Phatak Primality Test (PPT)
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Chapter 9 Quadratic Residues
Chapter 9 Quadratic Residues 9.1 Introduction Definition 9.1. We say that a 2 Z is a quadratic residue mod n if there exists b 2 Z such that a ≡ b2 mod n: If there is no such b we say that a is a quadratic non-residue mod n. Example: Suppose n = 10. We can determine the quadratic residues mod n by computing b2 mod n for 0 ≤ b < n. In fact, since (−b)2 ≡ b2 mod n; we need only consider 0 ≤ b ≤ [n=2]. Thus the quadratic residues mod 10 are 0; 1; 4; 9; 6; 5; while 3; 7; 8 are quadratic non-residues mod 10. Proposition 9.1. If a; b are quadratic residues mod n then so is ab. Proof. Suppose a ≡ r2; b ≡ s2 mod p: Then ab ≡ (rs)2 mod p: 9.2 Prime moduli Proposition 9.2. Suppose p is an odd prime. Then the quadratic residues coprime to p form a subgroup of (Z=p)× of index 2. Proof. Let Q denote the set of quadratic residues in (Z=p)×. If θ :(Z=p)× ! (Z=p)× denotes the homomorphism under which r 7! r2 mod p 9–1 then ker θ = {±1g; im θ = Q: By the first isomorphism theorem of group theory, × jkerθj · j im θj = j(Z=p) j: Thus Q is a subgroup of index 2: p − 1 jQj = : 2 Corollary 9.1. Suppose p is an odd prime; and suppose a; b are coprime to p. Then 1. 1=a is a quadratic residue if and only if a is a quadratic residue. -
Efficient Regular Modular Exponentiation Using
J Cryptogr Eng (2017) 7:245–253 DOI 10.1007/s13389-016-0134-5 SHORT COMMUNICATION Efficient regular modular exponentiation using multiplicative half-size splitting Christophe Negre1,2 · Thomas Plantard3,4 Received: 14 August 2015 / Accepted: 23 June 2016 / Published online: 13 July 2016 © Springer-Verlag Berlin Heidelberg 2016 Abstract In this paper, we consider efficient RSA modular x K mod N where N = pq with p and q prime. The private exponentiations x K mod N which are regular and con- data are the two prime factors of N and the private exponent stant time. We first review the multiplicative splitting of an K used to decrypt or sign a message. In order to insure a integer x modulo N into two half-size integers. We then sufficient security level, N and K are chosen large enough take advantage of this splitting to modify the square-and- to render the factorization of N infeasible: they are typically multiply exponentiation as a regular sequence of squarings 2048-bit integers. The basic approach to efficiently perform always followed by a multiplication by a half-size inte- the modular exponentiation is the square-and-multiply algo- ger. The proposed method requires around 16% less word rithm which scans the bits ki of the exponent K and perform operations compared to Montgomery-ladder, square-always a sequence of squarings followed by a multiplication when and square-and-multiply-always exponentiations. These the- ki is equal to one. oretical results are validated by our implementation results When the cryptographic computations are performed on which show an improvement by more than 12% compared an embedded device, an adversary can monitor power con- approaches which are both regular and constant time. -
A FEW FACTS REGARDING NUMBER THEORY Contents 1
A FEW FACTS REGARDING NUMBER THEORY LARRY SUSANKA Contents 1. Notation 2 2. Well Ordering and Induction 3 3. Intervals of Integers 4 4. Greatest Common Divisor and Least Common Multiple 5 5. A Theorem of Lam´e 8 6. Linear Diophantine Equations 9 7. Prime Factorization 10 8. Intn, mod n Arithmetic and Fermat's Little Theorem 11 9. The Chinese Remainder Theorem 13 10. RelP rimen, Euler's Theorem and Gauss' Theorem 14 11. Lagrange's Theorem and Primitive Roots 17 12. Wilson's Theorem 19 13. Polynomial Congruencies: Reduction to Simpler Form 20 14. Polynomial Congruencies: Solutions 23 15. The Quadratic Formula 27 16. Square Roots for Prime Power Moduli 28 17. Euler's Criterion and the Legendre Symbol 31 18. A Lemma of Gauss 33 −1 2 19. p and p 36 20. The Law of Quadratic Reciprocity 37 21. The Jacobi Symbol and its Reciprocity Law 39 22. The Tonelli-Shanks Algorithm for Producing Square Roots 42 23. Public Key Encryption 44 24. An Example of Encryption 47 References 50 Index 51 Date: October 13, 2018. 1 2 LARRY SUSANKA 1. Notation. To get started, we assume given the set of integers Z, sometimes denoted f :::; −2; −1; 0; 1; 2;::: g: We assume that the reader knows about the operations of addition and multiplication on integers and their basic properties, and also the usual order relation on these integers. In particular, the operations of addition and multiplication are commuta- tive and associative, there is the distributive property of multiplication over addition, and mn = 0 implies one (at least) of m or n is 0. -
Miller-Rabin Primality Test (Java)
Miller-Rabin primality test (Java) Other implementations: C | C, GMP | Clojure | Groovy | Java | Python | Ruby | Scala The Miller-Rabin primality test is a simple probabilistic algorithm for determining whether a number is prime or composite that is easy to implement. It proves compositeness of a number using the following formulas: Suppose 0 < a < n is coprime to n (this is easy to test using the GCD). Write the number n−1 as , where d is odd. Then, provided that all of the following formulas hold, n is composite: for all If a is chosen uniformly at random and n is prime, these formulas hold with probability 1/4. Thus, repeating the test for k random choices of a gives a probability of 1 − 1 / 4k that the number is prime. Moreover, Gerhard Jaeschke showed that any 32-bit number can be deterministically tested for primality by trying only a=2, 7, and 61. [edit] 32-bit integers We begin with a simple implementation for 32-bit integers, which is easier to implement for reasons that will become apparent. First, we'll need a way to perform efficient modular exponentiation on an arbitrary 32-bit integer. We accomplish this using exponentiation by squaring: Source URL: http://www.en.literateprograms.org/Miller-Rabin_primality_test_%28Java%29 Saylor URL: http://www.saylor.org/courses/cs409 ©Spoon! (http://www.en.literateprograms.org/Miller-Rabin_primality_test_%28Java%29) Saylor.org Used by Permission Page 1 of 5 <<32-bit modular exponentiation function>>= private static int modular_exponent_32(int base, int power, int modulus) { long result = 1; for (int i = 31; i >= 0; i--) { result = (result*result) % modulus; if ((power & (1 << i)) != 0) { result = (result*base) % modulus; } } return (int)result; // Will not truncate since modulus is an int } int is a 32-bit integer type and long is a 64-bit integer type. -
Number Theory Summary
YALE UNIVERSITY DEPARTMENT OF COMPUTER SCIENCE CPSC 467: Cryptography and Computer Security Handout #11 Professor M. J. Fischer November 13, 2017 Number Theory Summary Integers Let Z denote the integers and Z+ the positive integers. Division For a 2 Z and n 2 Z+, there exist unique integers q; r such that a = nq + r and 0 ≤ r < n. We denote the quotient q by ba=nc and the remainder r by a mod n. We say n divides a (written nja) if a mod n = 0. If nja, n is called a divisor of a. If also 1 < n < jaj, n is said to be a proper divisor of a. Greatest common divisor The greatest common divisor (gcd) of integers a; b (written gcd(a; b) or simply (a; b)) is the greatest integer d such that d j a and d j b. If gcd(a; b) = 1, then a and b are said to be relatively prime. Euclidean algorithm Computes gcd(a; b). Based on two facts: gcd(0; b) = b; gcd(a; b) = gcd(b; a − qb) for any q 2 Z. For rapid convergence, take q = ba=bc, in which case a − qb = a mod b. Congruence For a; b 2 Z and n 2 Z+, we write a ≡ b (mod n) iff n j (b − a). Note a ≡ b (mod n) iff (a mod n) = (b mod n). + ∗ Modular arithmetic Fix n 2 Z . Let Zn = f0; 1; : : : ; n − 1g and let Zn = fa 2 Zn j gcd(a; n) = 1g. For integers a; b, define a⊕b = (a+b) mod n and a⊗b = ab mod n. -
RSA Power Analysis Obfuscation: a Dynamic FPGA Architecture John W
Air Force Institute of Technology AFIT Scholar Theses and Dissertations Student Graduate Works 3-22-2012 RSA Power Analysis Obfuscation: A Dynamic FPGA Architecture John W. Barron Follow this and additional works at: https://scholar.afit.edu/etd Part of the Electrical and Computer Engineering Commons Recommended Citation Barron, John W., "RSA Power Analysis Obfuscation: A Dynamic FPGA Architecture" (2012). Theses and Dissertations. 1078. https://scholar.afit.edu/etd/1078 This Thesis is brought to you for free and open access by the Student Graduate Works at AFIT Scholar. It has been accepted for inclusion in Theses and Dissertations by an authorized administrator of AFIT Scholar. For more information, please contact [email protected]. RSA POWER ANALYSIS OBFUSCATION: A DYNAMIC FPGA ARCHITECTURE THESIS John W. Barron, Captain, USAF AFIT/GE/ENG/12-02 DEPARTMENT OF THE AIR FORCE AIR UNIVERSITY AIR FORCE INSTITUTE OF TECHNOLOGY Wright-Patterson Air Force Base, Ohio APPROVED FOR PUBLIC RELEASE; DISTRIBUTION UNLIMITED. The views expressed in this thesis are those of the author and do not reflect the official policy or position of the United States Air Force, Department of Defense, or the United States Government. This material is declared a work of the U.S. Government and is not subject to copyright protection in the United States. AFIT/GE/ENG/12-02 RSA POWER ANALYSIS OBFUSCATION: A DYNAMIC FPGA ARCHITECTURE THESIS Presented to the Faculty Department of Electrical and Computer Engineering Graduate School of Engineering and Management Air Force Institute of Technology Air University Air Education and Training Command In Partial Fulfillment of the Requirements for the Degree of Master of Science in Electrical Engineering John W. -
THE MILLER–RABIN PRIMALITY TEST 1. Fast Modular
THE MILLER{RABIN PRIMALITY TEST 1. Fast Modular Exponentiation Given positive integers a, e, and n, the following algorithm quickly computes the reduced power ae % n. • (Initialize) Set (x; y; f) = (1; a; e). • (Loop) While f > 1, do as follows: { If f%2 = 0 then replace (x; y; f) by (x; y2 % n; f=2), { otherwise replace (x; y; f) by (xy % n; y; f − 1). • (Terminate) Return x. The algorithm is strikingly efficient both in speed and in space. To see that it works, represent the exponent e in binary, say e = 2f + 2g + 2h; 0 ≤ f < g < h: The algorithm successively computes (1; a; 2f + 2g + 2h) f (1; a2 ; 1 + 2g−f + 2h−f ) f f (a2 ; a2 ; 2g−f + 2h−f ) f g (a2 ; a2 ; 1 + 2h−g) f g g (a2 +2 ; a2 ; 2h−g) f g h (a2 +2 ; a2 ; 1) f g h h (a2 +2 +2 ; a2 ; 0); and then it returns the first entry, which is indeed ae. 2. The Fermat Test and Fermat Pseudoprimes Fermat's Little Theorem states that for any positive integer n, if n is prime then bn mod n = b for b = 1; : : : ; n − 1. In the other direction, all we can say is that if bn mod n = b for b = 1; : : : ; n − 1 then n might be prime. If bn mod n = b where b 2 f1; : : : ; n − 1g then n is called a Fermat pseudoprime base b. There are 669 primes under 5000, but only five values of n (561, 1105, 1729, 2465, and 2821) that are Fermat pseudoprimes base b for b = 2; 3; 5 without being prime. -
Lecture 10: Quadratic Residues
Lecture 10: Quadratic residues Rajat Mittal IIT Kanpur n Solving polynomial equations, anx + ··· + a1x + a0 = 0 , has been of interest from a long time in mathematics. For equations up to degree 4, we have an explicit formula for the solutions. It has also been shown that no such explicit formula can exist for degree higher than 4. What about polynomial equations modulo p? Exercise 1. When does the equation ax + b = 0 mod p has a solution? This lecture will focus on solving quadratic equations modulo a prime number p. In other words, we are 2 interested in solving a2x +a1x+a0 = 0 mod p. First thing to notice, we can assume that every coefficient ai can only range between 0 to p − 1. In the assignment, you will show that we only need to consider equations 2 of the form x + a1x + a0 = 0. 2 Exercise 2. When will x + a1x + a0 = 0 mod 2 not have a solution? So, for further discussion, we are only interested in solving quadratic equations modulo p, where p is an odd prime. For odd primes, inverse of 2 always exists. 2 −1 2 −2 2 x + a1x + a0 = 0 mod p , (x + 2 a1) = 2 a1 − a0 mod p: −1 −2 2 Taking y = x + 2 a1 and b = 2 a1 − a0, 2 2 Exercise 3. solving quadratic equation x + a1x + a0 = 0 mod p is same as solving y = b mod p. The small amount of work we did above simplifies the original problem. We only need to solve, when a number (b) has a square root modulo p, to solve quadratic equations modulo p. -
Modular Exponentiation: Exercises
Modular Exponentiation: Exercises 1. Compute the following using the method of successive squaring: (a) 250 (mod 101) (b) 350 (mod 101) (c) 550 (mod 101). 2. Using an example from this lecture, compute 450 (mod 101) with no effort. How did you do it? 3. Explain how we could have predicted the answer to problem 1(a) with no effort. 4. Compute the following using the method of successive squaring: 50 58 44 (a) (3) in Z=101Z (b) (3) in Z=61Z (c)(4) in Z=51Z. 5000 5. Compute (78) in Z=79Z, and explain why this calculation is so very trivial. 4999 What is (78) in Z=79Z? 60 6. Fermat's Little Theorem says that (3) = 1 in Z=61Z. Use this fact to 58 compute (3) in Z=61Z (see problem 4(b) above) without using successive squaring, but by computing the inverse of (3)2 instead, for instance by the Euclidean algorithm. Explain why this works. 7. We may see later on that the set of all a 2 Z=mZ such that gcd(a; m) = 1 is a group. Let '(m) be the number of elements in this group, which is often × × denoted by (Z=mZ) . It turns out that for each a 2 (Z=mZ) , some power × of a must be equal to 1. The order of any a in the group (Z=mZ) is by definition the smallest positive integer e such that (a)e = 1. × (a) Compute the orders of all the elements of (Z=11Z) . × (b) Compute the orders of all the elements of (Z=17Z) . -
Consecutive Quadratic Residues and Quadratic Nonresidue Modulo P
Consecutive Quadratic Residues And Quadratic Nonresidue Modulo p N. A. Carella Abstract: Let p be a large prime, and let k log p. A new proof of the existence of ≪ any pattern of k consecutive quadratic residues and quadratic nonresidues is introduced in this note. Further, an application to the least quadratic nonresidues np modulo p shows that n (log p)(log log p). p ≪ 1 Introduction Given a prime p 2, a nonzero element u F is called a quadratic residue, equiva- ≥ ∈ p lently, a square modulo p whenever the quadratic congruence x2 u 0mod p is solvable. − ≡ Otherwise, it called a quadratic nonresidue. A finite field Fp contains (p + 1)/2 squares = u2 mod p : 0 u < p/2 , including zero. The quadratic residues are uniformly R { ≤ } distributed over the interval [1,p 1]. Likewise, the quadratic nonresidues are uniformly − distributed over the same interval. Let k 1 be a small integer. This note is concerned with the longest runs of consecutive ≥ quadratic residues and consecutive quadratic nonresidues (or any pattern) u, u + 1, u + 2, . , u + k 1, (1) − in the finite field F , and large subsets F . Let N(k,p) be a tally of the number of p A ⊂ p sequences 1. Theorem 1.1. Let p 2 be a large prime, and let k = O (log p) be an integer. Then, the ≥ finite field Fp contains k consecutive quadratic residues (or quadratic nonresidues or any pattern). Furthermore, the number of k tuples has the asymptotic formulas p 1 k 1 (i) N(k,p)= 1 1+ O , if k 1. -
Lecture 19 1 Readings 2 Introduction 3 Shor's Order-Finding Algorithm
C/CS/Phys 191 Shor’s order (period) finding algorithm and factoring 11/01/05 Fall 2005 Lecture 19 1 Readings Benenti et al., Ch. 3.12 - 3.14 Stolze and Suter, Quantum Computing, Ch. 8.3 Nielsen and Chuang, Quantum Computation and Quantum Information, Ch. 5.2 - 5.3, 5.4.1 (NC use phase estimation for this, which we present in the next lecture) literature: Ekert and Jozsa, Rev. Mod. Phys. 68, 733 (1996) 2 Introduction With a fast algorithm for the Quantum Fourier Transform in hand, it is clear that many useful applications should be possible. Fourier transforms are typically used to extract the periodic components in functions, so this is an immediate one. One very important example is finding the period of a modular exponential function, which is also known as order-finding. This is a key element of Shor’s algorithm to factor large integers N. In Shor’s algorithm, the quantum algorithm for order-finding is combined with a series of efficient classical computational steps to make an algorithm that is overall polynomial in the input size 2 n = log2N, scaling as O(n lognloglogn). This is better than the best known classical algorithm, the number field sieve, which scales superpolynomially in n, i.e., as exp(O(n1/3(logn)2/3)). In this lecture we shall first present the quantum algorithm for order-finding and then summarize how this is used together with tools from number theory to efficiently factor large numbers. 3 Shor’s order-finding algorithm 3.1 modular exponentiation Recall the exponential function ax. -
STORY of SQUARE ROOTS and QUADRATIC RESIDUES
CHAPTER 6: OTHER CRYPTOSYSTEMS and BASIC CRYPTOGRAPHY PRIMITIVES A large number of interesting and important cryptosystems have already been designed. In this chapter we present several other of them in order to illustrate other principles and techniques that can be used to design cryptosystems. Part I At first, we present several cryptosystems security of which is based on the fact that computation of square roots and discrete logarithms is in general unfeasible in some Public-key cryptosystems II. Other cryptosystems and groups. cryptographic primitives Secondly, we discuss one of the fundamental questions of modern cryptography: when can a cryptosystem be considered as (computationally) perfectly secure? In order to do that we will: discuss the role randomness play in the cryptography; introduce the very fundamental definitions of perfect security of cryptosystem; present some examples of perfectly secure cryptosystems. Finally, we will discuss, in some details, such very important cryptography primitives as pseudo-random number generators and hash functions . IV054 1. Public-key cryptosystems II. Other cryptosystems and cryptographic primitives 2/65 FROM THE APPENDIX MODULAR SQUARE ROOTS PROBLEM The problem is to determine, given integers y and n, such an integer x that y = x 2 mod n. Therefore the problem is to find square roots of y modulo n STORY of SQUARE ROOTS Examples x x 2 = 1 (mod 15) = 1, 4, 11, 14 { | } { } x x 2 = 2 (mod 15) = and { | } ∅ x x 2 = 3 (mod 15) = { | } ∅ x x 2 = 4 (mod 15) = 2, 7, 8, 13 { | } { } x x 2 = 9 (mod 15) = 3, 12 QUADRATIC RESIDUES { | } { } No polynomial time algorithm is known to solve the modular square root problem for arbitrary modulus n.