[QUOTE=kijinSeija;604214]By the way, do you know the condition for example 6*p+1 or 10*p+1 divides 2^p-1 ?[/QUOTE]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 2k[sup]th[/sup] 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 [i]not[/i] 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.

Within the limited sample of known Mersenne primes, there are 7 Mersenne primes, with prime exponents [I]p[/I] = 2, 5, 89, 9689, 21701, 859433, and 43112609, for which 2[I]p[/I] + 1 is a prime number. Obviously, [I]p[/I] mod 4 = 1 for [I]p[/I] = 5, 89, 9689, 21701, 859433, and 43112609.

Concerning [I]k[/I] = 5, there are 10 known Mersenne primes, with prime exponents [I]p[/I] = 3, 7, 13, 19, 31, 1279, 2203, 2281, 23209, and 44497, for which 2*5*[I]p[/I]+1 is a prime number. Here [I]p[/I] mod 6 = 1 for [I]p[/I] = 7, 13, 19, 31, 1279, 2203, 2281, 23209, and 44497.

[QUOTE=Dr Sardonicus;604242]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 2k[sup]th[/sup] 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 [i]not[/i] 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.[/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^p-1 right ?

[QUOTE=kijinSeija;604284]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 ?[/QUOTE]
Here is a counterexample: [I]p[/I] = 5, [I]p[/I] mod 4 = 1, and 2[I][SUP]p[/SUP][/I] - 1 = 6[I]p[/I] + 1 = 27×1[SUP]2[/SUP] + 2[SUP]2[/SUP] is the 3[SUP]rd[/SUP] Mersenne prime number [I]M[/I][SUB]5[/SUB].

[QUOTE=kijinSeija;604284]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 ?[/QUOTE]
Excluding [I]p[/I] = 5 for which 2[SUP]5[/SUP] - 1 = 6×5 + 1 = 27×1[SUP]2[/SUP] + 2[SUP]2[/SUP], there are no other counterexamples for [I]p[/I] mod 4 = 1 and [I]k[/I] = 3 within the limited sample of known Mersenne primes.

Starting from the composite Mersenne number for [I]p[/I] = 101, whenever [I]p[/I] mod 4 = 1 and [I]M[SUB]p[/SUB][/I] mod (6[I]p[/I] + 1) ≠ 0 there are no instances of 6[I]p[/I] + 1 = 27[I]a[/I][SUP]2[/SUP] + [I]b[/I][SUP]2[/SUP] for [I]p[/I] = 101, 107, 173, 181, 241, 257, 277, 293, 313,...

Starting from the composite Mersenne number for [I]p[/I] = 37, whenever [I]p[/I] mod 4 = 1 and [I]M[SUB]p[/SUB][/I] mod (6[I]p[/I] + 1) = 0 there are instances of 6[I]p[/I] + 1 = 27[I]a[/I][SUP]2[/SUP] + [I]b[/I][SUP]2[/SUP] for [I]p[/I] = 37, 73, 233, 397, 461, 557, 577, 601, 761,...

Therefore, the quoted statement could be considered as a possible conjecture for now.

[QUOTE=Dobri;604319]Here is a counterexample: [I]p[/I] = 5, [I]p[/I] mod 4 = 1, and 2[I][SUP]p[/SUP][/I] - 1 = 6[I]p[/I] + 1 = 27×1[SUP]2[/SUP] + 2[SUP]2[/SUP] is the 3[SUP]rd[/SUP] Mersenne prime number [I]M[/I][SUB]5[/SUB].[/QUOTE]

But 31 divides 31 for example. This is really a counterexample ?

Oh I see what you mean :grin:

[QUOTE=Dobri;604324]Excluding [I]p[/I] = 5 for which 2[SUP]5[/SUP] - 1 = 6×5 + 1 = 27×1[SUP]2[/SUP] + 2[SUP]2[/SUP], there are no other counterexamples for [I]p[/I] mod 4 = 1 and [I]k[/I] = 3 within the limited sample of known Mersenne primes.

Starting from the composite Mersenne number for [I]p[/I] = 101, whenever [I]p[/I] mod 4 = 1 and [I]M[SUB]p[/SUB][/I] mod (6[I]p[/I] + 1) ≠ 0 there are no instances of 6[I]p[/I] + 1 = 27[I]a[/I][SUP]2[/SUP] + [I]b[/I][SUP]2[/SUP] for [I]p[/I] = 101, 107, 173, 181, 241, 257, 277, 293, 313,...

Starting from the composite Mersenne number for [I]p[/I] = 37, whenever [I]p[/I] mod 4 = 1 and [I]M[SUB]p[/SUB][/I] mod (6[I]p[/I] + 1) = 0 there are instances of 6[I]p[/I] + 1 = 27[I]a[/I][SUP]2[/SUP] + [I]b[/I][SUP]2[/SUP] for [I]p[/I] = 37, 73, 233, 397, 461, 557, 577, 601, 761,...

Therefore, the quoted statement could be considered as a possible conjecture for now.[/QUOTE]

Thanks, this is interesting. How do you check that ? With Wolfram Alpha ?

[QUOTE=Dobri;604324]Starting from the composite Mersenne number for [I]p[/I] = 101, whenever [I]p[/I] mod 4 = 1 and [I]M[SUB]p[/SUB][/I] mod (6[I]p[/I] + 1) ≠ 0 there are no instances of 6[I]p[/I] + 1 = 27[I]a[/I][SUP]2[/SUP] + [I]b[/I][SUP]2[/SUP] for [I]p[/I] = 101, 107, 173, 181, 241, 257, 277, 293, 313,...[/QUOTE]
There is a typo, it has to be "... for [I]p[/I] = 101, [COLOR="Red"][B]137[/B][/COLOR], 173, 181, 241, 257, 277, 293, 313,..."

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.

[QUOTE=kijinSeija;604326]Thanks, this is interesting. How do you check that ? With Wolfram Alpha ?[/QUOTE]
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++;];
[/code]
and the obtained output
[code]
5 {{ac->1,bc->2}}
13 {}
17 {}
61 {}
2281 {}
44497 {}
3021377 {}
57885161 {}
82589933 {}
[/code]
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++;];
[/code]
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}}
...
[/code]