Announcements

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Oct 28, 2009

· Hints on HW 5

8.7.11

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!)

8.7.12

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

8.7.17

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

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