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Montgomery and Vaughan showed that the exceptional set of even numbers not expressible as the sum of two primes has a density zero, although the set is not proven to be finite. [9] The best current bounds on the exceptional set is E ( x ) < x 0.72 {\displaystyle E(x)<x^{0.72}} (for large enough x ) due to Pintz , [ 10 ] [ 11 ] and E ( x ) ≪ x ...
Example grid for a cross-figure puzzle with some answers filled in. A cross-figure (also variously called cross number puzzle or figure logic) is a puzzle similar to a crossword in structure, but with entries that consist of numbers rather than words, where individual digits are entered in the blank cells.
Some of them, like the 3rd problem, which was the first to be solved, or the 8th problem (the Riemann hypothesis), which still remains unresolved, were presented precisely enough to enable a clear affirmative or negative answer. For other problems, such as the 5th, experts have traditionally agreed on a single interpretation, and a solution to ...
In mathematics, Hilbert's second problem was posed by David Hilbert in 1900 as one of his 23 problems. It asks for a proof that arithmetic is consistent – free of any internal contradictions. Hilbert stated that the axioms he considered for arithmetic were the ones given in Hilbert (1900) , which include a second order completeness axiom.
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.
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 ...
If p is an odd prime and p − 1 = 2 s d with s > 0 and d odd > 0, then for every a coprime to p, either a d ≡ 1 (mod p) or there exists r such that 0 ≤ r < s and a 2 r d ≡ −1 (mod p). This result may be deduced from Fermat's little theorem by the fact that, if p is an odd prime, then the integers modulo p form a finite field , in which ...
The prime power solution can be applied with further exponentiation of prime numbers, resulting in very large room numbers even given small inputs. For example, the passenger in the second seat of the third bus on the second ferry (address 2-3-2) would raise the 2nd odd prime (5) to 49, which is the result of the 3rd odd prime (7) being raised ...