network profile
| bio | website | |
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| location | ||
| age | ||
| visits | member for | 1 year, 1 month |
| seen | Nov 21 '12 at 15:43 | |
| stats | profile views | 4 |
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Nov 21 |
accepted | Difficulty of breaking RSA for a given key size |
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Nov 21 |
accepted | Can RSA encryption produce collisions? |
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Nov 21 |
comment |
RSA finding the inverse of the public exponent You can say that Euler's theorem forms the basis for PKI. |
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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 |
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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? |
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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? |
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May 14 |
revised |
Difficulty of breaking RSA for a given key size edited body; edited title |
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May 14 |
comment |
Difficulty of breaking RSA for a given key size @PaŭloEbermann : I have edited it as AES.: |
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May 14 |
awarded | Editor |
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May 14 |
revised |
Difficulty of breaking RSA for a given key size added 67 characters in body |
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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? |
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May 14 |
awarded | Commentator |
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May 14 |
comment |
Exposing RSA private-key data… bad? What is 'd' that you have mentioned? |
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May 14 |
asked | Difficulty of breaking RSA for a given key size |
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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. |
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Apr 16 |
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Can RSA encryption produce collisions? @HenrickHellström : dude pleaseeeeeeeeeeeeeeeeeeeeeee!!!! |
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Apr 16 |
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Can RSA encryption produce collisions? @HenrickHellström : dude please tell me the logic behind GCD(m1-m2,N)!=1 |
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Apr 15 |
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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 |
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Apr 15 |
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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. |
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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 |
