mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Blogorrhea > Dobri

Reply
 
Thread Tools
Old 2022-04-18, 21:32   #12
Dr Sardonicus
 
Dr Sardonicus's Avatar
 
Feb 2017
Nowhere

583410 Posts
Default

Quote:
Originally Posted by kijinSeija View Post
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
Dr Sardonicus is offline   Reply With Quote
Old 2022-04-19, 15:52   #13
Dobri
 
"Καλός"
May 2018

1010000112 Posts
Default

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.
Dobri is offline   Reply With Quote
Old 2022-04-19, 16:25   #14
Dobri
 
"Καλός"
May 2018

17·19 Posts
Default

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.
Dobri is offline   Reply With Quote
Old 2022-04-19, 18:48   #15
kijinSeija
 
Mar 2021
France

2·3·5 Posts
Default

Quote:
Originally Posted by Dr Sardonicus View Post
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 ?
kijinSeija is offline   Reply With Quote
Old 2022-04-20, 05:14   #16
Dobri
 
"Καλός"
May 2018

17×19 Posts
Default

Quote:
Originally Posted by kijinSeija View Post
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.
Dobri is offline   Reply With Quote
Old 2022-04-20, 06:48   #17
Dobri
 
"Καλός"
May 2018

17·19 Posts
Default

Quote:
Originally Posted by kijinSeija View Post
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
Dobri is offline   Reply With Quote
Old 2022-04-20, 07:18   #18
kijinSeija
 
Mar 2021
France

2·3·5 Posts
Default

Quote:
Originally Posted by Dobri View Post
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
kijinSeija is offline   Reply With Quote
Old 2022-04-20, 07:54   #19
kijinSeija
 
Mar 2021
France

2·3·5 Posts
Default

Quote:
Originally Posted by Dobri View Post
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 ?
kijinSeija is offline   Reply With Quote
Old 2022-04-20, 07:58   #20
Dobri
 
"Καλός"
May 2018

5038 Posts
Default

Quote:
Originally Posted by Dobri View Post
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,..."
Dobri is offline   Reply With Quote
Old 2022-04-20, 08:44   #21
kijinSeija
 
Mar 2021
France

2·3·5 Posts
Default

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
kijinSeija is offline   Reply With Quote
Old 2022-04-20, 09:29   #22
Dobri
 
"Καλός"
May 2018

1010000112 Posts
Default

Quote:
Originally Posted by kijinSeija View Post
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}}
...
Dobri is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Repeating residues in LL tests of composite Mersenne numbers Viliam Furik Number Theory Discussion Group 22 2020-12-08 14:45
Mersenne number with exponent 333333367 is composite TheGuardian GPU Computing 25 2019-05-09 21:53
Factoring composite Mersenne numbers using Pollard Rho jshort Factoring 9 2019-04-09 16:34
2 holes in bizarre theorem about composite Mersenne Numbers wildrabbitt Math 120 2016-09-29 21:52
Factoring highly composite Mersenne numbers philmoore Factoring 21 2004-11-18 20:00

All times are UTC. The time now is 05:34.


Mon Jul 4 05:34:53 UTC 2022 up 81 days, 3:36, 0 users, load averages: 1.42, 1.26, 1.17

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔