a (mod m ) b is congruent to modulo a m b mod - PDF document

Number Theory and its Applications • Modular Exponentiation • Euclidean Algorithm for GCD • Solving Linear Congruences • Chinese Remainder Theorem and Application to Arithmetic with large numbers • Covered in Sections 3.6 and 3.7 Based on Rosen and slides by K. Busch 1 Modular Arithmetic Recap  Z   , m a b Z a  (mod m ) b “ is congruent to modulo ” a m b  mod mod a m b m Examples: 1   13 (mod 12 ) 0 (mod ) m m 11    5 (mod 6 ) 0 (mod ) k m m 2 1

Equivalent statements a  (mod m ) b  mod mod a m b m  | m a b     , k Z a b km 3  3 mod 8 3 0 7 1 3 6 2 3 5 4 Length of line represents number 4 2

 11 mod 8 3 0 7 1 11 6 2 3 5 4 Length of line represents number 5  19 mod 8 3 0 7 1 19 6 2 3 5 4 Length of line represents number 6 3

  3 11 19 (mod 8 ) 0 0 0 7 1 7 1 7 1 11 3 19 6 2 6 2 6 2 3 5 3 5 3 5 4 4 4 All lines terminate in same number 7 “Congruence class” of modulo : a m   { | (mod )} S a b a b m There are congruence classes: m  , , , S S S  0 1 m 1 8 4

Closure under addition: a  (mod m ) b    (mod m ) a c b d c  (mod m ) d Proof sketch: a    (mod m ) b a b sm     (  ) a c d b s t m c    (mod m ) d c d tm 9 Closure under multiplication: a  (mod m ) b    (mod m ) a c b d c  (mod m ) d Proof sketch: a    (mod m ) b a b sm     ( )( ) a c b sm d tm     bd ( ) m bt ds stm c    (mod m ) d c d tm 10 5

Closure under mod:  mod ( mod ) mod a m a m m (Follows from definition of mod)  ( 7 mod 5 ) 2   ( 7 mod 5 ) mod 5 2 mod 5 2 11 Useful results for arithmetic with large numbers:    ( ) mod (( mod ) ( mod )) mod a b m a m b m m  mod (( mod )( mod )) mod ab m a m b m m (Follows from previous slides) Example:   57 55 mod 50 (( 57 mod 50 )( 55 mod 50 )) mod 50   7 5 mod 50  35 12 6

Modular exponentiation b n mod Compute efficiently using m small numbers Binary expansion of n       k 1  k 1 2  2 2 2 n a a a a a a    b b b b b k 1 1 0 k 1 1 0 n mod b m   k 1 a 2 2 a  a  mod b b b m k 1 1 0  1  k    2 2 a  a a ((  mod ) ( mod ) ( mod )) mod b m b m b m m k 1 1 0 13  Example: 3 644 mod 645 36     9 7 2 644 1010000100 2 2 2  9  7  2  9 7 2 644 2 2 2 2 2 2 3 3 3 3 3 644 3 mod 645  9 7 2 2 2 2 ( 3 3 3 ) mod 645  9 7 2 2 2 2 (( 3 mod 645 )( 3 mod 645 )( 3 mod 645 ) mod 645 ) 14 7

Compute the powers of 3 efficiently   2 3 mod 645 9 mod 645 9   2  2     2 2 2 2 3 mod 645 3 mod 645 (( 3 mod 645 )( 3 mod 645 )) mod 645 ( 9 9 mod 645 ) 81   2 3  2  2 2    2 2 2 2 3 mod 645 3 mod 645 (( 3 mod 645 )( 3 mod 645 )) mod 645 81 81 mod 645 111  Use the powers of 3 to get result efficiently 644 3  9 7 2 2 2 2 ( 3 3 3 mod 645 )  9 7 2  9 7 2 2 2 2 2 ( 3 3 ( 3 mod 645 ) mod 645 ) ( 3 3 81 mod 645 )  9 7  9   9  2 2 2 2 ( 3 ((( 3 mod 645 ) 81 ) mod 645 ) mod 645 ) ( 3 (( 396 81 ) mod 645 ) mod 645 ) ( 3 471 mod 645 )  9     2 ((( 3 mod 645 ) 471 ) mod 645 ) 111 471 mod 645 36 15 Modular_Exponentiation( ) { , , b n m   ( ) n a a a a   1 2 1 0 2 n n  1 x  mod power b m for to {   0 1 i k  (  if ( )  ) mod 1 x x power m a i   ( ) mod power power power m } return n ( mod ) x b m } 16 8

Recall: Greatest Common Divisor  largest integer gcd( , ) a b d such that and d | d | a b  , a b Z   | | | | 0 a b  Examples: gcd( 24 , 36 ) 12 Common divisors of 24, 36: 1, 2, 3, 4, 6, 12  gcd( 17 , 22 ) 1 Common divisors of 17, 22: 1 17 Trivial cases:  gcd( , 1 ) 1 m   gcd( , 0 ) 0 m m m  If then are relatively prime gcd ( , ) 1 a , a b b and have no common factors a b Example: 21, 22 are relatively prime  gcd( 21 , 22 ) 1 18 9

How do we compute GCD efficiently? (Finding prime factorization is slow) 19    Theorem:  r  If a b q r 0 b then  gcd( , ) gcd( , ) a b b r Proof:   a  ( ) r d s tq d | d | a ds r b  b  d | d | b dt dt b Thus, and have ( , ) ( , ) a b b r the same set of common divisors End of proof 20 10

a  b  remainder divisions r r 0 1     0 r r q r r r 0 / r r 0 1 1 2 2 1 1     0 r r q r r r 1 / r r 1 2 2 3 3 2 2        0 2 / r r q r r r r r       1 n n n 2 n 1 n 1 n n n 1   / 0 r r r r q   1 n n 1 n n n first zero result     gcd( , ) gcd( , ) gcd( , ) gcd( , ) a b r r r r r r 0 1 1 2 2 3      gcd( , ) gcd( , ) gcd( , 0 ) r r r r r r    2 1 1 n n n n n n 21   662 414 a b       662 414 1 248 248 414 r r 2 1       414 248 1 166 166 248 r r 3 2       248 166 1 82 82 166 r r 4 3       166 82 2 2 2 82 r r 5 4    82 2 41 0 result   gcd( 662 , 414 ) gcd( 414 , 248 ) gcd( 248 , 166 )     gcd( 166 , 82 ) gcd( 82 , 2 ) gcd( 2 , 0 ) 2 22 11

Euclidean Algorithm for GCD gcd( ) { a , a b x  y  b  while ( ) { 0 y  mod r x y x  y y  r } return x } 23 Useful Result regarding GCDs s    Z if then there are such that , t Z a , b   gcd( , ) a b sa tb (i.e., gcd is a linear combination of a and b)       Example: gcd( 6 , 14 ) 2 ( 2 ) 6 1 14 24 12

The linear combination can be found by reversing the Euclidian algorithm steps      gcd( 252 , 198 ) 18 4 252 5 198    252 1 198 54    198 3 54 36    54 1 36 18    36 2 18 0  gcd( 252 , 198 ) 18         54 1 36 54 1 ( 198 3 54 )           4 54 1 198 4 ( 252 1 198 ) 1 198     4 252 5 198 25 Linear congruences We want to solve this equation for x   (mod m ) a x b x  ? (mod ) m 26 13

a  1 (mod ) Inverse of : a a m   (mod m ) a x b   (mod m ) a a x a b  mod a a m a  1 (mod ) a m    1 (mod ) a a x x m x  (mod m ) x x  (mod m ) a b 27 Theorem: If and are relatively prime a m then the inverse modulo exists a m  1   (linear combo Proof: gcd( , ) a m sa tm theorem) sa  (Def. of 1 (mod ) m mod) a  (Def. of inverse s mod m) End of proof 28 14

3  Example: solve equation 4 (mod 7 ) x    3 , 4 , 7 a b m Inverse of 3:          gcd( 3 , 7 ) 1 2 3 1 7 2 3 1 (mod ) m   2 a x  (mod m ) a b       2 4 (mod 7 ) 8 (mod 7 ) 6 mod 7 x 29 A Chinese Puzzle (by Sun-Tzu, 300-500 AD) I have some things whose number you don’t know. If divided by 3, the remainder is 2 If divided by 5, the remainder is 3 If divided by 7, the remainder is 2 How many things do I have? 30 15

Sun- Tzu’s Puzzle  2 (mod 3 ) x  3 (mod 5 ) x  2 (mod 7 ) x What is x ? 31 Chinese remainder theorem (CRT) :pairwise relatively prime  , , , m m m 1 2 n x  (mod ) a m 1 1 x  (mod ) a m 2 2  x  (mod ) a m n n Has unique solution for modulo  1  x  m m m m 2 n 32 16

Unique solution modulo :  1   m m m m 2 n      x a M y a M y a M y 1 1 1 2 2 2 n n n m M  where k m k :inverse of modulo y M m k k k 33 Explanation: :inverse of modulo y M m k k k  1 mod m M y m M  k k k k m k = 1:  1 mod M y m k 1 1 1 0 (mod ) 0 (mod ) m m 1 1      x a M y a M y a M y 1 1 1 2 2 2 n n n   0 (mod ) M k m (mod ) x a M y m  1 1 1 1 1 1 i.e., x satisfies 1 st equation  (mod ) x a m 1 1 Similar for any m j 34 17