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2022-04-18, 21:32   #12
Dr Sardonicus

Feb 2017
Nowhere

583410 Posts

Quote:
 Originally Posted by kijinSeija By the way, do you know the condition for example 6*p+1 or 10*p+1 divides 2^p-1 ?
The real problem is that for k > 1, the Galois group of the polynomial f(x) = x^(2*k) - 2 is non-Abelian. (It is however, a solvable group, meaning the equation f(x) = 0 is "solvable by radicals.")

The polynomial is irreducible (e.g. by Eisenstein's criterion). The 2k solutions to f(x) = 0 consist of the real 2k-th root of 2, multiplied by each of the 2k 2kth roots of unity.

The significance of the Galois group being non-Abelian 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 2022-04-19 at 13:08 Reason: Fix clumsy phrasing

 2022-04-19, 15:52 #13 Dobri   "Καλός" May 2018 1010000112 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.
 2022-04-19, 16:25 #14 Dobri   "Καλός" May 2018 17·19 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.
2022-04-19, 18:48   #15
kijinSeija

Mar 2021
France

2·3·5 Posts

Quote:
 Originally Posted by Dr Sardonicus The real problem is that for k > 1, the Galois group of the polynomial f(x) = x^(2*k) - 2 is non-Abelian. (It is however, a solvable group, meaning the equation f(x) = 0 is "solvable by radicals.") The polynomial is irreducible (e.g. by Eisenstein's criterion). The 2k solutions to f(x) = 0 consist of the real 2k-th root of 2, multiplied by each of the 2k 2kth roots of unity. The significance of the Galois group being non-Abelian 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.

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^p-1 right ?

2022-04-20, 05:14   #16
Dobri

"Καλός"
May 2018

17×19 Posts

Quote:
 Originally Posted by kijinSeija 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^p-1 right ?
Here is a counterexample: p = 5, p mod 4 = 1, and 2p - 1 = 6p + 1 = 27×12 + 22 is the 3rd Mersenne prime number M5.

2022-04-20, 06:48   #17
Dobri

"Καλός"
May 2018

17·19 Posts

Quote:
 Originally Posted by kijinSeija 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^p-1 right ?
Excluding p = 5 for which 25 - 1 = 6×5 + 1 = 27×12 + 22, there are no other counterexamples for p mod 4 = 1 and k = 3 within the limited sample of known Mersenne primes.

Starting from the composite Mersenne number for p = 101, whenever p mod 4 = 1 and Mp mod (6p + 1) ≠ 0 there are no instances of 6p + 1 = 27a2 + b2 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 Mp mod (6p + 1) = 0 there are instances of 6p + 1 = 27a2 + b2 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 2022-04-20 at 07:44

2022-04-20, 07:18   #18
kijinSeija

Mar 2021
France

2·3·5 Posts

Quote:
 Originally Posted by Dobri Here is a counterexample: p = 5, p mod 4 = 1, and 2p - 1 = 6p + 1 = 27×12 + 22 is the 3rd Mersenne prime number M5.
But 31 divides 31 for example. This is really a counterexample ?

Oh I see what you mean

2022-04-20, 07:54   #19
kijinSeija

Mar 2021
France

2·3·5 Posts

Quote:
 Originally Posted by Dobri Excluding p = 5 for which 25 - 1 = 6×5 + 1 = 27×12 + 22, there are no other counterexamples for p mod 4 = 1 and k = 3 within the limited sample of known Mersenne primes. Starting from the composite Mersenne number for p = 101, whenever p mod 4 = 1 and Mp mod (6p + 1) ≠ 0 there are no instances of 6p + 1 = 27a2 + b2 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 Mp mod (6p + 1) = 0 there are instances of 6p + 1 = 27a2 + b2 for p = 37, 73, 233, 397, 461, 557, 577, 601, 761,... Therefore, the quoted statement could be considered as a possible conjecture for now.
Thanks, this is interesting. How do you check that ? With Wolfram Alpha ?

2022-04-20, 07:58   #20
Dobri

"Καλός"
May 2018

5038 Posts

Quote:
 Originally Posted by Dobri Starting from the composite Mersenne number for p = 101, whenever p mod 4 = 1 and Mp mod (6p + 1) ≠ 0 there are no instances of 6p + 1 = 27a2 + b2 for p = 101, 107, 173, 181, 241, 257, 277, 293, 313,...
There is a typo, it has to be "... for p = 101, 137, 173, 181, 241, 257, 277, 293, 313,..."

 2022-04-20, 08:44 #21 kijinSeija   Mar 2021 France 2·3·5 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 2022-04-20 at 08:46
2022-04-20, 09:29   #22
Dobri

"Καλός"
May 2018

1010000112 Posts

Quote:
 Originally Posted by kijinSeija Thanks, this is interesting. How do you check that ? With Wolfram Alpha ?
I just connect a Raspberry Pi 4B device to the HDMI port of my 4K TV set and use Wolfram Mathematica for free.
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++;];
and the obtained output
Code:
5 {{ac->1,bc->2}}
13 {}
17 {}
61 {}
2281 {}
44497 {}
3021377 {}
57885161 {}
82589933 {}
followed by the Wolfram code for the composite Mersenne numbers
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++;];
and the obtained output
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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