Oct 28, 2009
· Hints on HW 5
For (n-1) case
You can find the generic expression for (n-1)^e and then apply the “mod n” expression all
The induction method I mentioned in the class can be used to prove that the same happens for (n-1)^m where m is odd – note e is odd and hence a special case (you have to show through discussion this!)
Here note that: e and d are public and private keys respectively
In the first part you have to show that
m = c^d mod n [this is the decryption process]
c = m^e mod n [this is the encryption process] -
You can replace the value of c in the decryption expression
An important observation is in RSA the following holds:
ed mod Totient(n) = 1
Hence you can say ed = k.Totien(n) + 1 for some integer k; use this in your steps to show that the decryption process works out
Also assume m < n.
Of course you have to use the properties of “mod n” operations that are in one of the slides
For example: [(a mod n) × (b mod n)] mod n = (a × b) mod n
This should be enough (if you take crypto you will do more on proving this).
The second part is to show that m = (m^d mod n)^e mod n
Eve can essentially do a man in the middle attack by changing keys in the server: Alice and Bob would be sending to each other without knowing that Eve is able to intercept the encrypted messages, decrypt it and re-encrypt it to forward to the receiver.
Hope this helps.
Oct 12, 2009
· HW 4 posted - due on Oct 21
· Midterm date has been changed to Nov 3 !!
Sept 20, 2009
· HW 2 posted - due on Sept 29
Sept 12, 2009
· Lab 1 is posted - due on Sept 25
Sept 8, 2009
· Lecture 2 posted
Sept 5, 2009
· HW1 Posted - check the assignment link
Sept 1, 2009
· Handout, Course Schedule documents have been added (Click the link in the main page)
· Lecture 1 has been added