6m Theorem for Prime numbers

We show that for any \(P= 6^{m+1}.N -1 \) is a prime number for any \(1 < N \le 13\) , \(N \ne 8\) and \(N \ne i^{m+1}Mod(6i+1) \) where \( i \in Z^+ \) and \( m \in \) \(odd\) \(Z^+ \) for \(1 < N \le 13\) and \(N \ne 8\) and also we further discussed that \(P= 6^{m+1}.N -1 \) is a prime numb...

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Bibliographic Details
Published in:arXiv.org 2018-09
Main Authors: Gandarawatta R W M P I S B, Perera, S P, Rathnayake R M L S
Format: Article
Language:English
Subjects:
Prime numbers
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Summary:We show that for any \(P= 6^{m+1}.N -1 \) is a prime number for any \(1 < N \le 13\) , \(N \ne 8\) and \(N \ne i^{m+1}Mod(6i+1) \) where \( i \in Z^+ \) and \( m \in \) \(odd\) \(Z^+ \) for \(1 < N \le 13\) and \(N \ne 8\) and also we further discussed that \(P= 6^{m+1}.N -1 \) is a prime number for \( N >13 \) if and only if , \(N \ne i^{m+1}Mod(6i+1) +(6i +1)a \) \( ; i,a \le Z^+ \)
ISSN:2331-8422