20220418, 21:32  #12  
Feb 2017
Nowhere
2·29·103 Posts 
Quote:
The polynomial is irreducible (e.g. by Eisenstein's criterion). The 2k solutions to f(x) = 0 consist of the real 2kth root of 2, multiplied by each of the 2k 2k^{th} roots of unity. The significance of the Galois group being nonAbelian is that the factorization of f(x) == 0 (mod q) for prime q can not be characterized in terms of congruences mod M for any integer M. For k = 3, congruences only get us part of the way. We can say that p == 1 (mod 4) to insure that q = 6*p + 1 is congruent to 7 (mod 8), hence a quadratic residue (mod 2). But 2 also has to be a cubic residue (mod q). There's no integer congruence condition for that. There is the condition that, with q = a^2 + 3*b^2, b is divisible by 3. But that's not much help. Last fiddled with by Dr Sardonicus on 20220419 at 13:08 Reason: Fix clumsy phrasing 

20220419, 15:52  #13 
"Καλός"
May 2018
3·11^{2} Posts 
Within the limited sample of known Mersenne primes, there are 7 Mersenne primes, with prime exponents p = 2, 5, 89, 9689, 21701, 859433, and 43112609, for which 2p + 1 is a prime number. Obviously, p mod 4 = 1 for p = 5, 89, 9689, 21701, 859433, and 43112609.

20220419, 16:25  #14 
"Καλός"
May 2018
16B_{16} Posts 
Concerning k = 5, there are 10 known Mersenne primes, with prime exponents p = 3, 7, 13, 19, 31, 1279, 2203, 2281, 23209, and 44497, for which 2*5*p+1 is a prime number. Here p mod 6 = 1 for p = 7, 13, 19, 31, 1279, 2203, 2281, 23209, and 44497.

20220419, 18:48  #15  
Mar 2021
France
41 Posts 
Quote:
So if we add than p == 1 (mod 4) and 6*p+1 = 27a^2+b^2 should be the two conditions for 6*p+1 divides 2^p1 right ? 

20220420, 05:14  #16 
"Καλός"
May 2018
3×11^{2} Posts 

20220420, 06:48  #17  
"Καλός"
May 2018
3·11^{2} Posts 
Quote:
Starting from the composite Mersenne number for p = 101, whenever p mod 4 = 1 and M_{p} mod (6p + 1) ≠ 0 there are no instances of 6p + 1 = 27a^{2} + b^{2} for p = 101, 107, 173, 181, 241, 257, 277, 293, 313,... Starting from the composite Mersenne number for p = 37, whenever p mod 4 = 1 and M_{p} mod (6p + 1) = 0 there are instances of 6p + 1 = 27a^{2} + b^{2} for p = 37, 73, 233, 397, 461, 557, 577, 601, 761,... Therefore, the quoted statement could be considered as a possible conjecture for now. Last fiddled with by Dobri on 20220420 at 07:44 

20220420, 07:18  #18 
Mar 2021
France
41 Posts 

20220420, 07:54  #19  
Mar 2021
France
41 Posts 
Quote:


20220420, 07:58  #20 
"Καλός"
May 2018
553_{8} Posts 
There is a typo, it has to be "... for p = 101, 137, 173, 181, 241, 257, 277, 293, 313,..."

20220420, 08:44  #21 
Mar 2021
France
41 Posts 
Like Mersenne composites, it seems than p == 3 (mod 4) and 6*p+1 = 27a^2+16b^2 should be the two condition for 6p+1 divides Wagstaff numbers (2^p+1)/3. (7, 47, 83, 107, 263, 271 ...) The sequence isn't in OEIS by the way.
Last fiddled with by kijinSeija on 20220420 at 08:46 
20220420, 09:29  #22  
"Καλός"
May 2018
3×11^{2} Posts 
Quote:
Here is the Wolfram code for the Mersenne primes Code:
MPData = {2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667, 42643801, 43112609, 57885161, 74207281, 77232917, 82589933}; Np = 51; ic = 1; While[ic <= Np, p = MPData[[ic]]; If[(Mod[p, 4] == 1) && (PrimeQ[2*3*p + 1] == True), Ms = FindInstance[{2*3*p + 1 == 27*ac^2 + bc^2}, {ac, bc}, PositiveIntegers]; Print[p, " ", Ms]; ]; ic++;]; Code:
5 {{ac>1,bc>2}} 13 {} 17 {} 61 {} 2281 {} 44497 {} 3021377 {} 57885161 {} 82589933 {} Code:
kc = 3; ic = 1; While[ic <= 1000, p = Prime[ic]; fc = 2*kc*p + 1; If[(Mod[p, 4] == 1) && (PrimeQ[fc] == True) && (MersennePrimeExponentQ[p] == False), Mn = 2^p  1; rc = FindInstance[{fc == 27*ac^2 + bc^2}, {ac, bc}, PositiveIntegers]; Print[p, " ", Mod[Mn, fc], " ", rc];]; ic++;]; Code:
37 0 {{ac>1,bc>14}} 73 0 {{ac>3,bc>14}} 101 209 {} 137 173 {} 173 897 {} 181 828 {} 233 0 {{ac>3,bc>34}} 241 703 {} 257 680 {} 277 1343 {} 293 507 {} 313 487 {} 373 294 {} 397 0 {{ac>9,bc>14}} 461 0 {{ac>7,bc>38}} 557 0 {{ac>9,bc>34}} 577 0 {{ac>11,bc>14}} 593 2122 {} 601 0 {{ac>3,bc>58}} 641 1891 {} 653 2748 {} 661 3077 {} 761 0 {{ac>13,bc>2}} ... 

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