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Introduction

Introduction . Why the course? Introduction to Number Theory The make-up of the seminar Interdisciplinary collaboration. 1. What’s in it for the Student. Enrichment The needs of an advanced student Further benefits. 2. Seminar Structure. Student ownership of the course

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Introduction

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

  2. What’s in it for the Student • Enrichment • The needs of an advanced student • Further benefits 2

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

  4. Hands-on and Collaboration • Technology • Excel • C++ programming • An example: Application of the Euclidean Algorithm 4

  5. Encryption/Decryption Examples 4.2

  6. Encryption: • To illustrate the encoding and decoding procedure with the primes p = 12554 and q = 13007. We multiply them together to get the modulus m = pq = 163276871, and record for future use: • (m) = (p – 1) (q – 1) = 163251312 • Choose a k that is relative prime to (m) so take k = 79921. 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. The number m is 9 digits long, so we break the message up in to 8 digit numbers: 30251215, 25282425, 30302512, 15 • Next we use the method of successive squaringto raise each of these numbers to the nth power modulo m. 3025121579921 149419241 (mod 163276871) 252824257992162721998 (mod 163276871) 3030251279921118084566 (mod 163276871) 157992140481382 (mod 163276871) 4.5

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

  10. Decryption: • 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 kthmodulo mcomputing method to solve the congruences. 4.7

  11. x78821 = 145387828 (mod 163276871) x = 30182523 x78821 = 47164891 (mod 163276871) x = 26292524 x78821 = 152020614 (mod 163276871) x = 19291924 x78821 = 27279275 (mod 163276871) x = 30282531 x78821 = 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. Benefits • Increases student involvement • Learning by exploration • Getting back to formal mathematics • Broadening the student’s mathematical horizon 5

  15. Theorem Suppose p > 3 is a prime number, Then 24 l p2 – 1 The Idea Behind the Proof: • p2 – 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

  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 jis divisible by 3, then j = 3k • p2 – 1= 8j(2j + 1) = 8(3k)(2j + 1) = 24(k(2j + 1)) • If 2j + 1 is divisible by 3, then 2j + 1 = 3k p2 – 1= 8j(2j + 1) = 8(j)(3k) = 24(jk) • Thus, if i is even then p2 – 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 • p2 – 1= 8(3k)(2j + 1) = 24(k(2j + 1)) • If j + 1 is divisible by 3, then j + 1 = 3k p2 – 1= 8(2j + 1)(3k) = 24(2j + 1)(k) • Thus, if i is odd then p2 – 1 is divisible by 24. 6.4

  19. Summary: • Case 1: • p2 – 1= 24(k(2j + 1)) or • p2 – 1= 24(jk) • Case 2: • p2 – 1= 24(k(2j + 1)) or • p2 – 1= 24(2j + 1)(k) Therefore, 24 l p2 – 1 6.5

  20. Modular Nature of the Seminar • Topics are interchangeable • Disciplines are interchangeable 7

  21. Questions?

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