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Android apps Developed on Google Playstore - Poll-o-mania and 8-Puzzle


Aug
1
awarded  Popular Question
Nov
21
accepted Difficulty of breaking RSA for a given key size
Nov
21
accepted Can RSA encryption produce collisions?
Nov
21
comment RSA finding the inverse of the public exponent
You can say that Euler's theorem forms the basis for PKI.
May
15
comment Difficulty of breaking RSA for a given key size
so you are saying that breaking a 1024 bit rsa key is less difficult than breaking a 128 bit symmetric key
May
15
comment Difficulty of breaking RSA for a given key size
@mikeazo : "By restricting your tests to odd primes" - for this you will have to check whether the number is a prime or not first of all. And then divide N by that number. For checking I guess you would have to do a primality test which is another overhead. So isn't dividing N by every number from 3-root(N) a better idea?
May
14
comment Difficulty of breaking RSA for a given key size
@mikeazo : you are right about the root(N) thing. But why check only every odd prime number?
May
14
revised Difficulty of breaking RSA for a given key size
edited body; edited title
May
14
comment Difficulty of breaking RSA for a given key size
@PaŭloEbermann : I have edited it as AES.:
May
14
awarded  Editor
May
14
revised Difficulty of breaking RSA for a given key size
added 67 characters in body
May
14
comment Difficulty of breaking RSA for a given key size
@RickyDemer : You are right about that. So then you see the complexity increases even further. Then how is 1024 bit rsa key equivalent to 128 bit des key?
May
14
awarded  Commentator
May
14
comment Exposing RSA private-key data… bad?
What is 'd' that you have mentioned?
May
14
asked Difficulty of breaking RSA for a given key size
Apr
18
comment Can RSA encryption produce collisions?
can, you atleast point me to the proof of what henrick hellstrom said - me1(modN)=me2(modN), because it will only happen if GCD(e,LCM(p−1,q−1))≠1.
Apr
16
comment Can RSA encryption produce collisions?
@HenrickHellström : dude please tell me the logic behind GCD(m1-m2,N)!=1
Apr
15
comment Can RSA encryption produce collisions?
@HenrickHellström : Or can you just give me the link where there is a proof or something that explains what you said - m^e1(modN)=m^e2(modN), because it will only happen if GCD(m1−m2,N)≠1
Apr
15
comment Can RSA encryption produce collisions?
@HenrickHellström : You don't reply to my comments. But at least tell something about poncho's answer. poncho says collisions are impossible.
Apr
15
comment Can RSA encryption produce collisions?
@HenrickHellström : I understand that if GCD(k,N)!=1 then k is either a multiple of p or q. But I don't undertand that m1-m2