4 characterizes twin primes as pairs where n(n+2) satisfies a modular condition.
- Theorems aim to prove there are infinitely many twin primes, relying on the ratio of primes increasing without bound as n increases without bound.
Given a document describing functions and relations:
- It defines four types of relations: one-to-one, many-to-one, one-to-many, and many-to-many. Relations can be represented using arrow diagrams, ordered pairs, or graphs.
- An example question is worked through to identify the domain, codomain, range, and type of relation for a given set of ordered pairs.
- Functions are represented by lowercase letters like f, g, h. The value of a function f at x is written as f(x).
- Examples demonstrate evaluating functions by substituting values into the function equation and solving composite functions like fg(x).
Bayesian Inference and Uncertainty Quantification for Inverse ProblemsMatt Moores
So-called “inverse” problems arise when the parameters of a physical system cannot be directly observed. The mapping between these latent parameters and the space of noisy observations is represented as a mathematical model, often involving a system of differential equations. We seek to infer the parameter values that best fit our observed data. However, it is also vital to obtain accurate quantification of the uncertainty involved with these parameters, particularly when the output of the model will be used for forecasting. Bayesian inference provides well-calibrated uncertainty estimates, represented by the posterior distribution over the parameters. In this talk, I will give a brief introduction to Markov chain Monte Carlo (MCMC) algorithms for sampling from the posterior distribution and describe how they can be combined with numerical solvers for the forward model. We apply these methods to two examples of ODE models: growth curves in ecology, and thermogravimetric analysis (TGA) in chemistry. This is joint work with Matthew Berry, Mark Nelson, Brian Monaghan and Raymond Longbottom.
This document discusses numerical methods for solving ordinary differential equations (ODEs), including Euler methods, Runge-Kutta methods, and their application to problems in math and physics. It provides examples of using Euler's method and Runge-Kutta methods to solve ODEs describing a block-spring system. Code implementations of the Euler and Runge-Kutta methods are presented along with comparisons of numerical solutions to exact solutions.
The document provides information about an engineering mathematics examination that will take place. It consists of 5 modules with multiple choice and long answer questions in each module. The exam will last 3 hours and students must answer 5 full questions by selecting at least 2 questions from each part. The document then lists the questions under each module.
Identification of the Mathematical Models of Complex Relaxation Processes in ...Vladimir Bakhrushin
The approach to solving the problem of complex relaxation spectra is presented.
Presentation for the XI International Conference on Defect interaction and anelastic phenomena in solids. Tula, 2007.
There are 5 possible pairs: (0,0), (0,9), (9,0), (1,8), (8,1)
The zero can be in any of the 5 positions.
The other two digits in the pair can be arranged in 2 ways.
Hence total numbers = 5 × 2 × 1000 = 10,000
Jawapan: 10,000
SOALAN A4
BM Diberi ABCD satu segiempat sama sisi dengan sisi AB = 4 cm. E dan F
dua titik di atas AB dan BC masing-masing dengan AE = 2 cm dan BF = 3
cm
Beta and gamma are the two most popular functions in mathematics. Gamma is a single variable function, whereas Beta is a two-variable function. The relation between beta and gamma function will help to solve many problems in physics and mathematics.
Este documento describe un concurso promocional en Twitter y Foursquare para promover un producto anticelulítico. El concurso ofrecerá 10 premios del producto y tendrá lugar del 15 al 31 de mayo en Twitter y del 21 al 27 de mayo en Foursquare. El concurso busca aumentar las ventas del producto, atraer nuevos clientes y mejorar la presencia en redes sociales.
This document discusses smart cities, including their architecture, types, key areas, and implementation challenges. It outlines a 4-layer smart city architecture consisting of user, service, infrastructure, and data layers. It also discusses types of smart cities and key areas like smart energy and smart transport. The document identifies challenges like data confidentiality and availability. It presents a risk assessment matrix to evaluate risks like those posed to smart energy, transport, and data. Finally, it concludes that cyberattacks, malware, and availability are major smart city threats and that applying controls to networks and communications poses the highest costs.
(Www.entrance exam.net)-sail placement sample paper 5SAMEER NAIK
This document provides a 75 question multiple choice test paper with questions ranging in topics from algebra, trigonometry, geometry, and calculus. The test is allotted 90 minutes and covers concepts such as functions, equations, properties of circles, ellipses, parabolas, and hyperbolas, trigonometric identities, and geometric shapes. Several questions involve finding lengths, areas, angles between lines, points of intersection, tangents, and properties of conic sections.
This document contains a 75 question test paper covering topics in mathematics including algebra, trigonometry, coordinate geometry, and calculus. The test has 90 minutes allotted and covers topics such as functions, quadratic equations, trigonometric identities, binomial coefficients, and geometric concepts like angles, triangles, and coordinate planes.
Solving the energy problem of helium final reportJamesMa54
The document discusses solving the ground state energy of a helium atom. It involves computing the Hamiltonian and overlap matrices (H and S) of the system by representing the wavefunction as a linear combination of basis functions. Computing the entries of H and S requires evaluating triple integrals over the internal coordinates of the atom. The main work is to derive a general closed form for these integrals. This involves repeatedly using integration by parts to reduce the exponents in the integrands, yielding sums of terms that can be directly evaluated or fed into computational software for further analysis. Solving these integrals is the crucial step to enable determining the ground state energy by solving the eigenvalue problem Hc = λSc.
The Solovay-Kitaev Theorem guarantees that for any single-qubit gate U and precision ε > 0, it is possible to approximate U to within ε using Θ(logc(1/ε)) gates from a fixed finite universal set of quantum gates. The proof involves first using a "shrinking lemma" to show that any gate in an ε-net can be approximated to within Cε using a constant number of applications of gates from the universal set. This is then iterated to construct an approximation of the desired gate U to arbitrary precision using a number of gates that scales as the logarithm of 1/ε.
Antwoorden fourier and laplace transforms, manual solutionsHildemar Alvarez
The document provides solutions to selected exercises from chapter 1 of a textbook. It first shows the steps to express the sum of two time-harmonic signals as another time-harmonic signal. It then defines several integrals involving trigonometric and exponential functions, and calculates their values. Finally, it summarizes properties of linear time-invariant systems and provides example responses to different input signals.
R package 'bayesImageS': a case study in Bayesian computation using Rcpp and ...Matt Moores
There are many approaches to Bayesian computation with intractable likelihoods, including the exchange algorithm, approximate Bayesian computation (ABC), thermodynamic integration, and composite likelihood. These approaches vary in accuracy as well as scalability for datasets of significant size. The Potts model is an example where such methods are required, due to its intractable normalising constant. This model is a type of Markov random field, which is commonly used for image segmentation. The dimension of its parameter space increases linearly with the number of pixels in the image, making this a challenging application for scalable Bayesian computation. My talk will introduce various algorithms in the context of the Potts model and describe their implementation in C++, using OpenMP for parallelism. I will also discuss the process of releasing this software as an open source R package on the CRAN repository.
(1) The document discusses various integration techniques including: review of integral formulas, integration by parts, trigonometric integrals involving products of sines and cosines, trigonometric substitutions, and integration of rational functions using partial fractions.
(2) Examples are provided to demonstrate each technique, such as using integration by parts to evaluate integrals of the form ∫udv, using trigonometric identities to reduce powers of trigonometric functions, and using partial fractions to break down rational functions into simpler fractions.
(3) The key techniques discussed are integration by parts, trigonometric substitutions to transform integrals involving quadratic expressions into simpler forms, and partial fractions to decompose rational functions for integration. Various examples illustrate the
It turns out that an equation x^p+y^q=z^w has a solution in Z+ if and only if at least one exponent p or q or w equals 2.DOI:10.13140/RG.2.2.15459.22567/7
This document summarizes key concepts in singular perturbation problems. It provides an example problem of determining concentration profiles in a catalyst pellet where the reaction rate is fast compared to diffusion. Singular perturbation problems require rescaling to identify dominant terms as the perturbation parameter approaches zero. Solutions in different regions must then be matched asymptotically. The example demonstrates obtaining outer and inner solutions, and matching them to solve the full problem.
Fast and efficient exact synthesis of single qubit unitaries generated by cli...JamesMa54
The document describes a presentation on an algorithm for exact synthesis of single qubit unitaries generated by Clifford and T gates. The algorithm reduces the problem of implementing a unitary to the problem of state preparation. It then uses a series of HT gates to iteratively decrease the smallest denominator exponent of the state entries until it reaches a base case that can be looked up. The algorithm runs in time linear in the initial smallest denominator exponent and provides an optimal sequence of H and T gates for implementing the input unitary exactly.
On Triplet of Positive Integers Such That the Sum of Any Two of Them is a Per...inventionjournals
In this article we discussed determination of distinct positive integers a, b, c such that a + b, a + c, b + c are perfect squares. We can determine infinitely many such triplets. There are such four tuples and from them eliminating any one number we obtain triplets with the specific property. We can also obtain infinitely many such triplets from a single triplet.
The document discusses several theorems related to twin prime conjectures:
- Theorem 1 states that a prime p can be written in the form 3k+1 or 3k-1, with k being even.
- Theorem 4 characterizes twin primes as pairs where n(n+2) satisfies a modular condition.
- Theorems aim to prove there are infinitely many twin primes, relying on the ratio of primes increasing without bound as n increases without bound.
Given a document describing functions and relations:
- It defines four types of relations: one-to-one, many-to-one, one-to-many, and many-to-many. Relations can be represented using arrow diagrams, ordered pairs, or graphs.
- An example question is worked through to identify the domain, codomain, range, and type of relation for a given set of ordered pairs.
- Functions are represented by lowercase letters like f, g, h. The value of a function f at x is written as f(x).
- Examples demonstrate evaluating functions by substituting values into the function equation and solving composite functions like fg(x).
Bayesian Inference and Uncertainty Quantification for Inverse ProblemsMatt Moores
So-called “inverse” problems arise when the parameters of a physical system cannot be directly observed. The mapping between these latent parameters and the space of noisy observations is represented as a mathematical model, often involving a system of differential equations. We seek to infer the parameter values that best fit our observed data. However, it is also vital to obtain accurate quantification of the uncertainty involved with these parameters, particularly when the output of the model will be used for forecasting. Bayesian inference provides well-calibrated uncertainty estimates, represented by the posterior distribution over the parameters. In this talk, I will give a brief introduction to Markov chain Monte Carlo (MCMC) algorithms for sampling from the posterior distribution and describe how they can be combined with numerical solvers for the forward model. We apply these methods to two examples of ODE models: growth curves in ecology, and thermogravimetric analysis (TGA) in chemistry. This is joint work with Matthew Berry, Mark Nelson, Brian Monaghan and Raymond Longbottom.
This document discusses numerical methods for solving ordinary differential equations (ODEs), including Euler methods, Runge-Kutta methods, and their application to problems in math and physics. It provides examples of using Euler's method and Runge-Kutta methods to solve ODEs describing a block-spring system. Code implementations of the Euler and Runge-Kutta methods are presented along with comparisons of numerical solutions to exact solutions.
The document provides information about an engineering mathematics examination that will take place. It consists of 5 modules with multiple choice and long answer questions in each module. The exam will last 3 hours and students must answer 5 full questions by selecting at least 2 questions from each part. The document then lists the questions under each module.
Identification of the Mathematical Models of Complex Relaxation Processes in ...Vladimir Bakhrushin
The approach to solving the problem of complex relaxation spectra is presented.
Presentation for the XI International Conference on Defect interaction and anelastic phenomena in solids. Tula, 2007.
There are 5 possible pairs: (0,0), (0,9), (9,0), (1,8), (8,1)
The zero can be in any of the 5 positions.
The other two digits in the pair can be arranged in 2 ways.
Hence total numbers = 5 × 2 × 1000 = 10,000
Jawapan: 10,000
SOALAN A4
BM Diberi ABCD satu segiempat sama sisi dengan sisi AB = 4 cm. E dan F
dua titik di atas AB dan BC masing-masing dengan AE = 2 cm dan BF = 3
cm
Beta and gamma are the two most popular functions in mathematics. Gamma is a single variable function, whereas Beta is a two-variable function. The relation between beta and gamma function will help to solve many problems in physics and mathematics.
Este documento describe un concurso promocional en Twitter y Foursquare para promover un producto anticelulítico. El concurso ofrecerá 10 premios del producto y tendrá lugar del 15 al 31 de mayo en Twitter y del 21 al 27 de mayo en Foursquare. El concurso busca aumentar las ventas del producto, atraer nuevos clientes y mejorar la presencia en redes sociales.
This document discusses smart cities, including their architecture, types, key areas, and implementation challenges. It outlines a 4-layer smart city architecture consisting of user, service, infrastructure, and data layers. It also discusses types of smart cities and key areas like smart energy and smart transport. The document identifies challenges like data confidentiality and availability. It presents a risk assessment matrix to evaluate risks like those posed to smart energy, transport, and data. Finally, it concludes that cyberattacks, malware, and availability are major smart city threats and that applying controls to networks and communications poses the highest costs.
This document provides an overview of the Harrogate Hub, a new community center being opened to provide support to marginalized individuals in Harrogate, UK. The speaker discusses their own journey from a successful career to experiencing hardship and finding purpose through their faith. They describe the vision for Harrogate Hub, which was revealed to founder Carol, as a place providing pastoral care and signposting people to support services. Voluteers will be trained to befriend and support those referred from organizations struggling to meet community needs. The goal is for Harrogate Hub to be a place where the living water of hope transforms lives, so that no one in the community feels excluded or uncared for.
Instituto tecnológico superior «vida nueva»jonathan vega
El documento discute el papel de las tecnologías de la información y la comunicación (TIC) en la educación. Explica que las TIC son cada vez más accesibles para los estudiantes y que la escuela debe integrar la nueva cultura digital en la enseñanza. Argumenta que los estudiantes están acostumbrados a usar la tecnología y que los docentes deben incorporar herramientas TIC para motivar a los estudiantes y enseñar de manera más atractiva. También reconoce que mientras las TIC han tenido un gran impacto
Este documento describe un concurso promocional en Twitter y Foursquare para promover un producto anticelulítico. El concurso ofrecerá 10 premios del producto y tendrá lugar del 15 al 31 de mayo en Twitter y del 21 al 27 de mayo en Foursquare. El concurso busca aumentar las ventas del producto, atraer nuevos clientes y mejorar la presencia en redes sociales.
Antología de artículos del padre Federico Salvador Ramón publicados en la Revista Esclava y Reina nº 38 de la Congregación de Esclavas de la Inmaculada Niña.
This document provides tips and strategies for developing a strong brand identity. It discusses researching target audiences and competitors, defining core values, and establishing consistent branding elements like color, typography, imagery and tone. It also covers developing marketing strategies for websites, email, social media, networking and packaging to enhance brand awareness and engagement.
O documento descreve vários acidentes comuns em aciarias e suas causas, além de propor soluções para prevenir esses acidentes e tornar a produção de aço menos perigosa. Os principais acidentes listados incluem explosões na panela de aço líquido, vazamentos de aço pela válvula gaveta, reações químicas explosivas entre o carbono no aço e a escória, e acidentes envolvendo água dentro de fornos elétricos. As soluções propostas envolvem eliminar práticas de alto
This thesis presents a model of the basal ganglia using a continuous time recurrent neural network (CTRNN) to simulate its pathways and learning. The model reproduces the response of globus pallidus neurons seen in previous work. An error function is used to calculate the difference between striatum and globus pallidus activity for weight updates between the subthalamic nucleus and globus pallidus. Experiments adding noise and using a sigmoidal update rule are also presented. The thesis discusses representing the model with equations versus code, noting tradeoffs between readability and reproducibility. A CTRNN library was implemented in Python to simulate the basal ganglia model.
This document summarizes Natividad Molina Jimenez and Francisco Cazalilla Luque's visit to schools in Joensuu, Finland as part of the ERASMUS+ program. It describes several schools they visited including their common spaces, classrooms, gyms, playgrounds and other facilities. It reflects on improving their own school, CEIP Medina Elvira, after seeing innovative learning environments in Finland and trying their best without "going crazy" in the process.
O documento discute doenças cardíacas, incluindo as principais causas de morte como angina, infarto agudo do miocárdio e aterosclerose. Ele também descreve fatores de risco, sintomas, diagnóstico e os principais tipos de cardiopatias como insuficiência cardíaca e doenças coronarianas.
1) O documento apresenta uma lista de exercícios de matemática básica sobre porcentagem, proporções e regra de três. 2) Os exercícios incluem cálculos de porcentagens, proporções, conversão de frações em porcentagem e problemas envolvendo velocidade, volume e distância. 3) A lista termina com um problema sobre os custos de impressão de relatórios usando diferentes cartuchos de tinta.
The document discusses encryption and decryption techniques using the RSA algorithm. It explains how RSA uses a public and private key system where a sender encrypts a message using the receiver's public key, and the receiver decrypts it using their private key. The document provides an example of encrypting the message "STOP" into blocks using RSA, and then decrypting the cipher text back into the original message. It also discusses the mathematical operations like exponentiation modulo n that are used in the RSA encryption and decryption processes.
The document defines and provides examples of matrices. The key points are:
1. A matrix is a rectangular array of numbers organized in rows and columns. It is denoted by its dimensions, such as a 3x2 matrix having 3 rows and 2 columns.
2. Matrices can be added, subtracted, or multiplied by a scalar. Operations between matrices are defined element-wise.
3. Special types of matrices include row matrices, column matrices, square matrices, and zero matrices. Matrix equality and solving systems of equations using matrices are also introduced.
1. The document provides definitions and identities relating to theoretical computer science topics like asymptotic analysis, series, recurrences, and discrete structures.
2. It includes definitions for big O, Omega, and Theta notation used to describe the asymptotic behavior of functions.
3. Recurrences, like the master method, are presented for analyzing the runtime of divide-and-conquer algorithms.
4. Discrete structures like combinations, Stirling numbers, and trees are also covered, along with their properties and relationships.
The Gaussian or normal distribution is one of the most widely used in statistics. Estimating its parameters using
Bayesian inference and conjugate priors is also widely used. The use of conjugate priors allows all the results to be
derived in closed form. Unfortunately, different books use different conventions on how to parameterize the various
distributions (e.g., put the prior on the precision or the variance, use an inverse gamma or inverse chi-squared, etc),
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Disclaimer: No one is perfect, so please mind that there might be mistakes and typos.
dtubbenhauer@gmail.com
Corrected slides: dtubbenhauer.com/talks.html
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Research Project Primus
1. Research Project Primus
Predrag Terzi´c
e-mail: pedja.terzic@hotmail.com
April 25 , 2016
1 Compositeness tests for N = k · bn
± c
Definition 1.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
,
where m and x are nonnegative integers.
Conjecture 1.1.
Let N = k · bn
− c such that b ≡ 0 (mod 2), n > bc, k > 0, c > 0
and c ≡ 1, 7 (mod 8)
Let Si = Pb(Si−1) with S0 = Pbk/2(Pb/2(6)), thus
If N is prime then Sn−1 ≡ P(b/2)· c/2 (6) (mod N)
Conjecture 1.2.
Let N = k · bn
− c such that b ≡ 0, 4, 8 (mod 12), n > bc, k > 0, c > 0
and c ≡ 3, 5 (mod 8).
Let Si = Pb(Si−1) with S0 = Pbk/2(Pb/2(6)), thus
If N is prime then Sn−1 ≡ P(b/2)· c/2 (6) (mod N)
Conjecture 1.3.
Let N = k · bn
− c such that b ≡ 2, 6, 10 (mod 12), n > bc, k > 0, c > 0
and c ≡ 3, 5 (mod 8).
Let Si = Pb(Si−1) with S0 = Pbk/2(Pb/2(6)), thus
If N is prime then Sn−1 ≡ −P(b/2)· c/2 (6) (mod N)
1
2. Conjecture 1.4.
Let N = k · bn
+ c such that b ≡ 0 (mod 2), n > bc, k > 0, c > 0
and c ≡ 1, 7 (mod 8)
Let Si = Pb(Si−1) with S0 = Pbk/2(Pb/2(6)), thus
If N is prime then Sn−1 ≡ P(b/2)· c/2 (6) (mod N)
Conjecture 1.5.
Let N = k · bn
+ c such that b ≡ 0, 4, 8 (mod 12), n > bc, k > 0, c > 0
and c ≡ 3, 5 (mod 8).
Let Si = Pb(Si−1) with S0 = Pbk/2(Pb/2(6)), thus
If N is prime then Sn−1 ≡ P(b/2)· c/2 (6) (mod N)
Conjecture 1.6.
Let N = k · bn
+ c such that b ≡ 2, 6, 10 (mod 12), n > bc, k > 0, c > 0
and c ≡ 3, 5 (mod 8).
Let Si = Pb(Si−1) with S0 = Pbk/2(Pb/2(6)), thus
If N is prime then Sn−1 ≡ −P(b/2)· c/2 (6) (mod N)
Proof attempt by mathlove
First of all,
Pb/2(6) = 2−b/2
6 − 4
√
2
b/2
+ 6 + 4
√
2
b/2
= 3 − 2
√
2
b/2
+ 3 + 2
√
2
b/2
= pb
+ qb
where p =
√
2 − 1, q =
√
2 + 1 with pq = 1.
From
S0 = Pbk/2(Pb/2(6)) = 2−bk/2
2pb bk/2
+ 2qb bk/2
= pb2k/2
+ qb2k/2
and Si = Pb(Si−1), we can prove by induction on i ∈ N that
Si = pbi+2k/2
+ qbi+2k/2
.
2
3. By the way,
pN+1
+ qN+1
=
N+1
i=0
N + 1
i
(
√
2)i
(−1)N+1−i
+ 1
=
(N+1)/2
j=0
N + 1
2j
2j+1
≡ 2 + 2(N+3)/2
(mod N)
≡ 2 + 4 · 2
N−1
2 (mod N) (1)
Also,
pN+3
+ qN+3
=
N+3
i=0
N + 3
i
(
√
2)i
(−1)N+3−i
+ 1
=
(N+3)/2
j=0
N + 3
2j
2j+1
≡ 2 +
N + 3
2
· 22
+
N + 3
N + 1
· 2
N+3
2 + 2
N+5
2 (mod N)
≡ 14 + 12 · 2
N−1
2 + 8 · 2
N−1
2 (mod N) (2)
For N ≡ ±1 (mod 8), since 2
N−1
2 ≡ 1 (mod N), from (1)(2), we can prove by induction
on i ∈ Z that
pN+2i−1
+ qN+2i−1
≡ p2i
+ q2i
(mod N) (3)
For N ≡ 3, 5 (mod 8), since 2
N−1
2 ≡ −1 (mod N), from (1)(2), we can prove by induction on
i ∈ Z that
pN+2i−1
+ qN+2i−1
≡ − p2i−2
+ q2i−2
(mod N) (4)
To prove (3)(4), we can use
pN+2(i+1)−1
+ qN+2(i+1)−1
≡ pN+2i−1
+ qN+2i−1
p2
+ q2
−
− pN+2(i−1)−1
+ qN+2(i−1)−1
(mod N)
and
pN+2(i−1)−1
+ qN+2(i−1)−1
≡ pN+2i−1
+ qN+2i−1
p−2
+ q−2
−
− pN+2(i+1)−1
+ qN+2(i+1)−1
(mod N)
Now, for N ≡ ±1 (mod 8), from (3), we can prove by induction on j ∈ N that
pj(N+2i−1)
+ qj(N+2i−1)
≡ p2ij
+ q2ij
(mod N) (5)
Also, for N ≡ 3, 5 (mod 8), from (4), we can prove by induction on j ∈ N that
pj(N+2i−1)
+ qj(N+2i−1)
≡ (−1)j
pj(2i−2)
+ qj(2i−2)
(mod N) (6)
3
4. To prove (5)(6), we can use
p(j+1)(N+2i−1)
+ q(j+1)(N+2i−1)
≡ pj(N+2i−1)
+ qj(N+2i−1)
pN+2i−1
+ qN+2i−1
−
− p(j−1)(N+2i−1)
+ q(j−1)(N+2i−1)
(mod N)
For conjecture 1.1, N ≡ ±1 (mod 8) follows from the conditions N = k · bn
− c such that b ≡ 0
(mod 2), n > bc, k > 0, c > 0 and c ≡ 1, 7 (mod 8). Then, we can say that conjecture 1.1 is
true because using (5) and setting c = 2d − 1 gives
Sn−1 = pbn+1k/2
+ qbn+1k/2
= p(b/2)(N+c)
+ q(b/2)(N+c)
= p(b/2)(N+2d−1)
+ q(b/2)(N+2d−1)
≡ p2·d·(b/2)
+ q2·d·(b/2)
(mod N)
≡ P(b/2)·d(6) (mod N)
≡ P(b/2)· c/2 (6) (mod N)
Q.E.D.
For conjecture 1.2, N ≡ 3, 5 (mod 8) follows from the conditions N = k·bn
−c such that b ≡
0, 4, 8 (mod 12), n > bc, k > 0, c > 0, and c ≡ 3, 5 (mod 8). Then, we can say that conjecture
1.2 is true because using (6) and setting c = 2d − 1 gives
Sn−1 = pbn+1k/2
+ qbn+1k/2
= p(b/2)(N+c)
+ q(b/2)(N+c)
= p(b/2)(N+2d−1)
+ q(b/2)(N+2d−1)
≡ (−1)b/2
p(b/2)·(2d−2)
+ q(b/2)·(2d−2)
(mod N)
≡ P(b/2)·(d−1)(6) (mod N)
≡ P(b/2)· c/2 (6) (mod N)
Q.E.D.
For conjecture 1.3, N ≡ 3, 5 (mod 8) follows from the conditions N = k·bn
−c such that b ≡
2, 6, 10 (mod 12), n > bc, k > 0, c > 0 , and c ≡ 3, 5 (mod 8). Then, we can say that conjecture
1.3 is true because using (6) and setting c = 2d − 1 gives
Sn−1 = pbn+1k/2
+ qbn+1k/2
= p(b/2)(N+c)
+ q(b/2)(N+c)
= p(b/2)(N+2d−1)
+ q(b/2)(N+2d−1)
≡ (−1)b/2
p(b/2)·(2d−2)
+ q(b/2)·(2d−2)
(mod N)
≡ −P(b/2)·(d−1)(6) (mod N)
≡ −P(b/2)· c/2 (6) (mod N)
Q.E.D.
4
5. For conjecture 1.4, N ≡ ±1 (mod 8) follows from the conditions N = k·bn
+c such that b ≡
0 (mod 2), n > bc, k > 0, c > 0 and c ≡ 1, 7 (mod 8). Then, we can say that conjecture 1.4 is
true because using (5) and setting c = 2d − 1 gives
Sn−1 = pbn+1k/2
+ qbn+1k/2
= p(b/2)(N−c)
+ q(b/2)(N−c)
= p(b/2)(N−2d+1))
+ q(b/2)(N−2d+1)
= p(b/2)(N+2(−d+1)−1)
+ q(b/2)(N+2(−d+1)−1)
≡ p2·(−d+1)·(b/2)
+ q2·(−d+1)·(b/2)
(mod N)
≡ q2·(d−1)·(b/2)
+ p2·(d−1)·(b/2)
(mod N)
≡ P(b/2)·(d−1)(6) (mod N)
≡ P(b/2)· c/2 (6) (mod N)
Q.E.D.
For conjecture 1.5, N ≡ 3, 5 (mod 8) follows from the conditions N = k·bn
+c such that b ≡
0, 4, 8 (mod 12), n > bc, k > 0, c > 0 , and c ≡ 3, 5 (mod 8). Then, we can say that conjecture
1.5 is true because using (6) and setting c = 2d − 1 gives
Sn−1 = pbn+1k/2
+ qbn+1k/2
= p(b/2)(N−c)
+ q(b/2)(N−c)
= p(b/2)(N−2d+1)
+ q(b/2)(N−2d+1)
= p(b/2)(N+2(−d+1)−1)
+ q(b/2)(N+2(−d+1)−1)
≡ (−1)b/2
p(b/2)·(2(−d+1)−2)
+ q(b/2)·(2(−d+1)−2)
(mod N)
≡ q(b/2)·2d
+ p(b/2)·2d
(mod N)
≡ P(b/2)·d(6) (mod N)
≡ P(b/2)· c/2 (6) (mod N)
Q.E.D.
For conjecture 1.6, N ≡ 3, 5 (mod 8) follows from the conditions N = k·bn
+c such that b ≡
2, 6, 10 (mod 12), n > bc, k > 0, c > 0 , and c ≡ 3, 5 (mod 8). Then, we can say that conjecture
1.6 is true because using (6) and setting c = 2d − 1 gives
Sn−1 = pbn+1k/2
+ qbn+1k/2
= p(b/2)(N−c)
+ q(b/2)(N−c)
= p(b/2)(N−2d+1)
+ q(b/2)(N−2d+1)
= p(b/2)(N+2(−d+1)−1)
+ q(b/2)(N+2(−d+1)−1)
≡ (−1)b/2
p(b/2)·(2(−d+1)−2)
+ q(b/2)·(2(−d+1)−2)
(mod N)
≡ − q(b/2)·2d
+ p(b/2)·2d
(mod N)
≡ −P(b/2)·d(6) (mod N)
≡ −P(b/2)· c/2 (6) (mod N)
Q.E.D.
5
6. 2 Primality tests for specific classes of N = k · 2m
± 1
Throughout this post we use the following notations: Z-the set of integers , N-the set of positive
integers , a
p
-the Jacobi symbol , (m, n)-the greatest common divisor of m and n , Sn(x)-the
sequence defined by S0(x) = x and Sk+1(x) = (Sk(x))2
− 2(k ≥ 0) .
Basic Lemmas and Theorems
Definition 2.1. For P, Q ∈ Z the Lucas sequence {Vn(P, Q)} is defined by V0(P, Q) = 2, V1(P, Q) =
P, Vn+1(P, Q) = PVn(P, Q) − QVn−1(P, Q)(n ≥ 1) Let D = P2
− 4Q . It is known that
Vn(P, Q) =
P +
√
D
2
n
+
P −
√
D
2
n
Lemma 2.1. Let P, Q ∈ Z and n ∈ N. Then
Vn(P, Q) =
[n/2]
r=0
n
n − r
n − r
r
Pn−2r
(−Q)r
Theorem 2.1. (Zhi-Hong Sun)
For m ∈ {2, 3, 4, ...} let p = k · 2m
± 1 with 0 < k < 2m
and k odd . If b, c ∈ Z ,
(p, c) = 1 and 2c+b
p
= 2c−b
p
= − c
p
then p is prime if and only if p | Sm−2(x), where
x = c−k
Vk(b, c2
) =
(k−1)/2
r=0
k
k − r
k − r
r
(−1)r
(b/c)k−2r
Lemma 2.2. Let n be odd positive number , then
−1
n
=
1, if n ≡ 1 (mod 4)
−1, if n ≡ 3 (mod 4)
Lemma 2.3. Let n be odd positive number , then
2
n
=
1, if n ≡ 1, 7 (mod 8)
−1, if n ≡ 3, 5 (mod 8)
Lemma 2.4. Let n be odd positive number , then case 1. (n ≡ 1 (mod 4))
3
n
=
1 if n ≡ 1 (mod 3)
0 if n ≡ 0 (mod 3)
−1 if n ≡ 2 (mod 3)
case 2. (n ≡ 3 (mod 4))
3
n
=
1 if n ≡ 2 (mod 3)
0 if n ≡ 0 (mod 3)
−1 if n ≡ 1 (mod 3)
6
7. Proof. Since 3 ≡ 3 (mod 4) if we apply the law of quadratic reciprocity we have two cases .
If n ≡ 1 (mod 4) then 3
n
= n
3
and the result follows . If n ≡ 3 (mod 4) then 3
n
= − n
3
and the result follows .
Lemma 2.5. Let n be odd positive number , then
5
n
=
1 if n ≡ 1, 4 (mod 5)
0 if n ≡ 0 (mod 5)
−1 if n ≡ 2, 3 (mod 5)
Proof. Since 5 ≡ 1 (mod 4) if we apply the law of quadratic reciprocity we have 5
n
= n
5
and the result follows .
Lemma 2.6. Let n be odd positive number , then case 1. (n ≡ 1 (mod 4))
−3
n
=
1 if n ≡ 1, 11 (mod 12)
0 if n ≡ 3, 9 (mod 12)
−1 if n ≡ 5, 7 (mod 12)
case 2. (n ≡ 3 (mod 4))
−3
n
=
1 if n ≡ 5, 7 (mod 12)
0 if n ≡ 3, 9 (mod 12)
−1 if n ≡ 1, 11 (mod 12)
Proof. −3
n
= −1
n
3
n
. Applying the law of quadratic reciprocity we have : if n ≡ 1
(mod 4) then 3
n
= n
3
. If n ≡ 3 (mod 4) then 3
n
= − n
3
. Applying the Chinese
remainder theorem in both cases several times we get the result .
Lemma 2.7. Let n be odd positive number , then case 1. (n ≡ 1 (mod 4))
7
n
=
1 if n ≡ 1, 2, 4 (mod 7)
0 if n ≡ 0 (mod 7)
−1 if n ≡ 3, 5, 6 (mod 7)
case 2. (n ≡ 3 (mod 4))
7
n
=
1 if n ≡ 3, 5, 6 (mod 7)
0 if n ≡ 0 (mod 7)
−1 if n ≡ 1, 2, 4 (mod 7)
Proof. Since 7 ≡ 3 (mod 4) if we apply the law of quadratic reciprocity we have two cases .
If n ≡ 1 (mod 4) then 7
n
= n
7
and the result follows . If n ≡ 3 (mod 4) then 7
n
= − n
7
and the result follows .
7
8. Lemma 2.8. Let n be odd positive number , then case 1. (n ≡ 1 (mod 4))
−6
n
=
1 if n ≡ 1, 5, 19, 23 (mod 24)
0 if n ≡ 3, 9, 15, 21 (mod 24)
−1 if n ≡ 7, 11, 13, 17 (mod 24)
case 2. (n ≡ 3 (mod 4))
−6
n
=
1 if n ≡ 7, 11, 13, 17 (mod 24)
0 if n ≡ 3, 9, 15, 21 (mod 24)
−1 if n ≡ 1, 5, 19, 23 (mod 24)
Proof. −6
n
= −1
n
2
n
3
n
. Applying the law of quadratic reciprocity we have : if n ≡ 1
(mod 4) then 3
n
= n
3
. If n ≡ 3 (mod 4) then 3
n
= − n
3
. Applying the Chinese
remainder theorem in both cases several times we get the result .
Lemma 2.9. Let n be odd positive number , then
10
n
=
1 if n ≡ 1, 3, 9, 13, 27, 31, 37, 39 (mod 40)
0 if n ≡ 5, 15, 25, 35 (mod 40)
−1 if n ≡ 7, 11, 17, 19, 21, 23, 29, 33 (mod 40)
Proof. 10
n
= 2
n
5
n
. Applying the law of quadratic reciprocity we have : 5
n
= n
5
.
Applying the Chinese remainder theorem several times we get the result .
The Main Result
Theorem 2.2. Let N = k · 2m
− 1 such that m > 2 , 3 | k , 0 < k < 2m
and
k ≡ 1 (mod 10) with m ≡ 2, 3 (mod 4)
k ≡ 3 (mod 10) with m ≡ 0, 3 (mod 4)
k ≡ 7 (mod 10) with m ≡ 1, 2 (mod 4)
k ≡ 9 (mod 10) with m ≡ 0, 1 (mod 4)
Let b = 3 and S0(x) = Vk(b, 1) , thus
N is prime iff N | Sm−2(x)
Proof. Since N ≡ 3 (mod 4) and b = 3 from Lemma 2.2 we know that 2−b
N
= −1 .
Similarly , since N ≡ 2 (mod 5) or N ≡ 3 (mod 5) and b = 3 from Lemma 2.5 we know that
2+b
N
= −1 . From Lemma 2.1 we know that Vk(b, 1) = x . Applying Theorem 2.1 in the case
c = 1 we get the result.
Q.E.D.
8
9. Theorem 2.3. Let N = k · 2m
− 1 such that m > 2 , 3 | k , 0 < k < 2m
and
k ≡ 3 (mod 42) with m ≡ 0, 2 (mod 3)
k ≡ 9 (mod 42) with m ≡ 0 (mod 3)
k ≡ 15 (mod 42) with m ≡ 1 (mod 3)
k ≡ 27 (mod 42) with m ≡ 1, 2 (mod 3)
k ≡ 33 (mod 42) with m ≡ 0, 1 (mod 3)
k ≡ 39 (mod 42) with m ≡ 2 (mod 3)
Let b = 5 and S0(x) = Vk(b, 1) , thus N is prime iff N | Sm−2(x)
Proof. Since N ≡ 3 (mod 4) and N ≡ 11 (mod 12) and b = 5 from Lemma 2.6 we know
that 2−b
N
= −1 . Similarly , since N ≡ 3 (mod 4) and N ≡ 1 (mod 7) or N ≡ 2 (mod 7) or
N ≡ 4 (mod 7) and b = 5 from Lemma 2.7 we know that 2+b
N
= −1 . From Lemma 2.1 we
know that Vk(b, 1) = x . Applying Theorem 2.1 in the case c = 1 we get the result.
Q.E.D.
Theorem 2.4. Let N = k · 2m
+ 1 such that m > 2 , 0 < k < 2m
and
k ≡ 1 (mod 42) with m ≡ 2, 4 (mod 6)
k ≡ 5 (mod 42) with m ≡ 3 (mod 6)
k ≡ 11 (mod 42) with m ≡ 3, 5 (mod 6)
k ≡ 13 (mod 42) with m ≡ 4 (mod 6)
k ≡ 17 (mod 42) with m ≡ 5 (mod 6)
k ≡ 19 (mod 42) with m ≡ 0 (mod 6)
k ≡ 23 (mod 42) with m ≡ 1, 3 (mod 6)
k ≡ 25 (mod 42) with m ≡ 0, 2 (mod 6)
k ≡ 29 (mod 42) with m ≡ 1, 5 (mod 6)
k ≡ 31 (mod 42) with m ≡ 2 (mod 6)
k ≡ 37 (mod 42) with m ≡ 0, 4 (mod 6)
k ≡ 41 (mod 42) with m ≡ 1 (mod 6)
Let b = 5 and S0(x) = Vk(b, 1) , thus N is prime iff N | Sm−2(x)
Proof. Since N ≡ 1 (mod 4) and N ≡ 5 (mod 12) and b = 5 from Lemma 2.6 we know
that 2−b
N
= −1 . Similarly , since N ≡ 1 (mod 4) and N ≡ 3 (mod 7) or N ≡ 5 (mod 7) or
N ≡ 6 (mod 7) and b = 5 from Lemma 2.7 we know that 2+b
N
= −1 . From Lemma 2.1 we
know that Vk(b, 1) = x . Applying Theorem 2.1 in the case c = 1 we get the result.
Q.E.D.
9
10. Theorem 2.5. Let N = k · 2m
+ 1 such that m > 2 , 0 < k < 2m
and
k ≡ 1 (mod 6) and k ≡ 1, 7 (mod 10) with m ≡ 0 (mod 4)
k ≡ 5 (mod 6) and k ≡ 1, 3 (mod 10) with m ≡ 1 (mod 4)
k ≡ 1 (mod 6) and k ≡ 3, 9 (mod 10) with m ≡ 2 (mod 4)
k ≡ 5 (mod 6) and k ≡ 7, 9 (mod 10) with m ≡ 3 (mod 4)
Let b = 8 and S0(x) = Vk(b, 1) , thus N is prime iff N | Sm−2(x)
Proof. Since N ≡ 1 (mod 4) and N ≡ 17 (mod 24) and b = 8 from Lemma 2.8 we know
that 2−b
N
= −1 . Similarly , since N ≡ 17 (mod 40) or N ≡ 33 (mod 40) and b = 8 from
Lemma 2.9 we know that 2+b
N
= −1 . From Lemma 2.1 we know that Vk(b, 1) = x . Applying
Theorem 2.1 in the case c = 1 we get the result.
Q.E.D.
3 Three prime generating recurrences
Prime number generator I
Let bn = bn−1 + lcm(
√
2 · n , bn−1) with b1 = 2 then an = bn+1/bn − 1 is either 1 or prime .
Conjecture 3.1. 1. Every term of this sequence ai is either prime or 1 .
2. Every prime of the form
√
2 · n is member of this sequence .
Prime number generator II
Let bn = bn−1 + lcm(
√
3 · n , bn−1) with b1 = 3 then an = bn+1/bn − 1 is either 1 or prime .
Conjecture 3.2. 1. Every term of this sequence ai is either prime or 1 .
2. Every prime of the form
√
3 · n is member of this sequence .
Prime number generator III
Let bn = bn−1 + lcm(
√
n3 , bn−1) with b1 = 2 then an = bn+1/bn − 1 is either 1 or prime .
Conjecture 3.3. 1. Every term of this sequence ai is either prime or 1 .
2. Every prime of the form
√
n3 is member of this sequence .
4 Some properties of Fibonacci numbers
Conjecture 4.1. If p is prime , not 5 , and M ≥ 2 then :
MFp
≡ M(p−1)(1−( p
5 ))/2
(mod Mp−1
M−1
)
Conjecture 4.2. If p is prime , and M ≥ 2 then :
M
Fp−( p
5 ) ≡ 1 (mod Mp−1
M−1
)
10
11. Corollary of Cassini’s formula
Corollary 4.1. For n ≥ 2 :
Fn =
√
Fn−1 · Fn+1 , if n is even
√
Fn−1 · Fn+1 , if n is odd
5 A modification of Riesel’s primality test
Definition 5.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x are
nonnegative integers .
Corollary 5.1. Let N = k ·2n
−1 such that n > 2 , k odd , 3 k , k < 2n
, and f is proper factor
of n − 2 .
Let Si = P2f (Si−1) with S0 = Pk(4) , thus
N is prime iff S(n−2)/f ≡ 0 (mod N)
6 Primality criteria for specific classes of N = k · 2n
+ 1
Definition 6.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x are
nonnegative integers .
Conjecture 6.1. Let N = 3 · 2n
+ 1 such that n > 2 and n ≡ 1, 2 (mod 4)
Let Si = P2(Si−1) with
S0 =
P3(32), if n ≡ 1 (mod 4)
P3(28), if n ≡ 2 (mod 4)
thus , N is prime iff Sn−2 ≡ 0 (mod N)
Conjecture 6.2. Let N = 5 · 2n
+ 1 such that n > 2 and n ≡ 1, 3 (mod 4)
Let Si = P2(Si−1) with
S0 =
P5(28), if n ≡ 1 (mod 4)
P5(32), if n ≡ 3 (mod 4)
thus , N is prime iff Sn−2 ≡ 0 (mod N)
Conjecture 6.3. Let N = 7 · 2n
+ 1 such that n > 2 and n ≡ 0, 2 (mod 4)
Let Si = P2(Si−1) with
S0 =
P7(8), if n ≡ 0 (mod 4)
P7(32), if n ≡ 2 (mod 4)
thus , N is prime iff Sn−2 ≡ 0 (mod N)
Conjecture 6.4. Let N = 9 · 2n
+ 1 such that n > 3 and n ≡ 2, 3 (mod 4)
Let Si = P2(Si−1) with
11
12. S0 =
P9(28), if n ≡ 2 (mod 4)
P9(32), if n ≡ 3 (mod 4)
thus, N is prime iff Sn−2 ≡ 0 (mod N)
Conjecture 6.5. Let N = 11 · 2n
+ 1 such that n > 3 and n ≡ 1, 3 (mod 4)
Let Si = P2(Si−1) with
S0 =
P11(8), if n ≡ 1 (mod 4)
P11(28), if n ≡ 3 (mod 4)
thus , N is prime iff Sn−2 ≡ 0 (mod N)
Conjecture 6.6. Let N = 13 · 2n
+ 1 such that n > 3 and n ≡ 0, 2 (mod 4)
Let Si = P2(Si−1) with
S0 =
P13(32), if n ≡ 0 (mod 4)
P13(8), if n ≡ 2 (mod 4)
thus , N is prime iff Sn−2 ≡ 0 (mod N)
7 Congruence only holding for primes
Theorem 7.1. (Wilson)
A natural number n > 1 is a prime iff:
(n − 1)! ≡ −1 (mod n).
Theorem 7.2. A natural number n > 2 is a prime iff:
n−1
k=1
k ≡ n − 1 (mod
n−1
k=1
k).
Proof
Necessity: If n is a prime, then
n−1
k=1
k ≡ n − 1 (mod
n−1
k=1
k).
If n is an odd prime, then by Theorem 7.1 we have
n−1
k=1
k ≡ n − 1 (mod n)
Hence, n | ((n − 1)! − (n − 1)) and therefore n | (n − 1)((n − 2)! − 1).
Since n | (n − 1) it follows n | ((n − 2)! − 1) , hence
n(n − 1)
2
| (n − 1)((n − 2)! − 1),
12
13. thus
n−1
k=1
k ≡ n − 1 (mod
n−1
k=1
k).
Sufficiency: If
n−1
k=1
k ≡ n − 1 (mod
n−1
k=1
k)
then n is a prime.
Suppose n is a composite and p is a prime such that p | n, then since
n−1
k=1
k =
n(n − 1)
2
it
follows p |
n−1
k=1
k . Since
n−1
k=1
k ≡ n − 1 (mod
n−1
k=1
k),
we have
n−1
k=1
k ≡ n − 1 (mod p).
However , since p ≤ n − 1 it divides
n−1
k=1
k, and so
n−1
k=1
k ≡ 0 (mod p),
a contradiction . Hence n must be prime.
Q.E.D.
8 Primality test for N = 2 · 3n
− 1
Definition 8.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x are
nonnegative integers .
Conjecture 8.1. Let N = 2 · 3n
− 1 such that n > 1 .
Let Si = P3(Si−1) with S0 = P3(a) , where
a =
6, if n ≡ 0 (mod 2)
8, if n ≡ 1 (mod 2)
thus , N is prime iff Sn−1 ≡ a (mod N)
13
14. 9 Compositeness tests for specific classes of generalized Fer-
mat numbers
Definition 9.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x are
nonnegative integers .
Conjecture 9.1. Let Fn(b) = b2n
+ 1 such that n > 1 , b is even , 3 b and 5 b .
Let Si = Pb(Si−1) with S0 = Pb/2(Pb/2(8)) , thus
If Fn(b) is prime then S2n−2 ≡ 0 (mod Fn(b))
Conjecture 9.2. Let Fn(6) = 62n
+ 1 such that n > 1 .
Let Si = P6(Si−1) with S0 = P3(P3(32)) , thus
If Fn(6) is prime then S2n−2 ≡ 0 (mod Fn(6))
10 Primality tests for specific classes of N = k · 6n
− 1
Definition 10.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x are
nonnegative integers .
Conjecture 10.1.
Let N = k · 6n
− 1 such that n > 2, k > 0,
k ≡ 2, 5 (mod 7) and k < 6n
.
Let Si = P6(Si−1) with S0 = P3k(P3(5)), thus
N is prime iff Sn−2 ≡ 0 (mod N)
Conjecture 10.2.
Let N = k · 6n
− 1 such that n > 2, k > 0,
k ≡ 3, 4 (mod 5) and k < 6n
.
Let Si = P6(Si−1) with S0 = P3k(P3(3)), thus
N is prime iff Sn−2 ≡ 0 (mod N)
Incomplete proof by mathlove
I’m going to prove that
if N is prime, then Sn−2 ≡ 0 (mod N)
for both conjectures.
(For the first conjecture)
First of all,
P3(5) = 2−3
· 5 −
√
21
3
+ 5 +
√
21
3
= 110
14
16. Thus, Sn−2 ≡ 0 (mod N).
Q.E.D.
(For the second conjecture)
P3(3) = 2−3
· 3 −
√
5
3
+ 3 +
√
5
3
= 18
S0 = P3k(P3(3)) = 2−3k
· 18 −
√
182 − 4
3k
+ 18 +
√
182 − 4
3k
= (9 − 4
√
5)3k
+ (9 + 4
√
5)3k
= c6k
+ d6k
where c =
√
5 − 2, d =
√
5 + 2 with cd = 1.
We can prove by induction that
Si = c6i+1k
+ d6i+1k
Thus,
Sn−2 = c
N+1
6 + d
N+1
6 =
√
5
2
−
1
2
N+1
2
+
√
5
2
+
1
2
N+1
2
=
= 2−N+1
2
√
5 − 1
N+1
2
+
√
5 + 1
N+1
2
.
By the way, for N prime,
√
5 − 1
N+1
+
√
5 + 1
N+1
=
N+1
i=0
N + 1
i
(
√
5)i
(−1)N+1−i
+ 1N+1−i
=
(N+1)/2
j=0
N + 1
2j
(
√
5)2j
· 2
=
(N+1)/2
j=0
N + 1
2j
5j
· 2
≡ 2 + 5
N+1
2 · 2 (mod N)
≡ 2 + (−5) · 2 (mod N)
≡ −8 (mod N)
This is because N ≡ 2, 3 (mod 5) implies that
5
N−1
2 ≡ −1 (mod N).
From this, since 2N−1
≡ 1 (mod N),
2N+1
S2
n−2 = (
√
5 − 1)N+1
+ (
√
5 + 1)N+1
+ 2 · 4
N+1
2
≡ −8 + 2 · 2N−1
· 4 (mod N)
≡ 0 (mod N)
Thus, Sn−2 ≡ 0 (mod N).
Q.E.D.
16
17. 11 Compositeness tests for specific classes of N = k · bn
− 1
Definition 11.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x
are nonnegative integers .
Conjecture 11.1. Let N = k · bn
− 1 such that n > 2 , k is odd , 3 k , b is even , 3 b , k < bn
.
Let Si = Pb(Si−1) with S0 = Pbk/2(Pb/2(4)) , thus
if N is prime then Sn−2 ≡ 0 (mod N)
Conjecture 11.2. Let N = k · bn
− 1 such that n > 2 , k < bn
and
k ≡ 3 (mod 30) with b ≡ 2 (mod 10) and n ≡ 0, 3 (mod 4)
k ≡ 3 (mod 30) with b ≡ 4 (mod 10) and n ≡ 0, 2 (mod 4)
k ≡ 3 (mod 30) with b ≡ 6 (mod 10) and n ≡ 0, 1, 2, 3 (mod 4)
k ≡ 3 (mod 30) with b ≡ 8 (mod 10) and n ≡ 0, 1 (mod 4)
Let Si = Pb(Si−1) with S0 = Pbk/2(Pb/2(18)) , thus
If N is prime then Sn−2 ≡ 0 (mod N)
Conjecture 11.3. Let N = k · bn
− 1 such that n > 2 , k < bn
and
k ≡ 9 (mod 30) with b ≡ 2 (mod 10) and n ≡ 0, 1 (mod 4)
k ≡ 9 (mod 30) with b ≡ 4 (mod 10) and n ≡ 0, 2 (mod 4)
k ≡ 9 (mod 30) with b ≡ 6 (mod 10) and n ≡ 0, 1, 2, 3 (mod 4)
k ≡ 9 (mod 30) with b ≡ 8 (mod 10) and n ≡ 0, 3 (mod 4)
Let Si = Pb(Si−1) with S0 = Pbk/2(Pb/2(18)) , thus
If N is prime then Sn−2 ≡ 0 (mod N)
Conjecture 11.4. Let N = k · bn
− 1 such that n > 2 , k < bn
and
k ≡ 21 (mod 30) with b ≡ 2 (mod 10) and n ≡ 2, 3 (mod 4)
k ≡ 21 (mod 30) with b ≡ 4 (mod 10) and n ≡ 1, 3 (mod 4)
k ≡ 21 (mod 30) with b ≡ 8 (mod 10) and n ≡ 1, 2 (mod 4)
Let Si = Pb(Si−1) with S0 = Pbk/2(Pb/2(3)) , thus
If N is prime then Sn−2 ≡ 0 (mod N)
12 Compositeness tests for specific classes of N = k · 3n
± 2
Definition 12.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x
are nonnegative integers .
Conjecture 12.1. Let N = k · 3n
− 2 such that n ≡ 0 (mod 2) , n > 2 , k ≡ 1 (mod 4) and
k > 0 .
Let Si = P3(Si−1) with S0 = P3k(4) , thus
If N is prime then Sn−1 ≡ P1(4) (mod N)
17
18. Conjecture 12.2. Let N = k · 3n
− 2 such that n ≡ 1 (mod 2) , n > 2 , k ≡ 1 (mod 4) and
k > 0 .
Let Si = P3(Si−1) with S0 = P3k(4) , thus
If N is prime then Sn−1 ≡ P3(4) (mod N)
Conjecture 12.3. Let N = k · 3n
+ 2 such that n > 2 , k ≡ 1, 3 (mod 8) and k > 0 .
Let Si = P3(Si−1) with S0 = P3k(6) , thus
If N is prime then Sn−1 ≡ P3(6) (mod N)
Conjecture 12.4. Let N = k · 3n
+ 2 such that n > 2 , k ≡ 5, 7 (mod 8) and k > 0 .
Let Si = P3(Si−1) with S0 = P3k(6) , thus
If N is prime then Sn−1 ≡ P1(6) (mod N)
13 Compositeness tests for N = bn
± b ± 1
Definition 13.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x are
nonnegative integers.
Conjecture 13.1. Let N = bn
− b − 1 such that n > 2, b ≡ 0, 6 (mod 8).
Let Si = Pb(Si−1) with S0 = Pb/2(6),
thus
if N is prime, then Sn−1 ≡ P(b+2)/2(6) (mod N).
Conjecture 13.2. Let N = bn
− b − 1 such that n > 2, b ≡ 2, 4 (mod 8).
Let Si = Pb(Si−1) with S0 = Pb/2(6),
thus
if N is prime, then Sn−1 ≡ −Pb/2(6) (mod N).
Conjecture 13.3. Let N = bn
+ b + 1 such that n > 2, b ≡ 0, 6 (mod 8).
Let Si = Pb(Si−1) with S0 = Pb/2(6),
thus
if N is prime, then Sn−1 ≡ Pb/2(6) (mod N).
Conjecture 13.4. Let N = bn
+ b + 1 such that n > 2, b ≡ 2, 4 (mod 8).
Let Si = Pb(Si−1) with S0 = Pb/2(6),
thus
if N is prime, then Sn−1 ≡ −P(b+2)/2(6) (mod N).
Conjecture 13.5. Let N = bn
− b + 1 such that n > 3, b ≡ 0, 2 (mod 8).
Let Si = Pb(Si−1) with S0 = Pb/2(6),
thus
if N is prime, then Sn−1 ≡ Pb/2(6) (mod N).
Conjecture 13.6. Let N = bn
− b + 1 such that n > 3, b ≡ 4, 6 (mod 8).
Let Si = Pb(Si−1) with S0 = Pb/2(6),
thus
if N is prime, then Sn−1 ≡ −P(b−2)/2(6) (mod N).
18
19. Conjecture 13.7. Let N = bn
+ b − 1 such that n > 3, b ≡ 0, 2 (mod 8).
Let Si = Pb(Si−1) with S0 = Pb/2(6),
thus
if N is prime, then Sn−1 ≡ P(b−2)/2(6) (mod N).
Conjecture 13.8. Let N = bn
+ b − 1 such that n > 3, b ≡ 4, 6 (mod 8).
Let Si = Pb(Si−1) with S0 = Pb/2(6),
thus
if N is prime, then Sn−1 ≡ −Pb/2(6) (mod N).
Proof attempt by mathlove
First of all,
S0 = Pb/2(6) = 2− b
2 · 6 − 4
√
2
b
2
+ 6 + 4
√
2
b
2
= 3 − 2
√
2
b
2
+ 3 + 2
√
2
b
2
=
√
2 − 1
b
+
√
2 + 1
b
= pb
+ qb
where p =
√
2 − 1, q =
√
2 + 1 with pq = 1.
Now, we can prove by induction that
Si = pbi+1
+ qbi+1
.
By the way,
pN+1
+ qN+1
=
N+1
i=0
N + 1
i
(
√
2)i
(−1)N+1−i
+ 1
=
(N+1)/2
j=0
N + 1
2j
2j+1
≡ 2 + 2(N+3)/2
(mod N)
≡ 2 + 4 · 2
N−1
2 (mod N) (1)
Also,
pN+3
+ qN+3
=
N+3
i=0
N + 3
i
(
√
2)i
(−1)N+3−i
+ 1
=
(N+3)/2
j=0
N + 3
2j
2j+1
≡ 2 +
N + 3
2
· 22
+
N + 3
N + 1
· 2
N+3
2 + 2
N+5
2 (mod N)
≡ 14 + 12 · 2
N−1
2 + 8 · 2
N−1
2 (mod N) (2)
19
20. Here, for N ≡ ±1 (mod 8), since 2
N−1
2 ≡ 1 (mod N), from (1)(2), we can prove by
induction that
pN+2i−1
+ qN+2i−1
≡ p2i
+ q2i
(mod N) (3)
For N ≡ 3, 5 (mod 8), since 2
N−1
2 ≡ −1 (mod N), from (1)(2), we can prove by induction
that
pN+2i−1
+ qN+2i−1
≡ − p2i−2
+ q2i−2
(mod N) (4)
To prove (3)(4), we can use
pN+2(i+1)−1
+ qN+2(i+1)−1
≡ pN+2i−1
+ qN+2i−1
p2
+ q2
−
− pN+2(i−1)−1
+ qN+2(i−1)−1
(mod N)
and
pN+2(i−1)−1
+ qN+2(i−1)−1
≡ pN+2i−1
+ qN+2i−1
p−2
+ q−2
−
− pN+2(i+1)−1
+ qN+2(i+1)−1
(mod N)
(Note that (3)(4) holds for **every integer** i (not necessarily positive) because of pq = 1.)
Conjecture 13.1 is true because from (3)
Sn−1 = pN+b+1
+ qN+b+1
≡ pb+2
+ qb+2
(mod N)
≡ P(b+2)/2(6) (mod N)
Conjecture 13.2 is true because from (4)
Sn−1 = pN+b+1
+ qN+b+1
≡ − pb
+ qb
(mod N)
≡ −Pb/2(6) (mod N)
Conjecture 13.3 is true because from (3)
Sn−1 = pN−b−1
+ qN−b−1
≡ p−b
+ q−b
(mod N)
≡ qb
+ pb
(mod N)
≡ Pb/2(6) (mod N)
Conjecture 13.4 is true because from (4)
Sn−1 = pN−b−1
+ qN−b−1
≡ − p−b−2
+ q−b−2
(mod N)
≡ − qb+2
+ pb+2
(mod N)
≡ −P(b+2)/2(6) (mod N)
20
21. Conjecture 13.5 is true because from (3)
Sn−1 = pN+b−1
+ qN+b−1
≡ pb
+ qb
(mod N)
≡ Pb/2(6) (mod N)
Conjecture 13.6 is true because from (4)
Sn−1 = pN+b−1
+ qN+b−1
≡ − pb−2
+ qb−2
(mod N)
≡ −P(b−2)/2(6) (mod N)
Conjecture 13.7 is true because from (3)
Sn−1 = pN−b+1
+ qN−b+1
≡ p−b+2
+ q−b+2
(mod N)
≡ qb−2
+ pb−2
(mod N)
≡ P(b−2)/2(6) (mod N)
Conjecture 13.8 is true because from (4)
Sn−1 = pN−b+1
+ qN−b+1
≡ − p−b
+ q−b
(mod N)
≡ − qb
+ pb
(mod N)
≡ −Pb/2(6) (mod N)
Q.E.D.
14 Primality test for N = 8 · 3n
− 1
Definition 14.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x
are nonnegative integers .
Conjecture 14.1. Let N = 8 · 3n
− 1 such that n > 1 .
Let Si = P3(Si−1) with S0 = P12(4)
thus ,
N is prime iff Sn−1 ≡ 4 (mod N)
Incomplete proof by David Speyer
Let’s unwind your formula.
Sn−1 = P4·3n (4) = (2 +
√
3)4·3n
+ (2 −
√
3)4·3n
= (2 +
√
3)4·3n
+ (2 +
√
3)−4·3n
= (2 +
√
3)(N+1)/2
+ (2 +
√
3)−(N+1)/2
.
21
22. You are testing whether or not Sn−1 ≡ 4 mod N or, on other words,
(2 +
√
3)(N+1)/2
+ (2 +
√
3)−(N+1)/2
≡ 4 mod N. (∗)
If N is prime: (This section is rewritten to use some observations about roots of unity. It may
therefore look a bit less motivated.) The prime N is −1 mod 24, so N2
≡ 1 mod 24 and the
finite field FN2 contains a primitive 24-th root of unity, call it η. We have (η + η−1
)2
= 2 +
√
3,
for one of the two choices of
√
3 in FN . (Since N ≡ −1 mod 12, we have 3
N
= 1.) Now,
η ∈ FN . However, we compute (η + η−1
)N
= ηN
+ η−N
= η−1
+ η, since N ≡ −1 mod 24. So
η + η−1
∈ FN and we deduce that 2 +
√
3 is a square in FN .
So (2 +
√
3)(N−1)/2
≡ 1 mod N and (2 +
√
3)(N+1)/2
≡ (2 +
√
3) mod N. Similarly,
(2 +
√
3)−(N+1)/2
≡ (2 +
√
3)−1
≡ 2 −
√
3 mod N and (∗) holds.
If N is not prime. Earlier, I said that I saw no way to control whether or not (∗) held when N
was composite. I said that there seemed to be no reason it should hold and that, furthermore, it
was surely very rare, because N is exponentially large, so it is unlikely for a random equality to
hold modulo N.
Since then I had a few more ideas about the problem, which don’t make it seem any easier,
but clarify to me why it is so hard. To make life easier, let’s assume that N = p1p2 · · · pj is square
free. Of course, (∗) holds modulo N if and only if it holds modulo every pi.
Let η be a primitive 24-th root of unity in an appropriate extension of Fpj
. The following
equations all take place in this extension of Fpj
. It turns out that (∗) factors quite a bit:
(2 +
√
3)(N+1)/2
+ (2 +
√
3)−(N+1)/2
= 4
(2 +
√
3)(N+1)/2
= 2 ±
√
3
(η + η−1
)(N+1)
= (η + η−1
)2
or (η5
+ η−5
)2
(η + η−1
)(N+1)/2
∈ {η + η−1
, η3
+ η−3
, η5
+ η−5
, η7
+ η−7
}. (†)
Here is what I would like to do at this point, to follow the lines of the Lucas-Lehmer test, but
cannot.
(1) I’d like to know that (η + η−1
)(N+1)/2
= η + η−1
, not one of the other options in (†).
(This is what actually occurs in the N prime case, as shown previously.) This would imply that
(η + η−1
)(N−1)/2
= 1 ∈ Fpj
.
(2) I’d like to know that the order of η + η−1
was precisely (N − 1)/2, not some divisor
thereof.
(3) I’d like to thereby conclude that the multiplicative group of Fpj
was of order divisible by
(N − 1)/2, and thus pj ≥ (N − 1)/2. This would mean that there was basically only room for
one pj, and we would be able to conclude primality.
Now, (1) isn’t so bad, because you could directly compute in the ring Z/(NZ)[η]/(η8
−η4
+1),
rather than trying to disguise this ring with elementary polynomial formulas. So, while I don’t
see that your algorithm checks this point, it wouldn’t be hard.
And (2) =⇒ (3) is correct.
22
23. But you have a real problem with (2). This way this works in the Lucas-Lehmer test is that
you are trying to prove that 2 +
√
3 has order precisely 2p
in the field F2p−1[
√
3] ∼= F(2p−1)2 . You
already know that (2 +
√
3)2p
= 1. So it is enough to check that (2 +
√
3)2p−1
= −1, not 1.
In the current situation, the analogous thing would be to check that (η + η−1
)(N−1)/(2q)
= 1
for every prime q dividing (N − 1)/2. But I have no idea which primes divide q! This seems like
an huge obstacle to a proof that (∗) implies N is prime.
To repeat: I think it may well be true that (∗) implies N is prime, simply because there is
no reason that (†) should hold once N = pj, and the odds of (†) happening by accident are
exponentially small. But I see no global principle implying this.
Q.E.D.
15 Generalization of Kilford’s primality theorem
Conjecture 15.1. Natural number n greater than two is prime iff :
n−1
k=1
bk
− a ≡
an
− 1
a − 1
(mod
bn
− 1
b − 1
)
where b > a > 1 .
16 Prime generating sequence
Definition 16.1. Let bn = bn−2 + lcm(n − 1, bn−2) with b1 = 2 , b2 = 2 and n > 2 .
Let an = bn+2/bn − 1
Conjecture 16.1. 1. Every term of this sequence ai is either prime or 1 .
2. Every odd prime number is member of this sequence .
3. Every new prime in sequence is a next prime from the largest prime already listed .
Incomplete proof by Markus Schepherd
This is the full argument for conjectures 2 and 3. First we need the general relation between
gcd (a, b) and lcm [a, b]: a · b = (a, b) · [a, b]. Then we note that the lowest common multiple
[n − 1, bn−2] is in particular a multiple of bn−2, say kbn−2 with 1 ≤ k ≤ n − 1. Hence we have
bn = bn−2(k + 1), so in every step the term bn gets a new factor between 2 and n which means in
particular that all prime factors of bn are less or equal to n. Now we rearrange an with the above
observation to an = n+1
(n+1,bn)
. Let p be a prime. Then (p, bp−1) = 1 since all prime factors of bp−1
are strictly smaller than p. But then ap−1 = p
(p,bp−1)
= p as claimed in conjecture 2. Further, we
have obviously an ≤ n + 1 for all n, so the first index for which the prime p can appear in the
sequence is p − 1 which immediately implies conjecture 3.
Q.E.D.
23
24. 17 Primality test using Euler’s totient function
Theorem 17.1. (Wilson)
A positive integer n is prime iff (n − 1)! ≡ −1 (mod n)
Theorem 17.2. A positive integer n is prime iff ϕ(n)! ≡ −1 (mod n) .
Proof
Necessity : If n is prime then ϕ(n)! ≡ −1 (mod n)
If n is prime then we have ϕ(n) = n − 1 and
by Theorem 17.1 : (n − 1)! = −1 (mod n) ,
hence ϕ(n)! ≡ −1 (mod n) .
Sufficiency : If ϕ(n)! ≡ −1 (mod n) then n is prime
For n = 2 and n = 6 :
ϕ(2)! ≡ −1 (mod 2) and 2 is prime .
ϕ(6)! ≡ −1 (mod 6) and 6 is composite .
For n = 2, 6 :
Suppose n is composite and p is the least prime such that p | n ,
then we have ϕ(n)! ≡ −1 (mod p) .
Since ϕ(n) ≥
√
n for all n except n = 2 and n = 6
and p ≤
√
n it follows p | ϕ(n)! , hence ϕ(n)! ≡ 0 (mod p)
a contradiction .
Therefore , n must be prime .
Q.E.D.
18 Primality tests for specific classes of Proth numbers
Theorem 18.1. Let N = k · 2n
+ 1 with n > 1 , k < 2n
, 3 | k , and
k ≡ 3 (mod 30), with n ≡ 1, 2 (mod 4)
k ≡ 9 (mod 30), with n ≡ 2, 3 (mod 4)
k ≡ 21 (mod 30), with n ≡ 0, 1 (mod 4)
k ≡ 27 (mod 30), with n ≡ 0, 3 (mod 4)
thus,
N is prime iff 5
N−1
2 ≡ −1 (mod N) .
Proof
Necessity : If N is prime then 5
N−1
2 ≡ −1 (mod N)
Let N be a prime , then by Euler criterion :
5
N−1
2 ≡ 5
N
(mod N)
If N is a prime then N ≡ 2, 3 (mod 5) and therefore : N
5
= −1 .
Since N ≡ 1 (mod 4) according to the law of quadratic reciprocity it follows that : 5
N
= −1 .
Hence , 5
N−1
2 ≡ −1 (mod N) .
24
25. Sufficiency : If 5
N−1
2 ≡ −1 (mod N) then N is prime
If 5
N−1
2 ≡ −1 (mod N) then by Proth’s theorem N is prime .
Q.E.D.
Theorem 18.2. Let N = k · 2n
+ 1 with n > 1 , k < 2n
, 3 | k , and
k ≡ 3 (mod 42), with n ≡ 2 (mod 3)
k ≡ 9 (mod 42), with n ≡ 0, 1 (mod 3)
k ≡ 15 (mod 42), with n ≡ 1, 2 (mod 3)
k ≡ 27 (mod 42), with n ≡ 1 (mod 3)
k ≡ 33 (mod 42), with n ≡ 0 (mod 3)
k ≡ 39 (mod 42), with n ≡ 0, 2 (mod 3)
thus , N is prime iff 7
N−1
2 ≡ −1 (mod N)
Proof
Necessity : If N is prime then 7
N−1
2 ≡ −1 (mod N)
Let N be a prime , then by Euler criterion :
7
N−1
2 ≡ 7
N
(mod N)
If N is prime then N ≡ 3, 5, 6 (mod 7) and therefore : N
7
= −1 .
Since N ≡ 1 (mod 4) according to the law of quadratic reciprocity it follows that : 7
N
= −1 .
Hence , 7
N−1
2 ≡ −1 (mod N) .
Sufficiency : If 7
N−1
2 ≡ −1 (mod N) then N is prime
If 7
N−1
2 ≡ −1 (mod N) then by Proth’s theorem N is prime .
Q.E.D.
Theorem 18.3. Let N = k · 2n
+ 1 with n > 1 , k < 2n
, 3 | k , and
k ≡ 3 (mod 66), with n ≡ 1, 2, 6, 8, 9 (mod 10)
k ≡ 9 (mod 66), with n ≡ 0, 1, 3, 4, 8 (mod 10)
k ≡ 15 (mod 66), with n ≡ 2, 4, 5, 7, 8 (mod 10)
k ≡ 21 (mod 66), with n ≡ 1, 2, 4, 5, 9 (mod 10)
k ≡ 27 (mod 66), with n ≡ 0, 2, 3, 5, 6 (mod 10)
k ≡ 39 (mod 66), with n ≡ 0, 1, 5, 7, 8 (mod 10)
k ≡ 45 (mod 66), with n ≡ 0, 4, 6, 7, 9 (mod 10)
k ≡ 51 (mod 66), with n ≡ 0, 2, 3, 7, 9 (mod 10)
k ≡ 57 (mod 66), with n ≡ 3, 5, 6, 8, 9 (mod 10)
k ≡ 63 (mod 66), with n ≡ 1, 3, 4, 6, 7 (mod 10)
thus,
N is prime iff 11
N−1
2 ≡ −1 (mod N)
Proof
25
26. Necessity : If N is prime then 11
N−1
2 ≡ −1 (mod N)
Let N be a prime , then by Euler criterion :
11
N−1
2 ≡ 11
N
(mod N)
If N is prime then N ≡ 2, 6, 7, 8, 10 (mod 11) and therefore : N
11
= −1 .
Since N ≡ 1 (mod 4) according to the law of quadratic reciprocity it follows that : 11
N
= −1 .
Hence , 11
N−1
2 ≡ −1 (mod N) .
Sufficiency : If 11
N−1
2 ≡ −1 (mod N) then N is prime
If 11
N−1
2 ≡ −1 (mod N) then by Proth’s theorem N is prime .
Q.E.D.
19 Generalization of Wilson’s primality theorem
Theorem 19.1. For m ≥ 1 number n greater than one is prime iff :
(nm
− 1)! ≡ (n − 1)
(−1)m+1
2
· n
nm−mn+m−1
n−1 (mod n
nm−mn+m+n−2
n−1 )
20 Primality test for Fermat numbers using quartic recurrence
equation
Let us define sequence Si as :
Si =
8 if i = 0;
(S2
i−1 − 2)2
− 2 otherwise .
Theorem 20.1. Fn = 22n
+ 1, (n ≥ 2) is a prime if and only if Fn divides S2n−1−1 .
Proof
Let us define ω = 4 +
√
15 and ¯ω = 4 −
√
15 and then define Ln to be ω22n
+ ¯ω22n
, we
get L0 = ω + ¯ω = 8 , and Ln+1 = ω22n+2
+ ¯ω22n+2
= (ω22n+1
)2
+ (¯ω22n+1
)2
= = (ω22n+1
+
¯ω22n+1
)2
− 2 · ω22n+1
· ¯ω22n+1
= = ((ω22n
+ ¯ω22n
)2
− 2 · ω22n
· ¯ω22n
)2
− 2 · ω22n+1
· ¯ω22n+1
=
= ((ω22n
+¯ω22n
)2
−2·(ω·¯ω)22n
)2
−2·(ω·¯ω)22n+1
and since ω·¯ω = 1 we get : Ln+1 = (L2
n−2)2
−2
Because the Ln satisfy the same inductive definition as the sequence Si , the two sequences must
be the same .
Proof of necessity
If 22n
+ 1 is prime then S2n−1−1 is divisible by 22n
+ 1
We rely on simplification of the proof of Lucas-Lehmer test by Oystein J. R. Odseth .First
notice that 3 is quadratic non-residue (mod Fn) and that 5 is quadratic non-residue (mod Fn) .
Euler’s criterion then gives us : 3
Fn−1
2 ≡ −1 (mod Fn) and 5
Fn−1
2 ≡ −1 (mod Fn) On the other
hand 2 is a quadratic-residue (mod Fn) , Euler’s criterion gives: 2
Fn−1
2 ≡ 1 (mod Fn)
Next define σ = 2
√
15 , and define X as the multiplicative group of {a + b
√
15|a, b ∈ ZFn }
.We will use following lemmas :
26
27. Lemma 1. (x + y)Fn
= xFn
+ yFn
(mod Fn)
Lemma 2. aFn
≡ a (mod Fn) (Fermat little theorem)
Then in group X we have :
(6+σ)Fn
≡ (6)Fn
+(σ)Fn
(mod Fn) = = 6+(2
√
15)Fn
(mod Fn) = = 6+2Fn
·15
Fn−1
2 ·
√
15
(mod Fn) = = 6+2·3
Fn−1
2 ·5
Fn−1
2 ·
√
15 (mod Fn) = = 6+2·(−1)·(−1)·
√
15 (mod Fn) =
= 6 + 2
√
15 (mod Fn) = (6 + σ) (mod Fn)
We chose σ such that ω = (6+σ)2
24
. We can use this to compute ω
Fn−1
2 in the group X :
ω
Fn−1
2 = (6+σ)Fn−1
24
Fn−1
2
= (6+σ)Fn
(6+σ)·24
Fn−1
2
≡ (6+σ)
(6+σ)·(−1)
(mod Fn) = −1 (mod Fn)
where we use fact that :
24
Fn−1
2 = (2
Fn−1
2 )3
· (3
Fn−1
2 ) ≡ (13
) · (−1) (mod Fn) = −1 (mod Fn)
So we have shown that :
ω
Fn−1
2 ≡ −1 (mod Fn)
If we write this as ω
22n
+1−1
2 = ω22n−1
= ω22n−2
· ω22n−2
≡ −1 (mod Fn) ,multiply both
sides by ¯ω22n−2
, and put both terms on the left hand side to write this as : ω22n−2
+ ¯ω22n−2
≡ 0
(mod Fn) ω22(2n−1−1)
+ ¯ω22(2n−1−1)
≡ 0 (mod Fn) ⇒ S2n−1−1 ≡ 0 (mod Fn)
Since the left hand side is an integer this means therefore that S2n−1−1 must be divisible by
22n
+ 1 .
Proof of sufficiency
If S2n−1−1 is divisible by 22n
+ 1 , then 22n
+ 1 is prime .
We rely on simplification of the proof of Lucas-Lehmer test by J. W. Bruce .If 22n
+ 1 is not
prime then it must be divisible by some prime factor F less than or equal to the square root of
22n
+ 1 . From the hypothesis S2n−1−1 is divisible by 22n
+ 1 so S2n−1−1 is also multiple of F
, so we can write : ω22(2n−1−1)
+ ¯ω22(2n−1−1)
= K · F , for some integer K . We can write this
equality as : ω22n−2
+ ¯ω22n−2
= K · F Note that ω · ¯ω = 1 so we can multiply both sides by
ω22n−2
and rewrite this relation as : ω22n−1
= K · F · ω22n−2
− 1 . If we square both sides we get
: ω22n
= (K · F · ω22n−2
− 1)2
Now consider the set of numbers a + b
√
15 for integers a and b
where a + b
√
15 and c + d
√
15 are considered equivalent if a and c differ by a multiple of F , and
the same is true for b and d . There are F2
of these numbers , and addition and multiplication can
be verified to be well-defined on sets of equivalent numbers. Given the element ω (considered as
representative of an equivalence class) , the associative law allows us to use exponential notation
for repeated products : ωn
= ω · ω · · · ω , where the product contains n factors and the usual rules
for exponents can be justified . Consider the sequence of elements ω, ω2
, ω3
... . Because ω has the
inverse ¯ω every element in this sequence has an inverse . So there can be at most F2
− 1 different
elements of this sequence. Thus there must be at least two different exponents where ωj
= ωk
with j < k ≤ F2
. Multiply j times by inverse of ω to get that ωk−j
= 1 with 1 ≤ k −j ≤ F2
−1
. So we have proven that ω satisfies ωn
= 1 for some positive exponent n less than or equal to
F2
− 1 . Define the order of ω to be smallest positive integer d such that ωd
= 1 . So if n is any
other positive integer satisfying ωn
= 1 then n must be multiple of d . Write n = q · d + r with
r < d . Then if r = 0 we have 1 = ωn
= ωq·d+r
= (ωd
)q
· ωr
= 1q
· ωr
= ωr
contradicting
the minimality of d so r = 0 and n is multiple of d . The relation ω22n
= (K · F · ω22n−2
− 1)2
shows that ω22n
≡ 1 (mod F) . So that 22n
must be multiple of the order of ω . But the relation
27
28. ω22n−1
= K · F · ω22n−2
− 1 shows that ω22n−1
≡ −1 (mod F) so the order cannot be any proper
factor of 22n
, therefore the order must be 22n
. Since this order is less than or equal to F2
− 1 and
F is less or equal to the square root of 22n
+ 1 we have relation : 22n
≤ F2
− 1 ≤ 22n
. This is
true only if 22n
= F2
− 1 ⇒ 22n
+ 1 = F2
. We will show that Fermat number cannot be square
of prime factor .
Theorem : Any prime divisor p of Fn = 22n
+ 1 is of the form k · 2n+2
+ 1 whenever n is
greater than one .
So prime factor F must be of the form k · 2n+2
+ 1 , therefore we can write : 22n
+ 1 =
(k · 2n+2
+ 1)2
22n
+ 1 = k2
· 22n+4
+ 2 · k · 2n+2
+ 1 22n
= k · 2n+3
· (k · 2n+1
+ 1)
The last equality cannot be true since k · 2n+1
+ 1 is an odd number and 22n
has no odd prime
factors so 22n
+1 = F2
and therefore we have relation 22n
< F2
−1 < 22n
which is contradiction
so therefore 22n
+ 1 must be prime .
Q.E.D.
21 Prime number formula
pn = 1 +
2·( n ln(n) +1)
k=1
1 −
1
n
·
k
j=2
3 −
j
i=1
j
i
j
i
j
22 Primality criterion for specific class of N = 3 · 2n
− 1
Definition 22.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x
are nonnegative integers .
Conjecture 22.1. Let N = 3 · 2n
− 1 such that n > 2 and n ≡ 2 (mod 4)
Let Si = P2(Si−1) with
S0 =
P3(32), if n ≡ 2 (mod 8)
P3(36), if n ≡ 6 (mod 8)
thus , N is prime iff Sn−2 ≡ 0 (mod N)
23 Probable prime tests for generalized Fermat numbers
Definition 23.1. Let Pm(x) = 2−m
· x −
√
x2 − 4
m
+ x +
√
x2 − 4
m
, where m and x are
nonnegative integers.
Theorem 23.1. Let Fn(b) = b2n
+ 1 such that n ≥ 2 and b is even number .
Let Si = Pb(Si−1) with S0 = Pb(6), thus if Fn(b) is prime, then S2n−1 ≡ 2 (mod Fn(b)).
28
29. The following proof appeared for the first time on MSE forum in August 2016 .
Proof by mathlove . First of all, we prove by induction that
Si = αbi+1
+ βbi+1
(1)
where α = 3 − 2
√
2, β = 3 + 2
√
2 with αβ = 1.
Proof for (1) :
S0 = Pb(6)
= 2−b
· 6 − 4
√
2
b
+ 6 + 4
√
2
b
= 2−b
· 2b
(3 − 2
√
2)b
+ 2b
(3 + 2
√
2)b
= αb
+ βb
Suppose that (1) holds for i. Using the fact that
(αm
+ βm
)2
− 4 = (βm
− αm
)2
we get
Si+1 = Pb(Si)
= 2−b
· αbi+1
+ βbi+1
− (αbi+1
+ βbi+1
)2 − 4
b
+ αbi+1
+ βbi+1
+ (αbi+1
+ βbi+1
)2 − 4
b
= 2−b
· αbi+1
+ βbi+1
− (βbi+1
− αbi+1
)2
b
+ αbi+1
+ βbi+1
+ (βbi+1
− αbi+1
)2
b
= 2−b
· 2αbi+1 b
+ 2βbi+1 b
= αbi+2
+ βbi+2
Let N := Fn(b) = b2n
+ 1. Then, from (1),
S2n−1 = αb2n
+ βb2n
= αN−1
+ βN−1
Since αβ = 1,
S2n−1 = αN−1
+ βN−1
= αβ(αN−1
+ βN−1
)
= β · αN
+ α · βN
= 3(αN
+ βN
) − 2
√
2 (βN
− αN
) (2)
So, in the following, we find αN + βN (mod N) and
√
2 (βN − αN ) (mod N).
Using the binomial theorem,
αN
+ βN
= (3 − 2
√
2)N
+ (3 + 2
√
2)N
=
N
i=0
N
i
3i
· ((−2
√
2)N−i
+ (2
√
2)N−i
)
=
(N+1)/2
j=1
N
2j − 1
32j−1
· 2(2
√
2)N−(2j−1)
29
30. Since N
2j−1 ≡ 0 (mod N) for 1 ≤ j ≤ (N − 1)/2, we get
αN
+ βN
≡
N
N
3N
· 2(2
√
2)0
≡ 2 · 3N
(mod N)
Now, by Fermat’s little theorem,
αN
+ βN
≡ 2 · 3N
≡ 2 · 3 ≡ 6 (mod N) (3)
Similarly,
√
2 (βN
− αN
) =
√
2 ((3 + 2
√
2)N
− (3 − 2
√
2)N
)
=
√
2
N
i=0
N
i
3i
· ((2
√
2)N−i
− (−2
√
2)N−i
)
=
√
2
(N−1)/2
j=0
N
2j
32j
· 2(2
√
2)N−2j
≡
√
2
N
0
30
· 2(2
√
2)N
(mod N)
≡ 2N+1
· 2(N+1)/2
(mod N)
≡ 4 · 2(N+1)/2
(mod N) (4)
By the way, since b is even with n ≥ 2,
N = b2n
+ 1 ≡ 1 (mod 8)
from which
2(N−1)/2
≡
2
N
≡ (−1)(N2−1)/8
≡ 1 (mod N)
follows where
q
p
denotes the Legendre symbol .
So, from (4),
√
2 (βN
− αN
) ≡ 4 · 2(N+1)/2
≡ 4 · 2 ≡ 8 (mod N) (5)
Therefore, finally, from (2)(3) and (5),
S2n−1 ≡ 3(αN
+ βN
) − 2
√
2 (βN
− αN
) ≡ 3 · 6 − 2 · 8 ≡ 2 (mod Fn(b))
as desired.
Q.E.D.
Theorem 23.2. Let En(b) = b2n
+1
2 such that n > 1, b is odd number greater than one .
Let Si = Pb(Si−1) with S0 = Pb(6), thus if En(b) is prime, then S2n−1 ≡ 6 (mod En(b)).
The following proof appeared for the first time on MSE forum in August 2016 .
Proof by mathlove . First of all, we prove by induction that
Si = p2bi+1
+ q2bi+1
(6)
where p =
√
2 − 1, q =
√
2 + 1 with pq = 1.
30
31. Proof for (6) :
S0 = Pb(6) = 2−b
· 6 − 4
√
2
b
+ 6 + 4
√
2
b
= (3 − 2
√
2)b
+ (3 + 2
√
2)b
= p2b
+ q2b
Supposing that (6) holds for i gives
Si+1 = Pb(Si)
= 2−b
· Si − S2
i − 4
b
+ Si + S2
i − 4
b
= 2−b
· p2bi+1
+ q2bi+1
− q2bi+1
− p2bi+1 2
b
+ p2bi+1
+ q2bi+1
+ q2bi+1
− p2bi+1 2
b
= 2−b
· p2bi+1
+ q2bi+1
− q2bi+1
− p2bi+1 b
+ p2bi+1
+ q2bi+1
+ q2bi+1
− p2bi+1 b
= 2−b
· 2p2bi+1 b
+ 2q2bi+1 b
= p2bi+2
+ q2bi+2
Let N := 2n − 1, M := En(b) = (bN+1 + 1)/2. From (6), we have
S2n−1 = SN
= p2bN+1
+ q2bN+1
= p2(2M−1)
+ q2(2M−1)
= p4M−2
+ q4M−2
= (pq)2
(p4M−2
+ q4M−2
)
= 3(p4M
+ q4M
) − 2
√
2 (q4M
− p4M
)
Now using the binomial theorem and Fermat’s little theorem,
p4M
+ q4M
= (17 − 12
√
2)M
+ (17 + 12
√
2)M
=
M
i=0
M
i
17i
((−12
√
2)M−i
+ (12
√
2)M−i
)
=
(M+1)/2
j=1
M
2j − 1
172j−1
· 2(12
√
2)M−(2j−1)
≡
M
M
17M
· 2(12
√
2)0
(mod M)
≡ 17 · 2 (mod M)
≡ 34 (mod M)
31
32. Similarly,
2
√
2 (q4M
− p4M
) = 2
√
2 ((17 + 12
√
2)M
− (17 − 12
√
2)M
)
= 2
√
2
M
i=0
M
i
17i
((12
√
2)M−i
− (−12
√
2)M−i
)
= 2
√
2
(M−1)/2
j=0
M
2j
172j
· 2(12
√
2)M−2j
=
(M−1)/2
j=0
M
2j
172j
· 4 · 12M−2j
· 2(M−2j+1)/2
≡
M
0
170
· 4 · 12M
· 2(M+1)/2
(mod M)
≡ 4 · 12 · 2 (mod M)
≡ 96 (mod M)
since 2(M−1)/2 ≡ (−1)(M2−1)/8 ≡ 1 (mod M) (this is because M ≡ 1 (mod 8) from b2 ≡ 1, 9
(mod 16))
It follows from these that
S2n−1 = 3(p4M
+ q4M
) − 2
√
2 (q4M
− p4M
)
≡ 3 · 34 − 96 (mod M)
≡ 6 (mod En(b))
as desired.
Q.E.D.
32