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

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

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

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

4.2

4.3

 In summary, p=12553, q = 13006, m = pq = , and k =  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:

4.5

 The encoded message is the list of numbers: , , ,

 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: , , , , Use the k th modulo m computing method to solve the congruences. 4.7

x = (mod ) x = x = (mod ) x = x = (mod ) x = x = (mod ) x = x = (mod ) x =

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

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

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

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

 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

 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

 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

 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

 Topics are interchangeable  Disciplines are interchangeable 7

 Questions?