Problem 437*
Find the term independent of in
.
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Last edited by Khallil; 04022014 at 22:50. 
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 05022014 00:56

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 05022014 02:05
Ok guys, I'm pretty sure there are maths genii (is that a word??) Among you. I need your help. I have dome up with a supposedly unique and overall different approach to the twin prime conjecture and now I have solved it . Here is my solution
http://twinprimeconjecture.blogspot.co.uk/?
[m=1 SIZE=1]Posted from TSR Mobile[/SIZE] 
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 05022014 13:53
(Original post by theuser77)
Ok guys, I'm pretty sure there are maths genii (is that a word??) Among you. I need your help. I have dome up with a supposedly unique and overall different approach to the twin prime conjecture and now I have solved it . Here is my solution
http://twinprimeconjecture.blogspot.co.uk/?
[m=1 SIZE=1]Posted from TSR Mobile[/SIZE]
As you may have infinitely many primes expressible in this form, and the result does not follow. For example:
31 = 6(2)(3)  2  3
37 = 6(2)(3)  2 + 3
41 = 6(2)(3) + 2 + 3
43 = 6(1)(6) + 1 + 6 
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 05022014 14:43
You have misunderstood the fifth lemma, it states that when k is of the form 6ab+a+b or 6ab+a+b, then 6k+1 will not be prime, sub all those values into 6k+1 and you will find that for all cases 6k+1 isnt prime.
Furthermore it is the modulus of 6ab+a+b..so a and b can be negative integers. Thanks for responding though, it is very difficult for an ordinary person to show off the solution to a famous problem to the world as people come highly skeptical as they rightly should be.
(Original post by DJMayes)
Your first three lemmas are fine. Haven't looked at the fourth thoroughly. Your fifth one doesn't make sense to me  I am not seeing how you have made the leap that infinitely many primes implies infinitely many integers not expressible as:
As you may have infinitely many primes expressible in this form, and the result does not follow. For example:
31 = 6(2)(3)  2  3
37 = 6(2)(3)  2 + 3
41 = 6(2)(3) + 2 + 3
43 = 6(1)(6) + 1 + 6Last edited by theuser77; 05022014 at 14:49. 
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 05022014 15:01
I have proved that all primes are in the form 6k+1 but the reverse is obviously not true. I have then shown that k is of the form 6ab+a+b...
if and only if 6k+1 is a false prime. Therefore when k is not of the fom 6ab+a+b... then k is a prime since this acts as a seive and cancels out all false primes leaving only the prime. but since there are an infinite number of primes, there must be an infinite number of k's not of the previous form. and we have shown that when k is not of the forms previously mentioned, then the equation
24k= x^2y^2 yields exactly one unique twin prime pair x,y, but since there are an infinite number of k's there must be an infinite number of
x,y 's to satisfy all these equations where x,y ae twin primes which leads to the conclusion that there are an infinite number of twin primes. yes that's right you first heard it here guys. the twin prime conjecture has been solved.
(Original post by DJMayes)
Your first three lemmas are fine. Haven't looked at the fourth thoroughly. Your fifth one doesn't make sense to me  I am not seeing how you have made the leap that infinitely many primes implies infinitely many integers not expressible as:
As you may have infinitely many primes expressible in this form, and the result does not follow. For example:
31 = 6(2)(3)  2  3
37 = 6(2)(3)  2 + 3
41 = 6(2)(3) + 2 + 3
43 = 6(1)(6) + 1 + 6 
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 2690
 05022014 15:22
(Original post by theuser77)
I have proved that all primes are in the form 6k+1 but the reverse is obviously not true. I have then shown that k is of the form 6ab+a+b...
if and only if 6k+1 is a false prime. Therefore when k is not of the fom 6ab+a+b... then k is a prime since this acts as a seive and cancels out all false primes leaving only the prime. but since there are an infinite number of primes, there must be an infinite number of k's not of the previous form. and we have shown that when k is not of the forms previously mentioned, then the equation
24k= x^2y^2 yields exactly one unique twin prime pair x,y, but since there are an infinite number of k's there must be an infinite number of
x,y 's to satisfy all these equations where x,y ae twin primes which leads to the conclusion that there are an infinite number of twin primes. yes that's right you first heard it here guys. the twin prime conjecture has been solved.
However if you are so sure, then email this to a proper mathematician (We are students after all and you would be better off checking with the real experts) to have it verified. If it were true you'd be in for a good sum of money in lecturing royalties, a few honorary degrees and probably a Fields Medal. 
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 05022014 15:29
here is a better way of phrasing what i mean
6(6ab+a+b)+ or 1 and 6(6ab+a+b)+or1 will always be a false prime.
that is what I have proved.
Lol at least this will give me something to talk about in my uni interview next year lol
(Original post by DJMayes)
10 is not of the form , and is certainly not prime. Similarly 11 = 6(2)(1)  2 + 1 and 6x11+1 = 67 is prime.
However if you are so sure, then email this to a proper mathematician (We are students after all and you would be better off checking with the real experts) to have it verified. If it were true you'd be in for a good sum of money in lecturing royalties, a few honorary degrees and probably a Fields Medal.Last edited by theuser77; 05022014 at 15:36. 
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 05022014 15:46
The link isn't working on my phone. Have you explicitly proved that the vector space for integer solutions is infinite or have you just assumed that because there are infinite primes? I've haven't studied NT extensively but I suspect that the main reason the conjecture is unsolved is because of something relating to that rather than because noone has noticed the general forms.

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 2693
 05022014 16:27
(Original post by theuser77)
Ok guys, I'm pretty sure there are maths genii (is that a word??) Among you. I need your help. I have dome up with a supposedly unique and overall different approach to the twin prime conjecture and now I have solved it . Here is my solution
http://twinprimeconjecture.blogspot.co.uk/?
[m=1 SIZE=1]Posted from TSR Mobile[/SIZE] 
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 2694
 05022014 16:30
I have exlicitly shown that there are exactly four forms of false primes, where false primes are of the form 6k+1, and that all primes are of the form 6k+1.
I then showed that a prime 6k+1 is a false prime if it can be broken down by factors of the form (6a+1)^n(6b+1)^m and i proved inductively that the expansion of all such prime factors leads to a false prime of the form
6k+1.
but by the expansions of the factors:
 (6t+1)(6u1) = 6(6tu+ut)1 hence all such forms of k yield a false prime. But by using a values of k not of the forms of the expansion
we can sub such a k into the formula
24k=x^2^2 which allows us to avoid all such numbers that are false primes.
and the equation above has solutions
6k+1 and 6k1
and since k is not of the forms previously mentioned
6k+1 and 6k1 will not prrimes, and their difference of 2 suggests twin primes.
all then that is required is proof that there are an infinite number of k's not of the forms mentioned and the proof is complete. this is just a summary, you can read the fullproof as it is more detailed.
these form of k in 6k+1 and 6k1 represent all such k's that lead to a false prime.
But if k is not an integer that is expressiible in these forms, we can sub k into the eqation
24k=x^2y^2
(Original post by Llewellyn)
The link isn't working on my phone. Have you explicitly proved that the vector space for integer solutions is infinite or have you just assumed that because there are infinite primes? I've haven't studied NT extensively but I suspect that the main reason the conjecture is unsolved is because of something relating to that rather than because noone has noticed the general forms. 
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 2695
 05022014 16:34
(Original post by james22)
Bottom of page 8 is wrong. It implies that numbers of the form 6k+/1 only have 2 prime factors which is clearly false.Last edited by theuser77; 05022014 at 16:41. 
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 2696
 05022014 16:36
(Original post by theuser77)
I have exlicitly shown that there are exactly four forms of false primes, where false primes are of the form 6k+1, and that all primes are of the form 6k+1.
I then showed that a prime 6k+1 is a false prime if it can be broken down by factors of the form (6a+1)^n(6b+1)^m and i proved inductively that the expansion of all such prime factors leads to a false prime of the form
6k+1.
but by the expansions of the factors:
 (6t+1)(6u1) = 6(6tu+ut)1 hence all such forms of k yield a false prime. But by using a values of k not of the forms of the expansion
we can sub such a k into the formula
24k=x^2y^2 which allows us to avoid all such numbers that are false primes.
and the equation above has solutions
6k+1 and 6k1
and since k is not of the forms previously mentioned
6k+1 and 6k1 will be prrimes, and their difference of 2 suggests twin primes.
all then that is required is proof that there are an infinite number of k's not of the forms mentioned and the proof is complete. this is just a summary, you can read the fullproof as it is more detailed.
these form of k in 6k+1 and 6k1 represent all such k's that lead to a false prime.
But if k is not an integer that is expressiible in these forms, we can sub k into the eqation
24k=x^2y^2Last edited by theuser77; 05022014 at 16:48. 
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 2697
 05022014 17:30
(Original post by theuser77)
here is a better way of phrasing what i mean
6(6ab+a+b)+ or 1 and 6(6ab+a+b)+or1 will always be a false prime.
that is what I have proved.
Lol at least this will give me something to talk about in my uni interview next year lol
is not false prime, it's prime. 
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 06022014 07:39

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 06022014 13:00
(Original post by theuser77)

RoyalBlue7
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 06022014 13:39
This problem is going to take a while.
Its should have a number
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