Lecture 20 Enoch Cheung November 7 2013 1 Let a Z and m N If gcd a m 1 then a 1 is unique modulo m Suppose gcd a m 1 with ba 1 mod m and ca 1 mod m so b and c are both multiplicative inverses of a Then because multiplication is commutative ac 1 mod m Therefore ba 1 mod m bac c mod m b ac c mod m b c mod m Therefore b c mod m so a 1 is unique modulo m 2 Lemma If n N and a1 a2 an N and p a1 a2 an then i N with 1 i n such that p ai Note that this is only true for prime p We prove this by induction on n Base case If p a1 then p a1 Inductive case Assume that for any a1 an N p a1 a2 an implies p ai for some 1 i n Now given a1 an an 1 N then if p an 1 then we are done Otherwise p an 1 and since p is prime this means that p an 1 are relatively prime By Euclid s Lemma Lemma 6 5 25 since p a1 an an 1 and p an 1 are relatively prime so p a1 an Therefore by inductive hypothesis p ai for some 1 i n 3 a 6x 1 mod 13 Note that 6 2 12 1 mod 13 so 6 2 1 mod 13 Therefore x 1 mod 13 By question 1 multiplicative inverses are unique so 2 is the only solution modulo m b 4x 3 1 mod 9 Note that 4 2 1 mod 9 Therefore 4x 3 1 mod 9 4x 2 mod 9 x 2 2 mod 9 x 4 mod 9 c 6x 4 12 mod 15 There are no such solutions x because 6 15 are not relatively prime so 6 has no inverse modulo 15 Suppose 6x 4 12 mod 15 for some x then 6x 16 1 mod 15 so x is the multiplicative inverse to 6 which does not exist 4 Let a b Z and m d N Assume d gcd a m Consider the linear congruence ax b mod m a If d b can there be any x Z satisfying this congruence No Suppose ax b mod m for some x then m b ax Since d m this means that d b ax Note that d a so d ax Therefore d b which is a contradiction b If d b why can we say that there are solutions to this congruence Since d gcd a m there are s t N such that as mt d so taken mod m this means that as d mod m s is something similar to an inverse of a but instead of 1 it gives d because a need not have an inverse if d 1 This s need not be unique modulo m Suppose a cd then c m are relatively prime since if c and m then d a and d m so d would not be the gcd Therefore c 1 exists mod m so ax b mod m cdx b mod m xd c 1 b mod m so we can let c 1 b yd then we are looking for solutions of xd yd mod m where d m Lemma If d m then xd yd mod m x y 1 mod m d m Proof Certainly if x y mod m d then d x y so m x y Now suppose xd yd mod m then m xd yd so mk x y d for some k N Therefore m d k x y so m x y d Therefore since d b using the lemma ax b mod m xd c 1 b mod m x c 1 so there are d solutions modulo m since if x t mod d 1 m are all solutions distinct mod m d m d b d mod m d 2m is a solution then t t m d t d t c 9x 12 mod 15 Note that 9 2 18 3 mod 15 so 9 2 4 12 mod 15 Therefore x 8 mod 15 is a solution Since gcd 9 15 3 there are 3 solutions and we have shown that 8 8 5 8 2 5 are all solutions so x 8 13 3 are all distinct solutions mod 15 5 Let a b c Z Prove that gcd a b gcd a cb b We wish to show that the set of common divisors of a b and a cb b are the same Suppose d is a common divisor of a b then d a and d b so d cb so d a cb Therefore d is a common divisor of a cb b Now suppose d is a common divisor of a cb b then d a cb and d b So d cb so d a cb cb so d a Therefore d is a common divisor of a b 2