Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a number $n\ge 3$, then one of these is true: \begin{equation} \begin{cases}2n = (6m-1)+P, \ \ \ P \in \mathbb P, \ 6m-1 \in \mathbb P, \ 6m+1 \in \mathbb P \ \ \ \ (1) \\ 2n-1 \in \mathbb P, \ 2n-1 = P + 6k,\ 6k+1 \in \mathbb P \ \ \ \ (2) \end{cases} \end{equation}

In other words, any even number $2n$ greater than $4$ can be written either as a sum of a prime and another prime which is lower of a pair of twin primes or, if this cannot be done, then $2n-1$ is a prime and it can be written as a sum of a prime and a number divisible by six which is one smaller than a prime.

My justification for this is the Goldbach's conjecture.

Assume every even number greater than $2$ can be written as a sum of two primes. Then list every possible way to write $2n+2$ as sum of two primes: $h = \{P_1+P_2,P_3+P4... \}$

Now $2n = 2n-2$. Then $2n$ can be written as a sum of two primes as well, so there must be either some $P_n$ which is a component of some sum in $h$, for which $P_n-2 \in \mathbb P$ or, if no such prime exists, but we have to get the primes for $2n$ from somewhere, so some $P_n$ must be 3 and if $2n+2=3+P_s$, $2n = 1+P_s$ and $P_s$ can be expressed as a sum of a prime and a number $N$, such that $N+1 \in \mathbb P.$

Assume then that what I stated at the beginning is true. Then $2n$ can be written in one of the two forms and thus $2n+2$ is automatically a sum of two primes.

The other form is necessary, take as an example the smallest number for which the $(1)$ fails: $$38 = 19+19 = 7+31 = 37+1. $$ None of these is a lower of a twin prime pair, but $37$ is a prime and $37 = 31 + 6$ and $6 +1 = 7$ is a prime.

share|improve this question
    
How can $2n=6n-1+P$? Do you mean a different variable, like $2n=6m-1+P$? –  Thomas Andrews Nov 26 '13 at 1:18
    
@ThomasAndrews yes, different variable. Will fix. –  Valtteri Nov 26 '13 at 1:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.