[QUOTE=LaurV;279064]Does "no problem" means "without factoring n=p*q into p and q"? I think you can not... [/QUOTE]
My formulation was a be misleading, I wanted to say: If the you know the prime decomposition of n then are not hard problems to compute sqrt(a) mod n. If you do not know it you can say that a is a QNR if (a|n) = -1, but there seems to be no short cut to compute sqrt(a) mod n if (a|n) = 1 if n is composite.

[QUOTE=science_man_88;279079]((6*4+6)*4+7) is the way to get to 127(2^7-1) in 3 steps ( bits in binary(7))

so that get to solve for 2^127-1 using binary(7) instead of binary(127) now to get a general form so I can use binary(127) or binary (7) to search for factors of MM127.[/QUOTE]

I've got one for MM127 just not a good one ( a really high number involved):

[CODE]((((((289431*289431+289431)*289431+289431)*289431+289431)*289431+289431)*289431+289431)*289426+53109184246303113940243237269392)[/CODE]

[QUOTE=fivemack;279075]There are perfectly good algorithms for square root mod a prime p=4k+1, the only problem is that they work in a quadratic extension field, so you need to find a quadratic non-residue. In practice you start at 2 and have found the quadratic non-residue in on-average-two quadratic-residuosity computations, but nobody is able to prove that there is not a number whose smallest quadratic non-residue is enormous.[/QUOTE]

There's a paper by Bernstein about this problem -- something like "finding quadratic nonresidues modulo worst-case integers".

[QUOTE=science_man_88;279093]I've got one for MM127 just not a good one ( a really high number involved):

[CODE]((((((289431*289431+289431)*289431+289431)*289431+289431)*289431+289431)*289431+289431)*289426+53109184246303113940243237269392)[/CODE][/QUOTE]

This number ist equal 2^127!

[QUOTE=kar_bon;279168]This number ist equal 2^127![/QUOTE]

[CODE](22:56)>((((((289431*289431+289431)*289431+289431)*289431+289431)*289431+289431)*289431+289431)*289426+53109184246303113940243237269392)
%597 = 170141183460469231731687303715884105728
(09:03)>2^127
%598 = 170141183460469231731687303715884105728[/CODE]

[QUOTE=science_man_88;279177][CODE](22:56)>((((((289431*289431+289431)*289431+289431)*289431+289431)*289431+289431)*289431+289431)*289426+53109184246303113940243237269392)
%597 = 170141183460469231731687303715884105728
(09:03)>2^127
%598 = 170141183460469231731687303715884105728[/CODE][/QUOTE]

the part I hate is I forgot to try:

[CODE](10:18)>solve(k=1,1000000,((((((k*k+k)*k+k)*k+k)*k+k)*k+k)*k+k)-2^127)
%599 = 289430.2986150829576189457661[/CODE] which would have sped up me figuring it out quite a bit.

got another weird coincidence I think:

[CODE]a=vector(#MeVec,x,MeVec[x]^2-2);for(y=1,#a,print(factor(a[y])"\n"factor(a[y])%8"\n"floor(sqrt(a[y]+2))"\n\n"))[/CODE]

the lowest factor is either the number itself, or 7 mod 8 all but 3 times ( 2 being one of them) I should make a extended version of MeVec to try the higher ones maybe but I don't think anything is to come out of this.

Forgive us for our impertinence, how to resolve by name limit tempdir ..

is there a way to use:

mersenne mod (mersenne-1) = mersenne ( where mersenne = A000225), I want to say yes but I sense 14=15-1 = s[SUB]1[/SUB] is not a good enough reason.

Mersenne primes
 

[QUOTE=science_man_88;235459]this is from a pm I sent (bet if ever pm I sent was deleted from people's inbox's the server would run faster lol)[/QUOTE]
I believe I have a proof for an infinite number of Mersenne primes.
Anybody interested in proof reading? Only one side of A4.

[QUOTE=Stan;280985]I believe I have a proof for an infinite number of Mersenne primes.
Anybody interested in proof reading? Only one side of A4.[/QUOTE]
Post the pdf or use the forum's [TEX]^TE^X[/TEX] feature to present it here. If you choose not to present it here in full, you may subject yourself to severe ridicule.