Didn't even find a useless factor of an uninteresting number! Take that!

[storflyt32]

http://factordb.com/index.php?id=1100000000820473763

http://factordb.com/index.php?id=1100000000820473569

The C289 in the first link has a P35 factor which was added a short while ago.

P35 = 39329081168732850267866544033284131

By means of trial division here.

But factoring the C289 in the first link really does make it official.

This will probably take some time.

The remaining number becomes a P255.

[storflyt32]

If you happen to be reading some of the discussion taking place somewhere else here, the story becomes as follows:

2^31 - 1 = 2147483647

2^2147483647 - 1 = ...

isprime(...) (?)

What is next, or is there something else that could be tried out instead?

[paulunderwood]

http://www.doublemersennes.org/mm31.php

"Of these, MMp is known to be prime for p = 2, 3, 5, 7; for p = 13, 17, 19, and 31, explicit factors have been found showing that the corresponding double Mersenne numbers are not prime." from the home page.

[storflyt32]

Thanks Paul.

Noted down that page for reference.

[storflyt32]

Always a difference between those idiots (not Paul) and any geniuses.

Remeber the chess robot being mentioned.

[storflyt32]

Paul, or someone else here, a question for you.

http://factordb.com/index.php?id=1100000000822420576

If you look at the P44 and P46 pair, one might get the impression that these factors became known as a result of the factorization of RSA-768 or the similar.

Now, smaller factors are being added as well in an attempt to factorize even larger numbers of the same type for which there currently are no known factors for, because these numbers may be having only two factors.

In the end you may actually be able to find these factors, but only because you are getting a lot of others which does not divide the number you are trying to find.

Is this now the only way of doing this, or should we still believe that "brute force" is the better way of solving these problems?

[VBCurtis]

Originally Posted by storflyt32

Paul, or someone else here, a question for you.

http://factordb.com/index.php?id=1100000000822420576

If you look at the P44 and P46 pair, one might get the impression that these factors became known as a result of the factorization of RSA-768 or the similar.

No, nobody else might get that impression. Why do you think these have anything to do with RSA-768? What is "or the similar" to an RSA number?

Originally Posted by storflyt32

Now, smaller factors are being added as well in an attempt to factorize even larger numbers of the same type for which there currently are no known factors for, because these numbers may be having only two factors.

What is "of the same type"? Same as what? In what way ("type") are things the same? Your posts are unintelligible largely because you use words in ways other math people do not. In this example, you make up "same type" without actually saying what makes two numbers similar to you.

Originally Posted by storflyt32

In the end you may actually be able to find these factors, but only because you are getting a lot of others which does not divide the number you are trying to find.

No, getting a lot of other factors of unrelated numbers does NOTHING to help you factor a particular number, regardless what "type" the particular number is. This is one of your biggest errors- you think the useless factors you post have something to do with RSA. They don't. At all. Ever.

Originally Posted by storflyt32

Is this now the only way of doing this, or should we still believe that "brute force" is the better way of solving these problems?

Is what the only way of doing what? You haven't solved anything, done anything, or explained what your method is. You're so far from understanding how to factor that I don't think you have an actual method in mind for "brute force"; if you do, it's almost certainly not the one we would use to factor a number.

What is your "brute force" method to factor a number? To be specific: If I gave you a 512-bit RSA key, and told you I was sure it met the RSA style (exactly two factors, each the same bit length), how would you factor it? Please be more specific than "yafu factor(numberyougiveme)".

[storflyt32]

I read the rest of it before posting.

http://factordb.com/index.php?id=1100000000822862072

Not with ecm and 2^21 curves here.

By the way, I happen to be curious about the possible P306.

How do you get this number?

I will give it a try.

http://factordb.com/index.php?id=1100000000823523157

Edit: Made a copy and paste error and found this P219.

[storflyt32]

Could you please refer me to the mentioned a(196)?

http://mersenneforum.org/showpost.ph...&postcount=111

Not able to find it right now.

[storflyt32]

http://factordb.com/index.php?id=1100000000822862072

Thanks for doing that.

Edit:

http://factordb.com/index.php?id=1100000000823542817

http://factordb.com/index.php?id=1100000000823541641

What a coincidence.

Still working on the C126.

[storflyt32]

Another queston for you.

Definitely 2^31-1, 2^31+1, 2^127-1 or 2^127+1 are easy numbers to factorize.

You have the same list of factors there, in progressively larger order, including

Therefore a number like 438074125819448964817444191295445945307799572735 is easy to factorize as well.

Slightly more diiffcult when adding 2 to this number.

Surprisingly, 438074125819448964817444191295445945307799572737 had a P38 being added to the FDB.

What if you rather say 2^6700417-1 or 2^6700417+1 instead?

The latter number is having factor P1 = 3.

In comparison, I have not checked a factor like P8 = 14231699 for such a thing.

Is this P8 a Mersenne factor, or is it a Fermat factor?

Anyway, was able to get these two results.


pfgw64 -q"(2^6400417-1)"
PFGW Version 3.7.7.64BIT.20130722.Win_Dev [GWNUM 27.11]

(2^6400417-1) is composite: RES64: [D136471B5C1D058B] (42756.7714s+0.0024s)


pfgw64 -q"((2^6400417+1)/3)"
PFGW Version 3.7.7.64BIT.20130722.Win_Dev [GWNUM 27.11]

((2^6400417+1)/3) is composite: RES64: [E4558F8BD2593646] (39807.7262s+0.0811s)