Lecture 21 Enoch Cheung November 11 2013 1 Lemma Let m1 m2 mr N be pairwise relatively prime If a b mod mi for each i r then a b mod m1 m2 mr Proof We will induct on r N Base case If r 1 then a b mod m1 a b mod m1 so we are done For convenience we will prove it for r 2 as well because we will use it explicitly later Consider m1 m2 N relatively prime and a b mod mi for i 1 2 so m1 b a and m2 b a Suppose m1 m2 b a then b a m1 m2 q r for some q Z and r N with 0 r m1 m2 Then r b a m1 m2 q then since m1 b a we have m1 r Similarly m2 r This is a contradiction because gcd m1 m2 1 lcm m1 m2 m1 m2 this uses the fact that gcd m1 m2 lcm m1 m2 m1 m2 so r cannot be a common divisor of m1 m2 Therefore a b mod m1 m2 Inductive case Assume for some r N for any m1 mr N pairwise relatively prime a b mod mi for each i r implies a b mod m1 m2 mr Now consider m1 mr mr 1 N pairwise relatively prime then note that m1 mr and mr 1 is relatively prime because if they share a prime factor p then by lemma proved in our last recitation p mi for some 1 i r and p mr 1 which contradicts the fact that they are supposed to be pairwise relatively prime By the inductive hypothesis a b mod m1 mr and a b mod mr 1 so by the case for two relatively prime numbers a b mod m1 mr mr 1 2 Let gcd a b d and suppose d c Further let x0 y0 be a solution to the diophantine equation ax by c a k Z x0 db k y0 ad k is also a solution Note that the pair x0 db k y0 ad k is a pair of integers due to our divisibility assumptions Observe that b a ab ab ab ab a x0 k b y0 k ax0 k by0 k ax0 by0 k k c d d d d d d b Suppose x y Z is a solution to ax by c Prove that k Z such that x x0 db k and y y0 ad k i e every solution has this form Since ax by c and ax0 by0 c and gcd a b d ax by ax0 by0 a x x0 b y0 y d d x x0 y0 y b a so let k db x x0 ad y0 y Q then x x0 db k and y y0 db k Now we need to show that k Z In other words we wish to show that a d y0 y Since d gcd a b by Tuesday s discussion we can write b ed where e a are relatively prime Therefore since a x x0 b y0 y 0 b y0 y mod a 0 ed y0 y mod a 0 d y0 y since e is invertible mod a Therefore k db x x0 ad y0 y Z as desired 3 Find all solution to x 3 mod 4 x 1 mod 5 x 2 mod 3 1 mod a We wish to consider number of the form 3 A 5 z vanishes mod 5 and mod 3 4 z 3 B vanishes mod 4 and mod 3 5 C 4 z vanishes mod 4 and mod 5 where 5 3 A 3 mod 4 4 3 B 1 mod 5 and 4 5 C 2 mod 3 By multiplying with the corresponding inverses we find A 1 mod 4 B 3 mod 5 and C 1 mod 3 to work By Chinese remainder theorem the solution is unique modulo 4 5 3 60 Therefore 5 3 1 4 3 3 4 5 1 15 36 20 11 mod 60 4 Show that if gcd m1 m2 a1 a2 then there are no solutions to the system of linear congruences x a1 mod m1 x a2 mod m2 Suppose there is such a solution x then m1 a1 x and m2 a2 x Let d gcd m1 m2 then d m1 and d m2 so d a1 x and d a2 x Therefore d a1 x a2 x so d a1 a2 Thus we have shown the contrapositive 2