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 November 13th, 2010, 11:52 AM #1 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 new conjecture on prime numbers .... i hope for any prime number p bigger or equal to 13 ;p>=13 there exists at least one k such that both p-6k and p+6k are prime numbers for example : if p=13 k=1 coz 13-6 =7 and 13+6=19 which r both primes p=31 we have two k : k=2 and k=3 coz 31-12=19 and 31+12=43 and 31-18=13 and 31+18=47 which r all primes actually i checked it for many cases it always worked u can try urself ....... any ideas how to prove this conjecture thanks in advance...
November 13th, 2010, 01:40 PM   #2
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Re: new conjecture on prime numbers .... i hope

Quote:
 Originally Posted by islam any ideas how to prove this conjecture
We don't have the mathematical technology to prove conjectures like this yet. It's easy to make them, hard to prove them. This one is undoubtedly true -- it holds for p < 100,000,000 and there should be many k that make a given p > 100,000,000 work. (For example, the smallest such p has 404,753 k that make it work.)

November 13th, 2010, 02:29 PM   #3
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Re: new conjecture on prime numbers .... i hope

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 This one is undoubtedly true -- it holds for p < 100,000,000
Actually, it has already been verified up to p < 10^17 ( or even more ) because it's a special case of the Goldbach Conjecture.

The GC states that every even integer n = 2m can be expressed as the sum of 2 primes ( at least once )
Example: 30 = 11 + 19 ( 30 is a Goldbach number )

Which could be equivalently restated as follows: for every integer m, there exists 2 primes that are equidistant from m ( m-k, m+k ), at least once .
Example: for m = 15, 11 and 19 are 2 primes equidistant from m ( where n = 2m = 30 is a Goldbach number ).

Islam's statement is: for every prime number p, there exists ( at least ) two primes that are equidistant from p, which is a special case ( m is prime ).
Example: p = 13, 6 and 19 are 2 primes equidistant from p ( where n = 2p = 26 is a Goldbach number ).

Since all primes ( except 2 and 3 ) are of the form 6k +- 1, then the distance between any 2 primes will always be 6k

November 13th, 2010, 07:14 PM   #4
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Re: new conjecture on prime numbers .... i hope

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 Originally Posted by Wissam Actually, it has already been verified up to p < 10^17 ( or even more ) because it's a special case of the Goldbach Conjecture.
The GC is why I said it's too hard to prove. But it seems to be neither obviously weaker nor obviously stronger.

Quote:
 Originally Posted by Wissam Since all primes ( except 2 and 3 ) are of the form 6k +- 1, then the distance between any 2 primes will always be 6k
7 - 5 = 2, which is not a multiple of 6.

November 14th, 2010, 02:05 AM   #5
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Re: new conjecture on prime numbers .... i hope

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 7 - 5 = 2, which is not a multiple of 6.
All prime numbers ( except 2 and 3 ) are either of the form 6k+1 or 6k-1, the distance between any 2 primes belonging to one of the 2 sets is 6k.
That's what I meant. I was referring to Islam's statement.

 November 14th, 2010, 07:44 AM #6 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 Re: new conjecture on prime numbers .... i hope well actually that isnt always true if both primes r of the form 6n +1 or both r of the form 6n-1 the distance between them will be of the form 6k however if one is of the form 6n+1 while the other of the form 6n-1 then the distance will be of the form 6n+-2 for example 19-5 =14 14 isnt of the form 6k
November 14th, 2010, 08:04 AM   #7
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Re: new conjecture on prime numbers .... i hope

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 if both primes r of the form 6n +1 or both r of the form 6n-1 the distance between them will be of the form 6k however if one is of the form 6n+1 while the other of the form 6n-1 then the distance will be of the form 6n+-2
That's why I said both primes should belong to one of the two sets... which is the case in your original statement.

You started with any prime p >= 13 ( actually it should be p >= 11 ), so either p = 6k + 1 or p = 6k -1

If p = 6k + 1, then p - 6k' = 6(k-k') + 1 = 6k'' + 1 & p + 6k' = 6(k+k') + 1 = 6k'' + 1
If p = 6k - 1, then p - 6k' = 6(k-k') - 1 = 6k'' - 1 & p + 6k' = 6(k+k') - 1 = 6k'' - 1

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