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For if every even number greater than 4 is the sum of two odd primes, adding 3 to each even number greater than 4 will produce the odd numbers greater than 7 (and 7 itself is equal to 2+2+3). In 2013, Harald Helfgott released a proof of Goldbach's weak conjecture. [2]
For instance, if m is odd, then n − m is also odd, and if m is even, then n − m is even, a non-trivial relation because, besides the number 2, only odd numbers can be prime. Similarly, if n is divisible by 3, and m was already a prime other than 3, then n − m would also be coprime to 3 and thus be slightly more likely to be prime than a ...
Paul Erdős gave a proof [11] that also relies on the fundamental theorem of arithmetic. Every positive integer has a unique factorization into a square-free number r and a square number s 2. For example, 75,600 = 2 4 3 3 5 2 7 1 = 21 ⋅ 60 2. Let N be a positive integer, and let k be the number of primes less than or equal to N. Call those ...
If one considers only the odd numbers in the sequence generated by the Collatz process, then each odd number is on average 3 / 4 of the previous one. [16] (More precisely, the geometric mean of the ratios of outcomes is 3 / 4 .) This yields a heuristic argument that every Hailstone sequence should decrease in the long run ...
2013 Ternary Goldbach conjecture: Every odd number greater than 5 can be expressed as the sum of three primes. 2014 Proof of Erdős discrepancy conjecture for the particular case C=2: every ±1-sequence of the length 1161 has a discrepancy at least 3; the original proof, generated by a SAT solver, had a size of 13 gigabytes and was later ...
The classical proof. It is sufficient to prove the theorem for every odd prime number p. This immediately follows from Euler's four-square identity (and from the fact that the theorem is true for the numbers 1 and 2). The residues of a 2 modulo p are distinct for every a between 0 and (p − 1)/2 (inclusive). To see this, take some a and define ...
The full statement of Vinogradov's theorem gives asymptotic bounds on the number of representations of an odd integer as a sum of three primes. The notion of "sufficiently large" was ill-defined in Vinogradov's original work, but in 2002 it was shown that 10 1346 is sufficiently large.
This part of the proof of the odd-order theorem takes over 100 journal pages. A key step is the proof of the Thompson uniqueness theorem , stating that abelian subgroups of normal rank at least 3 are contained in a unique maximal subgroup, which means that the primes p for which the Sylow p -subgroups have normal rank at most 2 need to be ...