Math 412: Number Theory Lecture 13 Applications of

Transcription

1 Math 412: Number Theory Lecture 13 Applications of Gexin Yu College of William and Mary

2 Partition of integers A partition λ of the positive integer n is a non increasing sequence of positive integers (λ 1, λ 2,..., λ r ) such that λ 1 + λ λ r = n. The integers λ 1, λ 2,..., λ r are called the parts of the partition λ. The number of different partitions of n is dented by p(n), which is called the partition function. We often study the partitions of n with restriction on the parts. We use p(n conditions) to count the partitions of n where the parts satisfy the conditions specified. p S (n) is the number of partitions of n into parts from S; p D (n) is the number of partitions of n into distinct parts; p m (n) is the number of partitions of n into parts each m.

3 Thm: The generating function for p(n) equals p(n)x n 1 = 1 x j. n=0 j=1

4 Thm: The generating function for p D (n) equals p D (n)x n = (1 + x j ). n=0 j=1

5 Thm: Let S be a subset of the set of positive integers. The generating function for p S (n) equals n=0 p S (n)x n = j S 1 1 x j.

6 Thm: Let S be a subset of the set of positive integers. The generating function for p S (n) equals n=0 p S (n)x n = j S The generating function for ps D (n) equals n=0 p D S (n)x n = j S 1 1 x j. (1 + x j ).

7 Euler Parity Theorem: there are the same number of partitions of n into odd parts as there are partitions of n into distinct parts.

8 Applications in cryptography

9 Private cryptosystems Caesar Cipher: a letter is shifted by a constant, i.e., C P + k (mod 26)

10 Private cryptosystems Caesar Cipher: a letter is shifted by a constant, i.e., C P + k (mod 26) Affine Transformation: C ap + b (mod 26), wehre (a, 26) = 1.

11 Solving the equation a k b (mod m) If (b, m) = 1 and (k, φ(m)) = 1, then b u is a solution, where ku φ(m)v = 1 and u, v > 0.

12 Solving the equation a k b (mod m) If (b, m) = 1 and (k, φ(m)) = 1, then b u is a solution, where ku φ(m)v = 1 and u, v > 0. What if (b, m) > 1? Make m = pq!

13 Solving the equation a k b (mod m) If (b, m) = 1 and (k, φ(m)) = 1, then b u is a solution, where ku φ(m)v = 1 and u, v > 0. What if (b, m) > 1? Make m = pq! So to solve it, we need to compute φ(m). If m = pq for primes p, q, it is easy. But it is very hard to break m into pq!!

14 Public Key Cryptosystem -RSA cryptosystem RSA cryptosystem (Rivest-Shamir-Adleman 1983; Clifford Cocks 1973) Encrption:

15 Public Key Cryptosystem -RSA cryptosystem RSA cryptosystem (Rivest-Shamir-Adleman 1983; Clifford Cocks 1973) Encrption: 1 Take m = pq and k so that (k, φ(m)) = 1.

16 Public Key Cryptosystem -RSA cryptosystem RSA cryptosystem (Rivest-Shamir-Adleman 1983; Clifford Cocks 1973) Encrption: 1 Take m = pq and k so that (k, φ(m)) = 1. 2 Encode a message with numbers (A=11, B=12, C=13, et al), and get a large number N.

17 Public Key Cryptosystem -RSA cryptosystem RSA cryptosystem (Rivest-Shamir-Adleman 1983; Clifford Cocks 1973) Encrption: 1 Take m = pq and k so that (k, φ(m)) = 1. 2 Encode a message with numbers (A=11, B=12, C=13, et al), and get a large number N. 3 Divide N into segments a 1, a 2,..., a r. And let b i ai k (mod m).

18 Public Key Cryptosystem -RSA cryptosystem RSA cryptosystem (Rivest-Shamir-Adleman 1983; Clifford Cocks 1973) Encrption: 1 Take m = pq and k so that (k, φ(m)) = 1. 2 Encode a message with numbers (A=11, B=12, C=13, et al), and get a large number N. 3 Divide N into segments a 1, a 2,..., a r. And let b i ai k (mod m). 4 send out number M = b 1 b 2... b r.

19 Public Key Cryptosystem -RSA cryptosystem RSA cryptosystem (Rivest-Shamir-Adleman 1983; Clifford Cocks 1973) Encrption: 1 Take m = pq and k so that (k, φ(m)) = 1. 2 Encode a message with numbers (A=11, B=12, C=13, et al), and get a large number N. 3 Divide N into segments a 1, a 2,..., a r. And let b i ai k (mod m). 4 send out number M = b 1 b 2... b r. Decryption:

20 Public Key Cryptosystem -RSA cryptosystem RSA cryptosystem (Rivest-Shamir-Adleman 1983; Clifford Cocks 1973) Encrption: 1 Take m = pq and k so that (k, φ(m)) = 1. 2 Encode a message with numbers (A=11, B=12, C=13, et al), and get a large number N. 3 Divide N into segments a 1, a 2,..., a r. And let b i ai k (mod m). 4 send out number M = b 1 b 2... b r. Decryption: 1 Divide M into b 1, b 2,..., b r

21 Public Key Cryptosystem -RSA cryptosystem RSA cryptosystem (Rivest-Shamir-Adleman 1983; Clifford Cocks 1973) Encrption: 1 Take m = pq and k so that (k, φ(m)) = 1. 2 Encode a message with numbers (A=11, B=12, C=13, et al), and get a large number N. 3 Divide N into segments a 1, a 2,..., a r. And let b i ai k (mod m). 4 send out number M = b 1 b 2... b r. Decryption: 1 Divide M into b 1, b 2,..., b r 2 Solve x k b i (mod m) using the above method to get a i s

22 Public Key Cryptosystem -RSA cryptosystem RSA cryptosystem (Rivest-Shamir-Adleman 1983; Clifford Cocks 1973) Encrption: 1 Take m = pq and k so that (k, φ(m)) = 1. 2 Encode a message with numbers (A=11, B=12, C=13, et al), and get a large number N. 3 Divide N into segments a 1, a 2,..., a r. And let b i ai k (mod m). 4 send out number M = b 1 b 2... b r. Decryption: 1 Divide M into b 1, b 2,..., b r 2 Solve x k b i (mod m) using the above method to get a i s 3 We get a 1, a 2,..., a r so N. Now decode the numbers into a message.