Difference between revisions of "Twin Prime Conjecture"

Revision as of 18:15, 19 May 2019

The Twin Prime Conjecture is a conjecture (i.e., not a theorem) that states that there are infinitely many pairs of twin primes, i.e. pairs of primes that differ by $2$ .

Failed Proofs

Using an infinite series

One possible strategy to prove the infinitude of twin primes is an idea adopted from the proof of Dirichlet's Theorem. If one can show that the sum

$B=\frac{1}{3}+\frac{1}{5}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\frac{1}{17}+\frac{1}{19}+\cdots$

of the reciprocals of twin primes diverges, this would imply that there are infinitely many twin primes. Unfortunately, it has been shown that this sum converges to a constant $B$ , known as Brun's constant. This could mean either that there are finitely many twin prime pairs or that they are spaced "too far apart" for that series to diverge.

Yitang Zhang approach

A weaker version of twin prime conjecture was proved by Yitang Zhang in 2013. This version stated that there are infinitely many pairs of primes that differ by a finite number. The number Yitang chose was 7,000,000. Terence Tao and other people has reduced that boundary to 246 more numbers.

Elementary proof

Let $P_s$ be the multiplication of the first s prime numbers. Let $p_s$ be the sth prime number Let $A_s$ be the set of numbers relatively prime to $P_s$ and less than $P_s$ .

$\{m_1P_s + a_1\}$ and $\{m_2P_s + a_2\}$ where $a$ in $A_s$ and $0 \leq m<P_s$ and $a_1+2=a_2$ Pair up numbers generated from two arithmetic progression where $m_1 = m_2$ If it is not possible to generate a non-prime in each pair then there exist a twin prime.

The base case for numbers which differ by $2$ in $A_3$ is $11$ and $13$ . Induction there will always be two numbers which differ by $2$ in $A_s$ $s \geq 3$ .

Let $0 \leq m < p_{s+1}$

$\{mP_s+a_1\}$ and $\{mP_s+a_2\}$ will propagate $p_{s+1} - 1$ pairs of elements in $A_{s+1}$ which differ by $d$ where $a_1+d=a_2$ and $d=np_{s+1}$ and $n>0$ because only the unique values $m_1P_s+a_1$ and $m_2 P_s+a_2$ in their respective arithmetic progression has the factor of $p_{s+1}$ when $m_1=m_2$

$\{mP_s+a_1\}$ and $\{mP_s+a_2\}$ will propagate $p_{s+1} - 2$ pairs of elements in $A_{s+1}$ which differ by $d$ where $a_1+d=a_2$ and $d \neq np_{s+1}$ and $n>0$ because only the unique values $m_1P_s+a_1$ and $m_2 P_s+a_2$ in their respective arithmetic progression has the factor of $p_{s+1}$ when $m_1 \neq m_2$

All non-primes numbers generated by $\{mP_s+a\}-\{1\}$ where $a$ in $A_s$ and $0 \leq m<P_s$ can also be found in $\{f \times g | 1<f<P_s, 1<g, f=2n+1, 1 \leq n\}$ Therefore removing all numbers from $\{m P_s+a\}-\{1\}$ with odd factors between and including $3$ to $P_s-1$ will either leave an empty set or a set only containing prime numbers.

Using the fact that there is a fix set of sequential numbers between numbers with the same factor f in arithmetic progression.

$\{mP_s + a_1\}$ where $0 \leq m$ and $a_1$ is an unique pick from $A_s$ if $f$ is a factor of a number in $\{m P_s + a_1\}$ then there is an unique value in $\{mP_s + a_1\}$ which $f$ is a factor when $n \leq m<n+f$ , $0 \leq n$

Mark possible non-prime in pair values generated from arithmetic progression $m_1P_s+a_1$ and $m_2P_s+a_2$ where values are paired if $m_1=m_2$ .

The largest factor to eliminate $P_s-1$ is smaller than the number of pairs elements generate by two arithmetic progressions in $\{mP_s+a\}$ where $0 \leq m<P_s$ and $a$ in $A_s$ Can guarantee there are $P_s-1-2$ elements without the factor of $P_s-1$ in a consecutive sequence of $P_s-1$ elements from arithmetic progression $\{mP_s+a\}$ where the two numbers with factor of $P_s-1$ are generated in two different arithmetic progression in two different pairs. Assume the remaining $P_s-1-2$ pairs without factor of $P_s-1$ are in a consecutive sequence eliminate the next smaller odd number which differs by $2$ . Assume the remaining $P_s-1-2-2$ pairs without factor of $P_s-1-2$ are in a consecutive sequence eliminate the next smaller odd number which differs by $2$ . Repeat until the number of elements in consecutive sequence is $P_s-1-2n =3$ . Removing numbers with factor of $3$ . There must be a pair of numbers where both of them are prime numbers. There must be infinite number of twin primes.

This article is a stub. Help us out by expanding it.

Revision as of 18:13, 19 May 2019 (view source) Zhangaik (talk \| contribs) (→‎Elementary proof) ← Older edit		Revision as of 18:15, 19 May 2019 (view source) Zhangaik (talk \| contribs) (→‎Elementary proof) Newer edit →
Line 33:		Line 33:


−	All non-primes numbers generated by <math>\{mP_s+a\}-\{1\}</math> where <math>a</math> in <math>A_s</math> and <math>0 \leq m<P_s</math> can also be found in <math>\{f \times g \| 1<f<P_s, 1<g, f=2n+1, 1<=n\}</math>	+	All non-primes numbers generated by <math>\{mP_s+a\}-\{1\}</math> where <math>a</math> in <math>A_s</math> and <math>0 \leq m<P_s</math> can also be found in <math>\{f \times g \| 1<f<P_s, 1<g, f=2n+1, 1 \leq n\}</math>
	Therefore removing all numbers from <math>\{m P_s+a\}-\{1\}</math> with odd factors between and including <math>3</math> to <math>P_s-1</math> will either leave an empty set or a set only containing prime numbers.		Therefore removing all numbers from <math>\{m P_s+a\}-\{1\}</math> with odd factors between and including <math>3</math> to <math>P_s-1</math> will either leave an empty set or a set only containing prime numbers.

Page

Toolbox

Search