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Not all Euclid numbers are prime. E 6 = 13# + 1 = 30031 = 59 × 509 is the first composite Euclid number.. Every Euclid number is congruent to 3 modulo 4 since the primorial of which it is composed is twice the product of only odd primes and thus congruent to 2 modulo 4.
Legendre's conjecture, proposed by Adrien-Marie Legendre, states that there is a prime number between and (+) for every positive integer. [1] The conjecture is one of Landau's problems (1912) on prime numbers, and is one of many open problems on the spacing of prime numbers.
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.
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.
Euclid offered a proof published in his work Elements (Book IX, Proposition 20), [1] which is paraphrased here. [2] Consider any finite list of prime numbers p 1, p 2, ..., p n. It will be shown that there exists at least one additional prime number not included in this list. Let P be the product of all the prime numbers in the list: P = p 1 p ...
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 ...
If either p n # + 1 or p n # − 1 is a primorial prime, it means that there are larger primes than the nth prime (if neither is a prime, that also proves the infinitude of primes, but less directly; each of these two numbers has a remainder of either p − 1 or 1 when divided by any of the first n primes, and hence all its prime factors are ...
let s > 0 and d odd > 0 such that n − 1 = 2 s d # by factoring out powers of 2 from n − 1 repeat k times: a ← random(2, n − 2) # n is always a probable prime to base 1 and n − 1 x ← a d mod n repeat s times: y ← x 2 mod n if y = 1 and x ≠ 1 and x ≠ n − 1 then # nontrivial square root of 1 modulo n return (“multiple of ...