Linear congruences: ax b (mod n ) for x Z a x = b in Z n (in - PDF document

Linear congruences: • ax ≡ b (mod n ) for x ∈ Z • a ⊙ x = b in Z n (in particular x ∈ { 0 , 1 , . . . , n − 1 } ) • [ a ] n ⊙ [ x ] n = [ b ] n in Z n

x ∈ Z satisfies ax ≡ b (mod n ) if and only if there is y ∈ Z such that ax + yn = b .

Theorem. Let n ∈ N , let a, b ∈ Z . Assume that gcd( a, n ) = Aa + Bn . (i) If b is not a multiple of gcd( a, n ), then the equation ax ≡ b (mod n ) does not have a solution. (ii) If gcd( a, n ) | b , then the equation ax ≡ b (mod n ) does have a soltuion and the set of all its solutions is � � b n gcd( a, n ) + k gcd( a, n ); k ∈ Z A .

Theorem. Let n ∈ N , let [ a ] n , [ b ] n ∈ Z n . Assume that gcd( a, n ) = Aa + Bn . (i) If b is not a multiple of gcd( a, n ), then the equation [ a ] n ⊙ [ x ] n = [ b ] n does not have a solution. (ii) If gcd( a, n ) | b , then the equation [ a ] n ⊙ [ x ] n = [ b ] n has gcd( a, n ) distinct solutions in Z n and the set of all these solutions is �� � � b n gcd( a, n ) + k n ; k = 0 , 1 , . . . , gcd( a, n ) − 1 A . gcd( a, n )

Theorem. Let n ∈ N , let a, b ∈ Z . Consider some solution x p of the congruence ax ≡ b (mod n ). Then x ∈ Z is a solution of this congruence if and only if x = x p + x h for some x h ∈ Z that solves the congruence ax ≡ 0 (mod n ).

Fact. Let n ∈ N , consider a ∈ Z n . The set of all solutions of the congruence a ⊙ x = 0 (mod n ) is � n � k gcd( a, n ); k ∈ Z .

Algorithm for solving the equation ax ≡ b (mod n ) in Z , or the equation [ a ] n ⊙ [ x ] n = [ b ] n in Z n , or the equation ax = b in Z n for a, b ∈ Z n . 0. Rewrite the equation as ax + ny = b . Using the extended Eu- clidean algorithm find gcd( a, n ) = Aa + Bn (or guess it). 1. If gcd( a, n ) divides b , then the equation has a solution. 2. If gcd( a, n ) dl b , rovnice m een. b a) Multiply gcd( a, n ) = Aa + Bn by the number gcd( a, n ), changing Ab Bb it into a gcd( a, n ) + n gcd( a, n ) = b , which has the same form as the Ab equation in step 0 . We see a particular solution x p = gcd( a, n ). b) Cancel the associated homogeneous equation ax + ny = 0 by the number gcd( a, n ), creating a ′ x + n ′ y = 0, this has a general solution x h = kn ′ , k ∈ Z . c) A general solution of the given equation is then x p + x h . Depending on how the question was given you get the following: • The set of all integer solutions of the congruence ax ≡ b (mod n ) is { x p + kn ′ ; k ∈ Z } , that is, x = x p + kn ′ , k ∈ Z . • The set of all solutions in Z n of the equation [ a ] n [ x ] n = [ b ] n is { [ x p + kn ′ ] n ; k = 0 , 1 , 2 , . . . , gcd( a, n ) − 1 } . • The set of solutions in Z n of the equation ax = b can be obtained by choosing a suitable representative of the class [ x p + kn ′ ] n for all k = 0 , 1 , . . . , gcd( a, n ) − 1. Formally, let k 0 be the least integer such that x p + kn ′ ≥ 0. Then the set of all solutions in Z n of the equation ax = b is { x p + kn ′ ; k = k 0 , k 0 + 1 , . . . , k 0 + gcd( a, n ) − 1 } .

Systems of linear congruences: Assume that moduli n 1 , . . . , n m ∈ N and right-hand sides b 1 , . . . , b m ∈ Z are given. We are looking for integers x such that x ≡ b 1 (mod n 1 ), x ≡ b 2 (mod n 2 ), . . . x ≡ b m (mod n m ).

Theorem. (Chinese remainder theorem) Let n 1 , n 2 , . . . , n m ∈ N , b 1 , b 2 , . . . , b m ∈ Z . Consider the system of equations x ≡ b 1 (mod n 1 ), x ≡ b 2 (mod n 2 ), . . . x ≡ b m (mod n m ). If the numbers n i are all pairwise coprime, then this system has a solution. This solution is unique up to modulo n = n 1 n 2 · · · n m , conversely, all elements from this congruence class modulo n are also solutions.

Algorithm for solving systems of congruences x ≡ b 1 (mod n 1 ), x ≡ b 2 (mod n 2 ) , . . . , x ≡ b m (mod n m ) in case that all numbers n i are pairwise coprime. 1. Denote n = n 1 n 2 · · · n m and N i = n n i for all i . 2. For every i find the inverse of N i with respect to multiplication modulo n i . m 3. Let x = � b i x i N i . The set of all solutions of the given system is i =1 { x + kn ; k ∈ Z } .