**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