- This is another attempt to prove Twin Prime Conjecture.

I am sure there might be flaws, please be gentle :-)

To Prove:

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There are infinitely many twin primes.

Assumtion:

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There are twin primes between the squares of each prime in a twin

prime pair.

Proof Method:

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If for every twin prime square there are twin primes, inductively,

there would be infinite twin primes.

Illustration:

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(Please check the files section for a program to calculate twin

primes).

Twin Pair Square Pair Primes in Square Range

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3,5 - 9,25 11,13 : 17,19

5,7 - 25,49 29,31 : 41,43

41,43 - 1681,1849 1697,1699 : 1721,1723 : 1787,1789

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Proof:

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Let p and p+2 be a twin pair. Then, p^2 and (p+2)^2 will be the their

square range.

If this twin exists in the primes less than p+3, then in the range of

numbers between p^2 and (p+2)^2, the prime density should be less

than the previous range for us to not have a twin prime in this

square range. This would invalidate our assumption.

The contrary will prove the conjecture.

We compare these prime densities using Prime Number Theorum.

For primes including p+2, we take p+3.

Primes less than p+3 can be given by p + 3 / log (p+3)

This can be written as p + 3 / log p + log 3

Ignoring log 3 we get, p + 3 / log p ---- 1

Primes between p^2 and (p+2)^2 will be:

p^2 + 4p + 4 / log (p+2) ^ 2 - p^2 / log p ^2

p^2 + 4p + 4 / (2 log (p+2)) - p^2 / 2 log p

p^2 + 4p + 4 / 2 log p + 2 log 2 - p^2 / 2 log p

Ignoring 2 log 2,

p^2 + 4p + 4 / 2 log p - p^2 / 2 log p

Which reduces to:

2(p+1)/log p ---- 2

Comparing 1 and 2

2(p+1) / log p : p + 3 / log p

Gives,

2p : p + 1

This clearly shows that unless p = 1, 2p > p + 1 for all p > 1.

Hence there will be atleast one twin prime pair (or more) in the twin

prime square range.

Humbly,

Suresh - Suresh Batta wrote:

> Can we now not say that, since both the ranges have comparable prime

Definitely no. The only "comparable prime quantities" are about the _expected_

> quantities, if a P+3 has a twin prime, then squared range

> ((p+2)^2, p^2) should have too have a twin prime?

number of primes based on unproven heuristics. Even if this expectation could be

proven reasonably accurate, it would not say anything about the number of twin

primes.

p/log p is the approximate number of primes below p.

The prime number theorem says it is asymptotically right. This means the

relative error (NOT the absolute error) tends to 0 when p tends to infinite.

However, as Decio notes, this and better known approximations are too poor to

say anything about the number of primes (let alone twin primes) from p to p+x

when x is much smaller than p.

It can only be used to say things like:

The _average_ number of primes in an interval of x numbers _around_ the size of

p is approximately x/log p.

Little is known about how large the deviation from such averages can be.

To illustrate possible deviations, here are the most extreme values known around

100 digits:

There are 11 primes (0.16 expected) among the 104-digit numbers p to p+36 for

p = 24698258*239# + 28606476153371, found by Norman Luhn and I.

There are 0 primes (30 expected) among the 93-digit numbers c to c+6378 for

c = 5629854038470321802219554908853741163682800524507382035301697914566243\

83980052820124370178769, found by Torbjörn Alm and I.

There probably exists far more extreme cases.

As far as twin primes go, there is not even anything known about averages.

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Jens Kruse Andersen