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]


Note that


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