2006-06-23, 20:07 | #1 |
Aug 2002
Buenos Aires, Argentina
2^{2}×337 Posts |
Distribution of Mersenne prime factors mod 6
Let M be a Mersenne number with prime exponent q.
Then 2^{q} = 1 (mod M) 2^{aq} = 1 (mod M) 2^{aq+1} = 2 (mod M) Since q is not multiple of 3, we can set a=-q mod 3, so the exponent is also multiple of 3, say 3r. 2^{3r} = 2 (mod M) (2^{r})^{3} = 2 (mod M) Let x = 2^{r}: x^{3} = 2 (mod M) This means that for every prime factor p of M we have: x^{3} = 2 (mod p) This congruence is always satisfied when p mod 6 = 5, but it is interesting to notice that this is not the case for p mod 6 = 1. Primes p up to 100000: p = 5 (mod 6): 4806 p = 1 (mod 6): 1559 Primes p up to 1000000: p = 5 (mod 6): 39231 p = 1 (mod 6): 13032 So it appears that it is more common to have prime factors of the form 6k+5 than from the form 6k+1. It should be interesting to computing the real distribution from the known prime factors of Mersenne numbers (say from Will Edgington database). Last fiddled with by alpertron on 2006-06-23 at 20:08 |
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