20190916, 07:56  #1 
Sep 2019
1000_{2} Posts 
Behold, "The World's Most Secure RSA key"®©™ : 2^12771
Since the
Code:
BEGIN PGP PUBLIC KEY BLOCK mKsEXX4CgwEE/R////////////////////////////////////////////////// //////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////// //////////////////////////////////8AAQG0GE1lcnNlbm5lIDEyNzcgPDJe MTI3Ny0xPojVBBMBAgAgAhsDBAsHCAkDFQgCBBYDAgECHgECF4AFAl1/L18CGQEA CgkQwOplJxxx/xFMOgTxAf////////////////////////////////////////// //////////////////////////////////////////////////////////////// ////////////////////////////////////////////////////////ADAhMAkG BSsOAwIaBQAEFEw6OaDNBA4AZdu9aCnhmS7hlB6vuKwEXX7xkgEE/R////////// //////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////// //////////8ACwT9iL4EGAECAAkFAl1+8ZICGwwACgkQwOplJxxx/xFItQTxAf// //////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////// ////////////////////////////////ADAhMAkGBSsOAwIaBQAEFEi1Kw2NUHLo 5j00UwH3YABnS0aA =A2gW END PGP PUBLIC KEY BLOCK Obviously, please don't encrypt anything to this key, because it will be forever undecryptable, and the no one will never be able to read your message. 
20190916, 08:14  #2 
Romulan Interpreter
"name field"
Jun 2011
Thailand
71·139 Posts 
We believe that the number in cause is quite near to be factored... Just wait and see...
You should have chose some RSA2048 or so, for the candidate (although, many "keys" can be given, with a random higher length, which are more difficult to crack). (edit, yeah, I know, there is nothing like a joke killer...) Last fiddled with by LaurV on 20190916 at 09:35 
20190916, 08:50  #3  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6331_{10} Posts 
Quote:
Are you sure it is more secure than RSA's 2048 bit "modulus"? That also has no factor up to 11. I'm confused!? 

20190916, 11:13  #4  
"Robert Gerbicz"
Oct 2005
Hungary
1,531 Posts 
Quote:


20190916, 15:09  #5  
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
2×5,569 Posts 
Quote:
If you can prove this statement please publish your proof. You will become famous. Otherwise, learn more. AFAIK, although it is widely believed that breaking RSA is as hard as integer factorization, there is no proof of this. Neither is there a proof that an efficient attack on breaking RSA yields a factorization of the modulus; the converse, of course, was proved a very long time ago. One way of breaking RSA which does not yield a factorization (AFAIK) is to encrypt each possible plaintext in turn until the ciphertext is produced. This is a mildly inferior algorithm to factoring the modulus by trial division in that the latter has an obvious optimization: test only prime divisors, which have a density O(N / log N) whereas trial plaintexts have density O(N). Last fiddled with by xilman on 20190916 at 15:10 

20190916, 16:27  #6 
"Robert Gerbicz"
Oct 2005
Hungary
1,531 Posts 
Learnt this, but there should be a better wording on this, beacuse if n is a Carmichael number (OK, poor choice for RSA) then actually you don't need to factorize n, although a factorization on these numbers is in polynomial time, but just one RSA decryption is faster.

20190916, 17:57  #7  
Bamboozled!
"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across
2×5,569 Posts 
Quote:


20190917, 02:53  #8  
Sep 2019
1000_{2} Posts 
Quote:
Quote:
But guys, I goofed. I wanted to make a real PGP key with n=M1277. But I got too excited, and made the key above with e=1277  and that's the one RSA exponent that does not work :( So here's the corrected key instead. This is an RSA key with n=M1277, e=65537: Code:
BEGIN PGP PUBLIC KEY BLOCK mKsEXYBAPgEE/R////////////////////////////////////////////////// //////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////// //////////////////////////////////8AAQG0GE1lcnNlbm5lIDEyNzcgPDJe MTI3Ny0xPojRBBMBAgAcAhsBAh4BAheABQJdgEEOBAsHCAkCFQgEFgMCAQAKCRBG eU+ty/qRMQ8cBPEB//////////////////////////////////////////////// //////////////////////////////////////////////////////////////// //////////////////////////////////////////////////8AMCEwCQYFKw4D AhoFAAQUDxwCfLqEuWPYnX3zd2kOycbgFUS4rQRdgEIwAQT9H/////////////// //////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////// /////wARAQABiL4EGAECAAkFAl2AQjACGwwACgkQRnlPrcv6kTGxZgTxAf////// //////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////// ////////////////////////////ADAhMAkGBSsOAwIaBQAEFLFmtP3VkQJB2MCk VEbJ3hrY7xiJ =f9Li END PGP PUBLIC KEY BLOCK 

20190917, 09:55  #9  
"Robert Gerbicz"
Oct 2005
Hungary
1531_{10} Posts 
Quote:
if gcd(65537,eulerphi(n)) != 1, then the decryption (in general) is not unique!!!!!! Not speaking about that it would be a way slower (in these cases), you need to find the original message in a haystack. 

20190918, 06:58  #10 
Sep 2019
8_{10} Posts 

20190918, 14:24  #11 
Random Account
Aug 2009
3731_{8} Posts 
The rule of digits I memorized, 2 * log(2) * 1277 indicates 768 digits. Is the implication here meaning this exponent has no factor, or is it saying that current factoring software cannot run it?

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