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A primality test is an algorithm for determining whether an input number is prime.Among other fields of mathematics, it is used for cryptography.Unlike integer factorization, primality tests do not generally give prime factors, only stating whether the input number is prime or not.
This occurs for example when n is a probable prime to base a but not a strong probable prime to base a. [20]: 1402 If x is a nontrivial square root of 1 modulo n, since x 2 ≡ 1 (mod n), we know that n divides x 2 − 1 = (x − 1)(x + 1); since x ≢ ±1 (mod n), we know that n does not divide x − 1 nor x + 1.
Fermat's little theorem states that if p is prime and a is not divisible by p, then a p − 1 ≡ 1 ( mod p ) . {\displaystyle a^{p-1}\equiv 1{\pmod {p}}.} If one wants to test whether p is prime, then we can pick random integers a not divisible by p and see whether the congruence holds.
The AKS primality test (also known as Agrawal–Kayal–Saxena primality test and cyclotomic AKS test) is a deterministic primality-proving algorithm created and published by Manindra Agrawal, Neeraj Kayal, and Nitin Saxena, computer scientists at the Indian Institute of Technology Kanpur, on August 6, 2002, in an article titled "PRIMES is in P". [1]
A prime number is a natural number that has exactly two distinct natural number divisors: the number 1 and itself. To find all the prime numbers less than or equal to a given integer n by Eratosthenes' method: Create a list of consecutive integers from 2 through n: (2, 3, 4, ..., n). Initially, let p equal 2, the smallest prime number.
If a and p are coprime numbers such that a p−1 − 1 is divisible by p, then p need not be prime. If it is not, then p is called a (Fermat) pseudoprime to base a. The first pseudoprime to base 2 was found in 1820 by Pierre Frédéric Sarrus: 341 = 11 × 31. [12] [13] A number p that is a Fermat pseudoprime to base a for every number a coprime ...
The Mersenne number M 3 = 2 3 −1 = 7 is prime. The Lucas–Lehmer test verifies this as follows. Initially s is set to 4 and then is updated 3−2 = 1 time: s ← ((4 × 4) − 2) mod 7 = 0. Since the final value of s is 0, the conclusion is that M 3 is prime. On the other hand, M 11 = 2047 = 23 × 89 is not prime
A definite bound on the prime factors is possible. Suppose P i is the i 'th prime, so that P 1 = 2, P 2 = 3, P 3 = 5, etc. Then the last prime number worth testing as a possible factor of n is P i where P 2 i + 1 > n; equality here would mean that P i + 1 is a factor.