183 reputation
6
bio website mrgeek.me
location London, United Kingdom
age 23
visits member for 11 months
seen Jan 16 at 2:54

Founder at Mr. Geek

The person in my display picture is Al Khwarizmi, master of algorithms.

I have an infinite curiosity for knowledge.


Jan
5
comment In this example, which is a premaster secret, and which is a master secret?
Okay, assume I meant premaster-secret and master-secret, are they related to the Diffie Helman in any way, because with just RSA as key exchange, there would be no premaster-secret or master-secret. At least this is what Wikipedia depicts in its hadnshake section, which I believe is using the Diffie Helman as Key exchange. Insight is valued.
Jan
5
comment In this example, which is a premaster secret, and which is a master secret?
I did, didn't get the hang of it. Waiting for my fellow Crypto users to answer.
Jan
5
comment In this example, which is a premaster secret, and which is a master secret?
Thanks for your reply but it doesn't answer my question. I think the master key is probably the 2 that's in the end and the pre-master the numbers 6 and 15, but I am not sure, hence the question was asked to solicit expertise.
Dec
9
comment How does this happen in RSA malleability?
Makes more sense if Wikipedia said congruent to instead of using the equality operator. I believe you're right.
Dec
8
comment How does this happen in RSA malleability?
I have read up but I am not getting this specific example, can you enlighten me how does this come around. I know what congruence is, it means 594 is in the same equivalence class is 34 w.r.t mod 35.
Dec
8
comment How does this happen in RSA malleability?
It's 34. But we have taken the results of c1 and c2 already, which gives us 594. Why should we mod again? @mikeazo At this stage it's like 594 = 34 which is weird.
Dec
8
comment How does this happen in RSA malleability?
Anyone folks? Anyone who can clarify this.
Dec
8
comment Why are these techniques not feasible to crack RSA?
@AFS: You can calculate phi(n) because N , the modulus is public, remember (N,e). So calculating that would be easy, well for smaller number perhaps ? :)
Dec
7
comment Why are these techniques not feasible to crack RSA?
Made some corrections to the formula. There was some trouble in the denominator. It correctly represents d more accurately.
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
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
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?