# Twin prime conjecture - please check proof. [closed]

Prove infinite twin primes using deduction:

L M R T(n)

11 12 13 1

17 18 19 2

23 24 25 3 (3+5n)

29 30 31 4

35 36 37 5 (5n, 5+7n)

41 42 43 6

47 48 49 7 (7n)

53 54 55 8 (8+11n)

59 60 61 9

65 66 67 10

71 72 73 11

77 78 79 12 (12+11n)

83 84 85 13

89 90 91 14

95 96 97 15

Possible twin primes appear from two consecutive odd numbers between an even number.

odd, even, odd.


There is always one divisor of three within three consecutive numbers. The two odd numbers cannot be divisor of three in order to be prime - 3, 4, & 5 being the only exception. Thus, probable twin primes appear from two consecutive odd number between a divisor of six.

odd, 6k, odd. (k is any possible integers)


Furthermore, all positive integers can be expressed as 6k + i for some integer k and for i = 0, 1, 2, 3, 4, 5. 2 is the divisor of 6k+0, 6k+2, 6k+4 and 3 is the divisor of 6k+3. Thus probable primes are 6k-1 (or 6k+5) and 6k+1.

The fundamental theorem of arithmetic (unique factorization) states that every integer greater than one either is prime itself or is the product of prime numbers. The process of Eratosthenes Sieve or algorithm cancel composite numbers from positive integers.

Examining the left column (L) and cancelling divisor of 5 (35, 65, 95, 125, 155, 185 …) means those numbers cancelled are not prime numbers. Cancelling divisor of 7 (35, 77, 119, 161, …) means those numbers cancelled are not prime numbers. Cancelling divisor of 11 (77, 143, …) means those numbers cancelled are not prime numbers. The process of cancelling divisor of 13, 17, 19, … does not stop to find infinite number of primes.

The right column (R) undergoes the similar cancellation process as the left column (L). Cancelling divisor of 5 (25, 55, 85, 115, 145, 175, …), divisor of 7 (49, 91, 133, 175, …), divisor of 11 (55, 121, 187, …), divisor of prime numbers means those numbers cancelled are not prime numbers.

Both left and right columns continue the process of cancelling divisor of prime numbers. Only those numbers that have not been cancelled are prime numbers. Twin primes are those rows where both numbers on the left and right columns that have not been cancelled.

Both left and right columns are transformed and mapped onto a new column T. Column L, R, & T start from 11, 13, & 1 respectively. Each row of Column L & R (6k-1 & 6k+1) is mapped onto Column T (k=2 or n=1).

Column L divisor 5 ---> Column T(n) 5n

Column L divisor 7 ---> Column T(n) 5 + 7n

Column L divisor 11 ---> Column T(n) 12 + 11n

...

Column R divisor 5 ---> Column T(n) 3 + 5n

Column R divisor 7 ---> Column T(n) 7n

Column R divisor 11 ---> Column T(n) 8 + 11n

...

Similarity only numbers (n) in column T that have not been cancelled by the linear functions are twin primes.

To cancel all positive integers in column T, five linear functions having coefficient of five are needed: 5n, 5n+1, 5n+2, 5n+3, & 5n+4.

To cancel all positive integers in column T, seven linear functions having coefficient of seven are needed: 7n, 7n+1, 7n+2, 7n+3, 7n+4, 7n+5, & 7n+6.

In general, p linear functions having coefficient of p are needed to cancel all positive integers in column T.

However, column T only has two linear functions having the same coefficient because column T has been transformed or mapped from two columns (L & R). It is also impossible to combine other linear function together evenly because their coefficients are coprime. Thus it is impossible to cancel all positive integers in column T.

There are infinite twin primes by deduction.

Leow (2013)

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"Both left and right columns are transformed and mapped onto a new column T" - Could you explain what the entries under $T(n)$ are? As in, what are the transformations and maps? –  ronno Dec 21 '13 at 8:16
What do you mean by "cancelled by the linear functions"? Also, note that we don't have to cancel all positive integers in column $T$ for the twin primes conjecture to be false. We need only cancel all but finitely many of them. –  Cameron Buie Dec 21 '13 at 8:20
–  AHH Dec 21 '13 at 8:21
Learn to use MathJax. –  Student Dec 21 '13 at 9:11
“This question does not show any research effort; it is unclear or not useful.”? –  k.stm Dec 21 '13 at 15:39
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## closed as off-topic by Keenan Kidwell, Sami Ben Romdhane, anorton, Eric Stucky, PaulDec 28 '13 at 4:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is not about mathematics, within the scope defined in the help center." – Keenan Kidwell, Sami Ben Romdhane, anorton
If this question can be reworded to fit the rules in the help center, please edit the question.

Before the introduction of $T(n)$, what has been done is to observe that the set $S_L$ of positive integers $k$ such that $6k-1$ is prime is infinite , and that the set $S_R$ of positive integers $k$ such that $6k+1$ is prime is infinite. You did not give a proof of those observations, but they are correct, as special cases of Dirichlet's theorem on primes in arithmetic progressions.

The goal is to prove that the intersection of $S_L$ and $S_R$ is infinite.

In the second part of the argument, about $T(n)$, you consider every case where $6n-1$ or $6n+1$ (or both) are not prime, and remove from the set of possible $n$'s, two arithmetic progressions (linear functions) corresponding to all cases where the smallest prime factor of $6n-1$ divides the Left number, or where the smallest prime divisor of $6n+1$ divides the Right number. Or maybe you remove several progressions, using all of the prime factors, it was not clear. In either case, this is a sieve process that will reduce the search space for twin primes, but the entire problem is to prove that there are infinitely many $n$ that are not removed by the sieve.

You want to prove this by analogy with the Erastosthenes sieve, in which infinitely many primes are left over. But this does not work, because we are sieving from two sides. The fact that the sieves on the Left and the Right are somehow "thinner" than the Erastosthenes sieve might be a way to prove that there are infinitely many unsieved numbers on the Left, and infinitely many unsieved on the Right, but unfortunately

this does not say anything about whether the combination of two thin sieves, working from both sides, is "thick" and removes all $n$ larger than some number.

To be concrete, maybe all $n > 10^{100}$ are sieved out on one side or the other. We believe that this will not happen, but you have not given any proof that it is impossible.

If the twin prime sieve is done correctly (so that the un-sieved numbers are primes, and not only some collection of numbers for which some of the possible divisors have been excluded) then it is the same as the usual description of the sieve: for every prime number $p$, remove all integers $x > p$ such that $p$ divides $x$ or $x+2$. This is an extremely well-studied process and much is known about it. The exclusion of $k = 2$ arithmetic progressions (the "two linear functions" in the question) instead of the $1$ used in Erastosthenes sieve for ordinary primes, means that the frequency of twin primes near a large number $n$ is much smaller; it is conjectured to be approximately the $k$-th power of the density of primes near $n$ (of order $\frac{1}{\log^k x}$ instead of the $\frac{1}{\log x}$ in the Prime Number Theorem). The infinite number of twin primes is very easy to prove from that conjecture, which is the "real" Twin Prime hypothesis as understood by mathematicians, but nobody has any good idea how to prove the conjecture after 200 years that it has been circulating.

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This is not actually important, but you might want to increase the concrete bound, since $3756801695685\times2^{666669}\pm 1$ are known to be prime. :) –  ronno Dec 21 '13 at 8:59