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Chapter 8 Introduction to Number Theory Contents Prime Numbers Fermat’s and Euler’s Theorems 2 Prime Numbers Primes numbers An integer p > 1 is a prime number if and only if it is divisible by only 1 and p. 3 Prime Numbers Integer factorization Any integer a > 1 can be factored in a unique way as a p1a1 p2a2 p3a3 ... ptat where p1 < p2 < … < pt are prime numbers and each ai is a positive integer. 91 = 7 × 13; 11101 = 7 × 112 ×13 4 Prime Numbers Another integer factorization If P is the set of all prime numbers, then any positive integer can be written uniquely in the following form: ap ap where each a p 0 pP The right side is the product over all possible prime numbers p. Most of the exponents ap will be 0. 3600 = 24×32×52×70×110×…. 5 Prime Numbers Another integer factorization The value of any given positive integer can be specified by listing all the nonzero exponents. The integer 12 =22×31 is represented by {a2=2, a3=1}. The integer 18 =21×32 is represented by {a2=1, a3=2}. The integer 91= 72×131 is represented by {a7= 2, a13= 1}. 6 Prime Numbers Multiplication Multiplication of two numbers is adding the corresponding exponents. k = 12 × 18 = 216 12 = 22 × 31 18 = 21 × 32 -----------------216 = 23 × 33 7 Prime Numbers Divisibility a|b → ap ≤ bp for all p a = 12; b= 36; 12|36 12 = 22×3; 36 = 22×32 a2 = 2 = b2 a3 = 1 ≤ 2 = b3 8 Prime Numbers GCD k = gcd (a, b) → kp = min(ap, bp) for all p 300 18 = 22×31×52 = 21×32×50 gcd (18, 300) = 21×31×50 = 6 9 Fermat’s and Euler’s Theorems Fermat’s theorem If p is prime and a is a positive integer not divisible by p, then ap-1 ≡ 1 (mod p) 10 Fermat’s and Euler’s Theorems Proof of Fermat’s theorem. Outline Show {1, 2, …, p-1}={a mod p, 2a mod p, …, (p-1)a mod p} Show ( p 1)! ( p 1)!a p 1 mod p . Since ( p 1)! is relatively prime to p, we multiply ( p 1)!-1 to both sides to get 1 a p 1 mod p . 11 Fermat’s and Euler’s Theorems Proof of Fermat’s theorem Show {1, 2, …, p-1}={a mod p, 2a mod p, …, (p-1)a mod p} Show ka mod p for any 1 ≤ k ≤ p-1 is in {1, 2, …, p-1} by showing that ka mod p ≠ k’a mod p for k≠ k’. Show ka mod p ≠ k’a mod p for 1 ≤ k≠ k’ ≤ p-1. Proof by contradiction Assume that ka ≡ k’a mod p for some 1 ≤ k≠ k’ ≤ p-1. Since a is relatively prime to p, we multiply a-1 to get k ≡ k’ mod p, which contradiction the fact that k≠ k’. 12 Fermat’s and Euler’s Theorems Proof of Fermat’s theorem Show ( p 1)! a p 1 ( p 1)! mod p . {1, 2, …, p-1} = {a mod p, 2a mod p, …, (p-1)a mod p} [1 2 ... ( p 1)] [( a mod p) (2a mod p ) ... (( p 1)a mod p)] mod p [1 2 ... ( p 1)] [a 2a ... ( p 1)a] mod p ( p 1)! ( p 1)! a p 1 mod p 1 a p 1 mod p 13 Fermat’s and Euler’s Theorems An alternative form of Fermat’s Theorem ap ≡ a mod p where p is prime and a is any positive integer. Proof If a and p are relatively prime, we get ap ≡ a mod p by multiplying a to each side of ap-1 ≡ 1 mod p. If a and p are not relatively prime, a = cp for some positive integer c. So ap ≡ (cp)p ≡ 0 mod p and a ≡ 0 mod p, which means ap ≡ a mod p. 14 Fermat’s and Euler’s Theorems An alternative form of Fermat’s Theorem ap ≡ a mod p where p is prime and a is any positive integer. p = 5, a = 3 35 = 243 ≡ 3 mod 5 p = 5, a = 10 105 = 100000 ≡ 10 mod 5 ≡ 0 mod 5 15 Fermat’s and Euler’s Theorems Euler’s Totient Function (n) The number of positive integers less than n and relatively prime to n. (37) = 36 37 is prime, so all the positive number from 1 to 36 are relatively prime to 37. (35) = 24 35 = 5×7 1, 2, 3, 4, 6, 8, 9,11, 12, 13, 16, 17, 18, 19, 22, 23, 24, 26, 27, 29, 31, 32, 33, 34 16 Fermat’s and Euler’s Theorems How to compute (n) In general, 1 (n) n (1 ), where p runs over all the primes dividing n p For a prime n, (Zn = {1,2,…, n-1}) ( n) n 1 For n = pq, p and q are prime numbers and p≠ q (n) ( p 1) (q 1) 17 Fermat’s and Euler’s Theorems Proof of (n) ( p 1) (q 1) (n) is the number of positive integers less than pq that are relatively prime to pq. (n) can be computed by subtract from pq – 1 the number of positive integers in {1, …, pq – 1} that are not relatively prime to pq. The positive integers that are not relatively prime to pq are a multiple of either p or q. { p, 2p,…,(q – 1)p}, {q, 2q, …,(p – 1)q} There is no same elements in the two sets. So, there are p + q – 2 elements that are not relatively prime to pq. Hence, (n) = pq – 1– (p + q – 2) = pq – p – q +1 = (p – 1)(q – 1) 18 Fermat’s and Euler’s Theorems Φ(21) = Φ(3)×Φ(7) = (3-1)×(7-1) = 2 ×6 = 12 Z21={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20} Φ(3)={3,6,9,12,15,18} Φ(7)={7,14} where the 12 integers are {1,2,4,5,8,10,11,13,16,17,19,20} 19 Fermat’s and Euler’s Theorems Euler’s theorem For every a and n that are relatively prime: a ( n ) 1 mod n a = 3; a = 2; n = 10; n = 11; Φ(10) = 4; Φ(11) = 10; 34 = 81 ≡ 1 mod 10 210 = 1024 ≡ 1 mod 11 20 Fermat’s and Euler’s Theorems Proof of Euler’s theorem a ( n ) 1 mod n If n is prime, it holds due to Fermat’s theorem. a n 1 1 mod n Otherwise (If n is not prime), define two sets R and S. show the sets R and S are the same. then, show a ( n ) 1 mod n 21 Fermat’s and Euler’s Theorems Proof of Euler’s theorem Set R The elements are positive integers less than n and relatively prime to n. The number of elements is (n) R={x1, x2,…, xΦ(n)} where x1< x2<…< xΦ(n) Set S Multiplying each element of R by a∈R modulo n S ={(ax1 mod n), (ax2 mod n),…(axΦ(n) mod n)}. 22 Fermat’s and Euler’s Theorems Proof of Euler’s theorem The sets R and S are the same. We show S has all integers less than n and relatively prime to n. S ={(ax1 mod n), (ax2 mod n),…(axΦ(n) mod n)} 1. 2. All the elements of S are integers less than n that are relatively prime to n because a is relatively prime to n and xi is relatively prime to n, axi must also be relatively prime to n. There are no duplicates in S. If axi mod n = axj mod n, then xi = xj. by cancellation law. 23 Fermat’s and Euler’s Theorems Proof of Euler’s theorem Since R and S are the same sets, (n) (n) (ax mod n) x i i i 1 i 1 (n) (n) ax x (mod n) i i 1 i i 1 (n) (n) a ( n ) xi xi (mod n) i 1 i 1 a ( n ) 1(mod n) 24 Fermat’s and Euler’s Theorems Alternative form of the theorem a ( n )1 a(mod n) If a and n are relatively prime, it is true due to Euler’s theorem. Otherwise, …. 25 Fermat’s and Euler’s Theorem The validity of RSA algorithm Given 2 prime numbers p and q, and integers n = pq and m, with 0<m<n, the following relationship holds. m ( n )1 m ( p 1)( q 1)1 m mod n If m and n are re relatively prime, it holds by Euler’s theorem. If m and n are not relatively prime, m is a multiple of either p or q. 26 Fermat’s and Euler’s Theorem Case 1: m is a multiple of p m=cp for some positive integer c. gcd(m, q)=1, otherwise, m is a multiple of p and q and yet m<pq because gcd(m, q)=1, Euler’s theorem holds mq 1 1 mod q by the rules of modular arithmetic, [m q 1 ] p 1 1 mod q m ( n ) 1 mod q m ( n ) 1 kq, for some integer k Multiplying each side by m=cp m ( n )1 m kcpq m kcn m ( n )1 m mod n 27 Fermat’s and Euler’s Theorem Case 2: m is a multiple of q prove similarly. Thus, the following equation is proved. m ( n )1 m ( p 1)( q 1)1 m mod n 28 Fermat’s and Euler’s Theorem An alternative form of this corollary is directly relevant to RSA. m k ( n )1 [( m ( n ) k m1 ] mod n [(1) k m] mod n, m mod n by Euler' s theorem 29