Search results
Results From The WOW.Com Content Network
Let a be an integer that is not a square number and not −1. Write a = a 0 b 2 with a 0 square-free. Denote by S(a) the set of prime numbers p such that a is a primitive root modulo p. Then the conjecture states S(a) has a positive asymptotic density inside the set of primes. In particular, S(a) is infinite.
f has degree at most p − 2 (since the leading terms cancel), and modulo p also has the p − 1 roots 1, 2, ..., p − 1. But Lagrange's theorem says it cannot have more than p − 2 roots. Therefore, f must be identically zero (mod p), so its constant term is (p − 1)! + 1 ≡ 0 (mod p). This is Wilson's theorem.
However, it does not contain all the prime numbers, since the terms gcd(n + 1, a n) are always odd and so never equal to 2. 587 is the smallest prime (other than 2) not appearing in the first 10,000 outcomes that are different from 1. Nevertheless, in the same paper it was conjectured to contain all odd primes, even though it is rather inefficient.
If really is prime, it will always answer yes, but if is composite then it answers yes with probability at most 1/2 and no with probability at least 1/2. [132] If this test is repeated n {\displaystyle n} times on the same number, the probability that a composite number could pass the test every time is at most 1 / 2 ...
In computational number theory, the Lucas test is a primality test for a natural number n; it requires that the prime factors of n − 1 be already known. [ 1 ] [ 2 ] It is the basis of the Pratt certificate that gives a concise verification that n is prime.
This is a list of articles about prime numbers.A prime number (or prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. By Euclid's theorem, there are an infinite number of prime numbers.
Get AOL Mail for FREE! Manage your email like never before with travel, photo & document views. Personalize your inbox with themes & tabs. You've Got Mail!
John Selfridge has conjectured that if p is an odd number, and p ≡ ±2 (mod 5), then p will be prime if both of the following hold: 2 p−1 ≡ 1 (mod p), f p+1 ≡ 0 (mod p), where f k is the k-th Fibonacci number. The first condition is the Fermat primality test using base 2. In general, if p ≡ a (mod x 2 +4), where a is a quadratic non ...