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bio website mrgeek.me
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Founder at Mr. Geek

I have an infinite curiosity for knowledge.


Dec
7
answered How do I calculate the private key in RSA?
Dec
6
awarded  Editor
Dec
6
revised Why are these techniques not feasible to crack RSA?
added 29 characters in body
Dec
6
asked How do we arrive to the equation for solving D in RSA?
Dec
6
comment How do I calculate the private key in RSA?
I figured it out and I'll answer my question soon
Dec
6
comment Why are these techniques not feasible to crack RSA?
Haha, I know what you mean :)
Dec
6
comment Why are these techniques not feasible to crack RSA?
I mean to say, my technique works most of the time because initially I have this: de ≡ 1 mod(phi(n)) which in turn becomes de mod(phi(n)) = 1 , solving for d we get d = 1/e mod(phi(n)) and then once I get the answer, I simply inverse it. What am I doing wrong here, as it has worked for me all the time?
Dec
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awarded  Supporter
Dec
6
comment Why are these techniques not feasible to crack RSA?
You mean to say finding out the phi of n is as hard as finding out its factors. I believe finding out the factors would be a bit more harder computationally but I might be wrong because RSA hasn't been cracked yet with an efficient mathematical solution either. By the way, am I correct in saying that d is an inverse of 1/e mod phi(n) ? Because if we compute d = 1/e mod phi(n) we get a fraction so we must inverse d to get an integer, just clearing up buddy.
Dec
6
comment Why are these techniques not feasible to crack RSA?
You can solve for phi(n) by doing a quick computation in Wolfram Alpha and it tells you the totient or phi of n. How is that hard?
Dec
6
comment Why are these techniques not feasible to crack RSA?
How about solving for d when you know n and e as I explained in my question, using the inverse operation. I don't see how that is hard. I can't digest it, maybe I am imagining smaller numbers.
Dec
6
asked Why are these techniques not feasible to crack RSA?
Dec
5
asked How do I calculate the private key in RSA?
Dec
5
comment Get RSA PlainText without Knowing Private Key
@mikeazo Interesting. The description in wikipedia says E(mt) which I believe is the same thing you're trying to do here, which is right. So for instance if we compute the RSA-Crack() with the ciphertext 100 times, we will divide it by 100, which is 'i' This gives us a probability of 1.0 which serves as an answer. Correct?
Dec
5
comment Get RSA PlainText without Knowing Private Key
@CodesInChaos, can you tell me why is Mike trying to divide the answer by i to get m1. Confused a bit. Why isn't c2 written as c2 = c1 . m^e mod n?
Dec
5
awarded  Autobiographer