1.  Why the course?  Introduction to Number Theory  The make-up of the seminar  Interdisciplinary collaboration 1

2.  Enrichment  The needs of an advanced student  Further benefits 2

3.  Student ownership of the course  Student-teaching-students  Team oriented  Presentations, discussions, assignments, mini-projects 3

4.  Technology ◦ Excel ◦ C++ programming  An example: Application of the Euclidean Algorithm 4

5. 4.2

6. 4.3

7.  In summary, p=12553, q = 13006, m = pq = 163276871, and k = 79921.  Now to send the message “To be or not to be”, set the cipher shift A = 11, B = 12, and so on, and this message becomes the string of digits: 30251215252824253030251215 4.4

8. 4.5

9.  The encoded message is the list of numbers: 148419241, 62721998, 118084566, 40481382 4.6

10.  Now let’s try decoding a new message. It’s midnight, there’s a knock at our door, and a mysterious messenger delivers the following cryptic missive: 145387828, 47164891, 152020614, 27279275, 35356191 Use the k th modulo m computing method to solve the congruences. 4.7

11. x 78821 = 145387828 (mod 163276871) x = 30182523 x 78821 = 47164891 (mod 163276871) x = 26292524 x 78821 = 152020614 (mod 163276871) x = 19291924 x 78821 = 27279275 (mod 163276871) x = 30282531 x 78821 = 35356191 (mod 163276871) x = 122215 4.8

12.  This gives you the string of digits: 301525232629252419291924302825311122215  And now you use the number to letter substitution table for the final decoding step.  THOMPSONISINTROUBLE 4.9

13.  Supplying the obvious word breaks and punctuation, you read:  “Thompson is in trouble”  And off you go to the rescue 4.10

14.  Increases student involvement  Learning by exploration  Getting back to formal mathematics  Broadening the student’s mathematical horizon 5

15. The Idea Behind the Proof:  p 2 – 1 = (p + 1) (p – 1)  p + 1 and p – 1 are consecutive even numbers  (p + 1) (p – 1) = 2(i) * 2 (i + 1)  i is either even or odd integer 6.1 Suppose p > 3 is a prime number, Then 24 l p 2 – 1

16.  All prime numbers > 3 are odd, so p + 1 and p – 1 are consecutive even numbers and can be written as:  (p – 1) (p + 1) = 2 (i) * 2 (i + 1)  p-1, p and p+1 are three consecutive integers. One of three consecutive integers must be divisible by 3.  p is not divisible by 3, since p is a prime number, therefore either p + 1, or p – 1 is divisible by 3. 6.2

17.  Case 1:  (p + 1) (p – 1) = 2(i) * 2 (i + 1)  If i is even, let i = 2j, j = integer, then (p – 1)(p + 1) = 2(2j) *2((2j) + 1) = 8j(2j + 1)  Either j or 2j+ 1 is divisible by 3  If j is divisible by 3, then j = 3k  p 2 – 1= 8j(2j + 1) = 8(3k)(2j + 1) = 24(k(2j + 1))  If 2j + 1 is divisible by 3, then 2j + 1 = 3k p 2 – 1= 8j(2j + 1) = 8(j)(3k) = 24(jk)  Thus, if i is even then p 2 – 1 is divisible by 24. 6.3

18.  Case 2:  (p + 1) (p – 1) = 2(i) * 2 (i + 1)  If i is odd, let i = 2j + 1, j = integer, then (p – 1) (p + 1) = 2(2j + 1) *2((2j + 1) + 1) = 2(2j + 1) *4(j + 1) = 8(2j + 1)(j + 1)  Either 2j + 1 or j + 1 is divisible by 3  If 2j + 1 is divisible by 3, then 2j + 1 = 3k  p 2 – 1= 8(3k)(2j + 1) = 24(k(2j + 1))  If j + 1 is divisible by 3, then j + 1 = 3k p 2 – 1= 8(2j + 1)(3k) = 24(2j + 1)(k)  Thus, if i is odd then p 2 – 1 is divisible by 24. 6.4

19.  Summary:  Case 1: ◦ p 2 – 1= 24(k(2j + 1)) or ◦ p 2 – 1= 24(jk)  Case 2: ◦ p 2 – 1= 24(k(2j + 1)) or ◦ p 2 – 1= 24(2j + 1)(k) Therefore, 24 l p 2 – 1 6.5

20.  Topics are interchangeable  Disciplines are interchangeable 7

21.  Questions?