Reading assignment:
Sections 3.7, and 3.8 of Chapter 3 ( = Sections 2.6, and 2.7 of Chapter 2 of 5th Edition)

Written assignment:

1. (5 points) Exercise 17 Sect. 3.7, pages 244, 245 ( = Exercise 17 Sect. 2.6, pages 194, 195 of 5th Ed.)
   Note: As I mentioned in class, for part c), you can use Theorem 2 on page 234 ( = page 184 of 5th Ed.) to successively cancel each of the factors of (p-1)! from each side of the congruence. But it is legal (and easier) to use Wilson's Theorem instead.
2. (4 points) Exercise 27 Sect. 3.7, page 245 ( = Exercise 27 Sect. 2.6, page 195 of 5th Ed.)
3. (3 points) Exercise 46 Sect. 3.7, page 246 ( = Exercise 46 Sect. 2.6, page 196 of 5th Ed.)
   Note: You can do this "by hand", but there is a powerful calculator program, bc, on the Unix machines (ub, bulldog, and the csdev machines) that can do the necessary calculations very fast and easily. Here is how to encrypt as in Example 11 on pages 192-193. Here e = 13 and we need to raise 1819 and 1415 to the power 13 mod 2537:

   ub(17): bc       <-- run bc
   (1819^13)%2537   <-- enter the first calculation
   2081             <-- the result of the first calculation
   (1415^13)%2537   <-- enter the second calculation
   2182             <-- the result of the second calculation
   quit             <-- exit bc
   ub(18): 
   

4. (3 points) Exercise 47 Sect. 3.7, page 246 ( = Exercise 47 Sect. 2.6, page 196 of 5th Ed.)
   Again, you can do this "by hand", but it is grim, and bc can do it for you almost instantly.
   Here is how you can decrypt the previous message, 2081 2182, using d = 937 ( = the inverse of 13 mod (43*58) ):

   ub(19): bc
   (2081^937)%2537
   1819
   (2182^937)%2537
   1415
   quit
   ub(20):