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I've listed the 4 equations for a Mersenne prime number (M_r) where r is also prime.
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(only one of these needs to be true!)
p = 4k+1, q = 2p+3 (both prime) [(M_r)^p-p] mod q == -1, or
p = 4k+3, q = 2p+3 (again,... ), then [(M_r)^p-p] mod q == -1 (can't be +1)
p = 4k+1, q = 2p+1 (again, both prime), then [(M_r)^p-p] mod q == p
p = 4k+3, q = 2p+1 (again, both prime) [(M_r)^p-p] mod q == p+2
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let's have another look!!!
let 2^127 -1 = 170141183460469231731687303715884105727
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C_5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1
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let p = 3, q = 7 such that [(C_5)^3 -3] mod 7 = N; and 2^(27) mod 7 == 1 (by mere chance!!!),
so then...
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(1)^(left over exponent)*2 -1 ==
(1)^(2*49*19*43*73*127*337*5419*92737*649657*77158673929)*2^1 -1 == (1)*2 -1 = 2 -1
and (1)^3 -3 = 1 -3 = -2 and [(C_5)^3 -3] mod 7 == 5
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thus, if [(C_5)^3 -3] mod 7 == 5, or p +2 = 3 +2 = 5, then C_5 must be prime!
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C_5 is definitely prime, if my study is correct.
ЦитироватьВ целом, всё верно, только вот... Это что получается, сублинейный тест простоты найден?
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let 2^127 -1 = 170141183460469231731687303715884105727
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C_5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1
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let p = 5, and q = 13 such that [(C_5)^5 -5] mod 13 = N, but 2^(2*27) mod 13 == 12 == (-1)
(as noticed by shear discovery!!!), so then...
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(-1)^(odd power)*2 -1 ==
(-1)^(49*19*43*73*127*337*5419*92737*649657*77158673929)*2 -1 == (-1)*2 -1 = -2 -1,
and (-3)^5 -5 = -248 and [(C_5)^5 -5] mod 13 == 12.
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thus, if [(C_5)^5 -5] mod 13 == 12, and 13 mod 12 == 1, then C_5 must be prime!
Цитировать
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C_5 must be prime! let's take a little excursion...
remember... C_5 is a 51,217,599,719,369,681,875,006,054,625,051,616,350-digit number!
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if p= 4*k +1, and q= 2*p +3 are both prime, then if [(M_r)^p -p] mod q == N, and q mod N ==
+/-1, then (M_r), the base... is prime. also, if (M_r) mod p = 1, then choose a different 'p' or if N
is a square, then (M_r) is prime. (someone would have to prove this conjecture.)
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