Miller`s primality test - Mathematisch Instituut Leiden
... p - 1 is divisible by at least the same power of 2 as 4- 1 is,Putd=p4ifp1 andq- 1 areinfact divisible by the same power of 2, and d =p otheNvise. Notice that cl f 1 mod 4. Denote by a the smallest positive integer for which the Jacobi symbol (a/d) equals -1. From (3) it is obvious that a is a prime ...
... p - 1 is divisible by at least the same power of 2 as 4- 1 is,Putd=p4ifp1 andq- 1 areinfact divisible by the same power of 2, and d =p otheNvise. Notice that cl f 1 mod 4. Denote by a the smallest positive integer for which the Jacobi symbol (a/d) equals -1. From (3) it is obvious that a is a prime ...
Full text
... digit length n one of two things must happen; either in the sequence of iterates we find one with fewer than n digits, which property will then persist, or else the sequence becomes periodic eventually with all the numbers in the period having n digits. Within a period, the period-length p, is the n ...
... digit length n one of two things must happen; either in the sequence of iterates we find one with fewer than n digits, which property will then persist, or else the sequence becomes periodic eventually with all the numbers in the period having n digits. Within a period, the period-length p, is the n ...
Problem Fields in Elementary Arithmetic
... Another problem field with opposite handling could be the working with sums of two or three consecutive numbers. We start with 1+2 = 3, 2+3 = 5, 3+4 = 7, 4+5 = 9 and can discover the following statement: “Only odd numbers do appear as results for all conceivable sums of two consecutive numbers”. A r ...
... Another problem field with opposite handling could be the working with sums of two or three consecutive numbers. We start with 1+2 = 3, 2+3 = 5, 3+4 = 7, 4+5 = 9 and can discover the following statement: “Only odd numbers do appear as results for all conceivable sums of two consecutive numbers”. A r ...
No Slide Title
... 9. The sum of two consecutive even integers is 106. Find the integers. 10. The sum of two consecutive odd integers is –40. Find the integers. 11. Find three consecutive integers if twice the largest is 16 less than three times the smallest. Complete solutions follow this slide, so work these first b ...
... 9. The sum of two consecutive even integers is 106. Find the integers. 10. The sum of two consecutive odd integers is –40. Find the integers. 11. Find three consecutive integers if twice the largest is 16 less than three times the smallest. Complete solutions follow this slide, so work these first b ...
Lecture Notes on Primality Testing
... This algorithm is obviously correct. However, because the for-loop has O( n) iterations, the algorithm does not have running time polynomial in the number of input bits. (Consider the case where n is an integer ...
... This algorithm is obviously correct. However, because the for-loop has O( n) iterations, the algorithm does not have running time polynomial in the number of input bits. (Consider the case where n is an integer ...
Euler Totient Function ø(n)
... Discrete Logarithms or Indices • the inverse problem to exponentiation is to find the discrete logarithm of a number modulo p • that is to find x where ax = b mod p • written as x=loga b mod p or x=inda,p(b) • if a is a primitive root then always exists, otherwise may not – x = log3 4 mod 13 (x st ...
... Discrete Logarithms or Indices • the inverse problem to exponentiation is to find the discrete logarithm of a number modulo p • that is to find x where ax = b mod p • written as x=loga b mod p or x=inda,p(b) • if a is a primitive root then always exists, otherwise may not – x = log3 4 mod 13 (x st ...
Collatz conjecture
The Collatz conjecture is a conjecture in mathematics named after Lothar Collatz, who first proposed it in 1937. The conjecture is also known as the 3n + 1 conjecture, the Ulam conjecture (after Stanisław Ulam), Kakutani's problem (after Shizuo Kakutani), the Thwaites conjecture (after Sir Bryan Thwaites), Hasse's algorithm (after Helmut Hasse), or the Syracuse problem; the sequence of numbers involved is referred to as the hailstone sequence or hailstone numbers (because the values are usually subject to multiple descents and ascents like hailstones in a cloud), or as wondrous numbers.Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called ""Half Or Triple Plus One"", or HOTPO) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has also been called oneness.Paul Erdős said about the Collatz conjecture: ""Mathematics may not be ready for such problems."" He also offered $500 for its solution.