Breaking RSA if you know m^k = 1 mod n

  • Thread starter nudan
  • Start date
  • #1
1
0
Hello I'm looking for some guidance:
 
Last edited:

Answers and Replies

  • #2
1,056
0
You seem to assume that everyone knows what you are talking about--why?

You certainly have not defined terms, but when I look into this, it is my impression that you have no understanding of modular arithmetic. Why not check that out?
 
  • #3
2
0
Look, RSA is built on the fact in the private key (d,n) and the public key (e,n) d and e are exponential inverses of each other mod n. If x^d or x^e are congruent to 1 then d = phi(n) or e = phi(n) respectively... Take a look at Euler's Theorem which is a generalize form of Fermat's Little Theorem. This is why in RSA there is so much care taken in picking keys.
 

Related Threads on Breaking RSA if you know m^k = 1 mod n

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
11
Views
6K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
7
Views
3K
Top