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ok thank you anyway...!!!
make L that round to the closer integer ie if it is 1.2 the result should be 1!!!
please tell me that it is true!!!!
at the start i was very proud about this but i asked my bigger brother an he told me that i must be more serious!!!!!!:D
There are infinitely many primes p such that p + 2 is also prime.
proof
We assume that the biggest pair is p and p+2. We can know create the function :
L(n,p)= n(1/p(p+2))
Where p+2 is the largest asummed prime number of the pair p and p+2
and n is the a positive integer greater than p+2.
The function L gives the number of integers that have as factor p and p+2(up to n).
If we suppose that there exist a larger pair of the form p+2k and p+2(k+1) Than the set of numbers that have as factors p+2k and p+2(k+1) must be equal to 0 (since they do not exist).
L(n,p+2k)= n/((p+2k)(p+2(k+1))
Or
k-->oo
. Reductio ad absurdum!.
We supposed that p+2k and p+2(k+1) were integers. The numbers p+2k and p+2(k+1) does not exist unless L(n,p+2k) is not equal to 0 !! So there will always be a fraction of the set Z that will have p+2k and p+2(k+1) in their unique prime factorization. So there is an ifinite number of primes p such that p + 2 is also prime.
please tell me is all wrong???
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