Prime Numbers Prime Numbers Prime number : an integer p>1 that - PowerPoint PPT Presentation

Prime Numbers Prime Numbers  Prime number : an integer p>1 that is divisible only by 1 and itself, ex. 2, 3,5, 7, 11, 13, 17…  Composite number : an integer n>1 that is not prime p g p Prime Numbers  Fact : there are infinitely many prime numbers. (by Euclid) pf:  on the contrary, assume a n is the largest prime number  on the contrary assume a is the largest prime number pf:  let the finite set of prime numbers be {a 0 , a 1 , a 2 , …. a n } 密碼學與應用  the n mber b  the number b = a 0 *a 1 *a 2 *…*a n + 1 is not divisible by any a i a *a *a * *a + 1 is not di isible b an a 海洋大學資訊工程系 i.e. b does not have prime factors  a n 丁培毅 丁培毅  if b h  if b has a prime factor d, b>d> a n , then “d is a prime i f t d b>d> th “d i i 2 2 cases: number that is larger than a n ” … contradiction  if b does not have any prime factor less than b, then b is a  if b does not have any prime factor less than b then “b is a prime number that is larger than a n ” … contradiction 1 2 Prime Number Theorem Prime Number Theorem Factors Factors  Prime Number Theorem : e Nu be eo e :  Let  (x) be the number of primes less than x  Every composite number can be expressible as a  Then  Then x x  (x)  product aꞏb of integers with 1 < a, b< n ln x in the sense that the ratio  (x) / (x/ln x)  1 as x   in the sense that the ratio  (x) / (x/ln x)  1 as x    Every positive integer has a unique representation x x  (x)  1.10555 ln x  (x)  1 10555  (x)   (x)   Also, and for x  17, and for x  17  Also as a product of prime numbers raised to different ln x powers. p  Ex: number of 100-digit primes  Ex: number of 100 digit primes  Ex. 504 = 2 3 ꞏ 3 2 ꞏ 7, 1125 = 3 2 ꞏ 5 3 10 100 10 99  (10 100 )  (10 99 )   (10 100 ) -  (10 99 )  ln 10 100  3 9  10 97  3.9  10 - ln 10 99 3 4

Factors Factors Factorization into primes Factorization into primes  Theorem: Every positive integer is a product of primes.  Lemma: p is a prime number and p | a b p | a or p | b,  Lemma: p is a prime number and p | aꞏb p | a or p | b This factorization into primes is unique, up to more generally, p is a prime number and p | aꞏbꞏ...ꞏz reordering of the factors. • Empty product equals 1. p must divide one of a b p must divide one of a, b, …, z z • Prime is a one factor product. P i i f t d t  Proof: product of primes  proof:  assume there exist positive integers that are not product of primes  let n be the smallest such integer  let n be the smallest such integer  case 1: p | a  case 1: p | a  since n can not be 1 or a prime, n must be composite, i.e. n = aꞏb  case 2: p | a,  since n is the smallest, both a and b must be products of primes.  n = aꞏb must also be a product of primes contradiction  n = a b must also be a product of primes, contradiction  p | a and p is a prime number  gcd(p a) = 1  1 = a x + p y  p | a and p is a prime number  gcd(p, a) = 1  1 = a x + p y  Proof: uniqueness of factorization  multiply both side by b, b = b a x + b p y as = r 1 c 1 r 2 c 2 ꞏꞏꞏr k c k p 1 a 1 p 2 a 2 ꞏꞏꞏp s c 1 r 2 c 2 ꞏꞏꞏr k c k q 1 b 1 q 2 b 2 ꞏꞏꞏq t bt  p | a b  p | b  assume n = r 1 where p i , q j are all distinct primes.  In general: if p | a then we are done, if p | a then p | bc…z, continuing c 1 r 2 c 2 ꞏꞏꞏr k c k )  let m = n / (r 1 this way, we eventually find that p divides one of the factors of the  consider p 1 for example, since p 1 divide m = q 1 q 1 ..q 1 q 2 …q t , p 1 must product product 1 1 1 1 1 2 t 1 divide one of the factors q j , contradict the fact that “p i , q j are distinct primes” 5 6 (“Fair-MAH”) Fermat’s Little Theorem Fermat s Little Theorem Fermat s Little Theorem Fermat’s Little Theorem  Ex: 2 10 = 1024  1 (mod 11)  If p is a prime p | a then a p-1  1 (mod p) ( )  1 (mod p)  If p is a prime, p | a then a 2 53 = (2 10 ) 5 2 3  1 5 2 3  8 (mod 11) * ), define  (x)  a ꞏ x (mod p) be Proof:  let S = {1, 2, 3, …, p-1} (Z p i.e. 2 53  2 53 mod 10  2 3  8 (mod 11) ( ) a mapping  : S  Z a mapping  : S  Z   x  S,  (x)  0 (mod p)   x  S,  (x)  S, i.e.  : S  S if  (x)  a ꞏ x  0 (mod p)  x  0 (mod p) since gcd(a, p) = 1  ( ) ( p) ( p) g ( , p)  if n is prime then 2 n-1  1 (mod n)  1 (mod n)  if n is prime, then 2   x, y  S, if x  y then  (x)   (y) since i.e. if 2 n-1  1 (mod n) then n is not prime  (  ) if  (x)   (y)  a ꞏ x  a ꞏ y  x  y since gcd(a, p) = 1 usually if 2 n-1  1 (mod n) then n is prime  1 (mod n), then n is prime usually, if 2  from the above two observations,  (1),  (2),...  (p-1) are  exceptions: 2 561-1  1 (mod 561) although 561 = 3 ꞏ 11 ꞏ 17 distinct elements of S  1ꞏ2 ꞏ... ꞏ(p-1)   (1)ꞏ  (2)ꞏ...ꞏ  (p-1)  (aꞏ1)ꞏ(aꞏ2)ꞏ…ꞏ(aꞏ(p-1)) 2 1729-1  1 (mod 1729) although 1729 = 7 ꞏ 13 ꞏ 19 1729 1  1 2 ( 1) (1) (2) ( 1) ( 1) ( 2) ( ( 1))  a p-1 (1ꞏ2 ꞏ... ꞏ(p-1)) (mod p)  (  ) is a quick test for eliminating composite number  since gcd(j, p) = 1 for j  S, we can divide both side by 1, 2,  since gcd(j p) = 1 for j  S we can divide both side by 1 2 3, … p-1, and obtain a p-1  1 (mod p) 7 8

Euler’s Totient Function  (n) Euler s Totient Function  (n) How large is  (n)? How large is  (n)?   (n)  n ꞏ 6/  2 as n goes large   (n): the number of integers 1  a<n s.t. gcd(a,n) = 1  ex. n = 10,  (n) = 4 the set is {1,3,7,9}  Probability that a prime number p is a factor of a random  properties of  (•)  ( ) p p number r is 1/p u be s /p   (p) = p-1, if p is prime   (p r ) = p r - p r-1= (1-1/p) ꞏ p r if p is prime   (p ) = p - p = (1-1/p) p , if p is prime p p 2p 3p 4p 2p 3p 4p   (nꞏm) =  (n) ꞏ  (m) if gcd(n,m)=1 排容原理  Probability that two independent random numbers r 1 and r 2 n m - (n-  (n)) m - (m-  (m)) n + (n-  (n)) (m-  (m)) =  (n)  (m) (n  (n)) m (m  (m)) n + (n  (n)) (m  (m)) =  (n)  (m) n m both have a given prime number p as a factor is 1/p 2 b h h i i b i / 2   (nꞏm) =  The probability that they do not have p as a common factor  ((d /d /d ) 2 )ꞏ  (d 3 )ꞏ  (d 3 )ꞏ  (n/d /d )ꞏ  (m/d /d )  ((d 1 /d 2 /d 3 ) )  (d 2 )  (d 3 )  (n/d 1 /d 2 )  (m/d 1 /d 3 ) is thus 1 – 1/p 2 if gcd(n,m)=d 1 , gcd(n/d 1 ,d 1 )=d 2 , gcd(m/d 1 ,d 1 )=d 3  The probability that two numbers r 1 and r 2 have no common p y p|n p|n   (n) = n  (1-1/p)  ( )  (1 1/ ) 1 2 prime factor is P = (1-1/2 2 )(1-1/3 2 )(1-1/5 2 )(1-1/7 2 )…  ex.  (10)=(2-1)ꞏ(5-1)=4  (120)=120(1-1/2)(1-1/3)(1-1/5)=32 9 10 Pr{ r and r relatively prime } Pr{ r 1 and r 2 relatively prime } How large is  (n)? How large is  (n)?   (n) is the number of integers less than n that are relative  Equalities: 1 1 = 1+x+x 2 +x 3 +… prime to n 1-x   (n)/n is the probability that a randomly chosen integer is  ( ) 1 + 1/2 2 + 1/3 2 + 1/4 2 + 1/5 2 + 1/6 2 + p y y g 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + …  /6 =  2 /6 relatively prime to n  P = (1-1/2 2 )(1-1/3 2 )(1-1/5 2 )(1-1/7 2 ) ꞏ ...  Therefore,  (n)  n ꞏ 6/  2  Therefore,  (n) n 6/  = ((1+1/2 2 +1/2 4 +...)(1+1/3 2 +1/3 4 +...) ꞏ ...) 1 ) -1 ((1+1/2 2 +1/2 4 + )(1+1/3 2 +1/3 4 + )  P n = Pr { n random numbers have no common factor } = (1+1/2 2 +1/3 2 +1/4 2 +1/5 2 +1/6 2 +…) -1  n independent random numbers all have a given prime p as a  n independent random numbers all have a given prime p as a = 6/  2 factor is 1/p n  0.61 0.61  They do not all have p as a common factor 1 – 1/p n  They do not all have p as a common factor 1 – 1/p  P n = (1+1/2 n +1/3 n +1/4 n +1/5 n +1/6 n +…) -1 is the Riemann zeta each positive number has a unique prime number factorization 45 2 = 3 4 ꞏ 5 2 function  (n) http://mathworld.wolfram.com/RiemannZetaFunction.html function  (n) http://mathworld.wolfram.com/RiemannZetaFunction.html ex. 45 = 3 ex 5  Ex. n=4,  (4) =  4 /90  0.92 11 12