Originally posted by 37818
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Since all primes > 2 are of the form 2n+1,
2P-1 {mod 6}
= 22n+1-1 {mod 6}
= 2(22n)-1 {mod 6}
= 2(4n)-1 {mod 6}
= 2((3+1)n)-1 {mod 6}
= 2(3n + k1.3n-1 + k2.3n-2 + ... + kn-1.3 + 1) -1 {mod 6}, where kx are the coefficients of the binomial expansion of n
= 2(3.something + 3.something + ... + 3.something + 1) -1 {mod 6}
= 6.something + 6.something + ... + 6.something + 2 -1 {mod 6}
= 1 {mod 6}
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