CS 161 Computer Security Fall 2005 Joseph/Tygar/Vazirani/Wagner
... large number seems intractable. However, the closely related problem of testing a number for primality is easy, since Fermat’s theorem forms the basis of a kind of litmus test which helps decide whether a number is prime or not, without explicitly indentifying its factors. The idea is that to test i ...
... large number seems intractable. However, the closely related problem of testing a number for primality is easy, since Fermat’s theorem forms the basis of a kind of litmus test which helps decide whether a number is prime or not, without explicitly indentifying its factors. The idea is that to test i ...
29 APPROXIMATION EXPONENTS FOR FUNCTION
... We only need to show that other roots β of the Riccati equation cannot come arbitrarily close to the root α. Fix two other roots γ and δ. We use the well-known fact, easy to verify, that the cross-ratio of any 4 roots of the Riccati equation is a constant function, to deduce that the cross ratio (α ...
... We only need to show that other roots β of the Riccati equation cannot come arbitrarily close to the root α. Fix two other roots γ and δ. We use the well-known fact, easy to verify, that the cross-ratio of any 4 roots of the Riccati equation is a constant function, to deduce that the cross ratio (α ...
Transcendence of Periods: The State of the Art
... The known measures of linear independence of logarithms of algebraic numbers (lower bounds for linear combinations, with algebraic coefficients, of logarithms of algebraic numbers — see for instance [59]) imply that the absolute value of a nonzero integral of the form (6) is explicitly bounded from ...
... The known measures of linear independence of logarithms of algebraic numbers (lower bounds for linear combinations, with algebraic coefficients, of logarithms of algebraic numbers — see for instance [59]) imply that the absolute value of a nonzero integral of the form (6) is explicitly bounded from ...
Full text
... We refrain from describing our initial guesses in these cases, believing instead that the reader is ready to see some results. 3. Results We wish to consider nontrivial sequences (an ) of integers that satisfy the recurrence relation an+1 = ban + an−1 for some positive integer b, but whose initial t ...
... We refrain from describing our initial guesses in these cases, believing instead that the reader is ready to see some results. 3. Results We wish to consider nontrivial sequences (an ) of integers that satisfy the recurrence relation an+1 = ban + an−1 for some positive integer b, but whose initial t ...
1. Problems and Results in Number Theory
... infinitely many values of n for which all the integers n +a„ i = 1, . . . , k are primes is that for no prime p should the set a,, . . . , a k form a complete set of residues mod p . The condition is clearly necessary ; the proof of the sufficiency seems hopeless at present since e .g . the conjectu ...
... infinitely many values of n for which all the integers n +a„ i = 1, . . . , k are primes is that for no prime p should the set a,, . . . , a k form a complete set of residues mod p . The condition is clearly necessary ; the proof of the sufficiency seems hopeless at present since e .g . the conjectu ...
Solutions to Homework 1
... Proof. a). Since every polynomial of degree n ∈ N has at most n distinct roots in C, to show that the set of all algebraic numbers is countable, it suffices to show that there are countably many polynomials with integer coefficients. For each k ∈ N, we consider the number of polynomials an z n + an− ...
... Proof. a). Since every polynomial of degree n ∈ N has at most n distinct roots in C, to show that the set of all algebraic numbers is countable, it suffices to show that there are countably many polynomials with integer coefficients. For each k ∈ N, we consider the number of polynomials an z n + an− ...
40(1)
... Articles should be submitted using the format of articles in any current issues of THE FIBONACCI QUARTERLY. They should be typewritten or reproduced typewritten copies, that are clearly readable, double spaced with wide margins and on only one side of the paper. The full name and address of the auth ...
... Articles should be submitted using the format of articles in any current issues of THE FIBONACCI QUARTERLY. They should be typewritten or reproduced typewritten copies, that are clearly readable, double spaced with wide margins and on only one side of the paper. The full name and address of the auth ...
The Riddle of the Primes - Singapore Mathematical Society
... women can devote their entire lives to. But chess is finite. There are only so many games of chess. The number on the human scale is very large, but after you have listed all of the possible games of chess, there are still an infinite number of positive integers to go. Computers today can play very ...
... women can devote their entire lives to. But chess is finite. There are only so many games of chess. The number on the human scale is very large, but after you have listed all of the possible games of chess, there are still an infinite number of positive integers to go. Computers today can play very ...
Chapter 4
... Definition. Suppose that a, b ∈ Z and a 6= 0. Then we say that a divides b, denoted by a | b, if there exists c ∈ Z such that b = ac. In this case, we also say that a is a divisor of b, or b is a multiple of a. Example 4.1.1. For every a ∈ Z \ {0}, a | a and a | −a. Example 4.1.2. For every a ∈ Z, 1 ...
... Definition. Suppose that a, b ∈ Z and a 6= 0. Then we say that a divides b, denoted by a | b, if there exists c ∈ Z such that b = ac. In this case, we also say that a is a divisor of b, or b is a multiple of a. Example 4.1.1. For every a ∈ Z \ {0}, a | a and a | −a. Example 4.1.2. For every a ∈ Z, 1 ...
patterns in continued fraction expansions
... expansions in different bases the only thing that changes is how we represent those integers. Whether or not the expansion is finite or infinite does not change, even if we do change the base. For example, in base 10, 31/25 has continued fraction expansion [1,4,6], the expansion of 1/3 is [0,3], and ...
... expansions in different bases the only thing that changes is how we represent those integers. Whether or not the expansion is finite or infinite does not change, even if we do change the base. For example, in base 10, 31/25 has continued fraction expansion [1,4,6], the expansion of 1/3 is [0,3], and ...
Continued Fractions and the Euclidean Algorithm
... Proof. What is slightly strange about this important result is that while the {pr } and the {qr } are defined by the front end recursions, albeit double recursions, (2) and (3), the symbol [t1 , . . . , tr ] is defined by the back end recursion (1). The proof begins with the comment that the right- ...
... Proof. What is slightly strange about this important result is that while the {pr } and the {qr } are defined by the front end recursions, albeit double recursions, (2) and (3), the symbol [t1 , . . . , tr ] is defined by the back end recursion (1). The proof begins with the comment that the right- ...
Balancing sequence contains no prime number
... o Sub case 1: (a − 1) = 2 2 and (a + 1) = 2 p 2 . Solving this sub case we get p 2 = 3 , which is an absurd. o Sub case 2: (a − 1) = 2 p and (a + 1) = 2 2 p . Solving this sub case we get p = 1 , which is an absurd. o Sub case 3: (a − 1) = p 2 and (a + 1) = 23 . This sub case is not possible, since ...
... o Sub case 1: (a − 1) = 2 2 and (a + 1) = 2 p 2 . Solving this sub case we get p 2 = 3 , which is an absurd. o Sub case 2: (a − 1) = 2 p and (a + 1) = 2 2 p . Solving this sub case we get p = 1 , which is an absurd. o Sub case 3: (a − 1) = p 2 and (a + 1) = 23 . This sub case is not possible, since ...
Full text
... Suppose now that j divides both d and (c + d)/2. Then j will also divide 2(c + d)/2 − d = c. Because gcd(c, d) = 1, we have that j = 1. So both sides of Equation (9) are reduced fractions; hence, d = Fk , an odd Fibonacci number, and (c + d)/2 = Fk+1 , which gives c = 2Fk+1 − d. So Condition (i) mus ...
... Suppose now that j divides both d and (c + d)/2. Then j will also divide 2(c + d)/2 − d = c. Because gcd(c, d) = 1, we have that j = 1. So both sides of Equation (9) are reduced fractions; hence, d = Fk , an odd Fibonacci number, and (c + d)/2 = Fk+1 , which gives c = 2Fk+1 − d. So Condition (i) mus ...
An Unusual Continued Fraction
... where si = 2ν2 (i+1) and νp (x) is the p-adic valuation of x (the exponent of the largest power of p dividing x). To see that s is 2-regular, notice that every sequence in the 2-kernel is a linear combination of s itself and the constant sequence (1, 1, 1, . . .). The corresponding real number σ has ...
... where si = 2ν2 (i+1) and νp (x) is the p-adic valuation of x (the exponent of the largest power of p dividing x). To see that s is 2-regular, notice that every sequence in the 2-kernel is a linear combination of s itself and the constant sequence (1, 1, 1, . . .). The corresponding real number σ has ...