The Chinese Remainder Theorem

The Chinese Remainder Theorem 8-3-2014 The Chinese Remainder Theorem gives solutions to systems of congruences with relatively prime moduli The solution to a system of congruences with relatively prime moduli may be produced using a formula by computing modular inverses, or using an iterative procedure involving successive substitution The Chinese Remainder Theorem says that certain systems of simultaneous congruences with different moduli have solutions The idea embodied in the theorem was apparently known to Chinese mathematicians a long time ago hence the name I ll begin by collecting some useful lemmas Lemma 1 Let m and a 1,, a n be positive integers If m is relatively prime to each of a 1,, a n, then it is relatively prime to their product a 1 a n Proof If (m,a 1 a n ) 1, then there is a prime p which divides both m and a 1 a n Now p a 1 a n, so p must divide a i for some i But p divides both m and a i, so (m,a i ) 1 This contradiction implies that (m,a 1 a n ) = 1 Example 6 is relatively prime to 25, to 7, and to 11 25 7 11 = 1925, and (6,1925) = 1: a q 1925-6 320 5 1 1 5 I showed earlier that the greatest common divisor (a,b) of a and b is greatest in the sense that it is divisible by any common divisor of a and b The next result is the analogous statement for least common multiples Lemma 2 Let m and a 1,, a n be positive integers If m is a multiple of each of a 1,, a n, then m is a multiple of [a 1,,a n ] Proof By the Division Algorithm, there are unique numbers q and r such that m = q [a 1,,a n ]+r, where 0 r < [a 1,,a n ] Now a i divides both m and [a 1,,a n ], so a i divides r Since this is true for all i, r is a common multiple of the a i smaller than the least common multiple [a 1,,a n ] This is only possible if r = 0 Then m = q [a 1,,a n ], ie m is a multiple of [a 1,,a n ] Example 88 is a multiple of 4 and 22 The least common multiple of 4 and 22 is 44, and 88 is also a multiple of 44 1

Lemma 3 Let a 1,, a n be positive integers If a 1,, a n are pairwise relatively prime (that is, (a i,a j ) = 1 for i j), then [a 1,,a n ] = a 1 a n Proof Induct on n The statement is trivially true for n = 1, so I ll start with n = 2 The statement for n = 2 follows from the equation xy = [x,y](x,y): [a 1,a 2 ] = a 1a 2 (a 1,a 2 ) = a 1a 2 = a 1 a 2 1 Now assume n > 2, and assume the result is true for n I will prove that it holds for n+1 Claim: [[a 1,,a n ],a n+1 ] = [a 1,,a n,a n+1 ] (Some people take this as an iterative definition of [a 1,,a n,a n+1 ]) [a 1,,a n,a n+1 ] is a multiple of each of a 1,, a n, so by Lemma 2 it s a multiple of [a 1,,a n ] It s also a multiple of a n+1, so On the other hand, for i = 1,,n, Therefore, Obviously, [[a 1,,a n ],a n+1 ] [a1,,a n,a n+1 ] a i [a1,,a n ] and [a 1,,a n ] [[a1,,a n ],a n+1 ] a i [[a 1,,a n ],a n+1 ] a n+1 [[a 1,,a n ],a n+1 ] Thus, [[a 1,,a n ],a n+1 ]isacommonmultipleofallthea i s Since[a 1,,a n,a n+1 ]istheleastcommon multiple, Lemma 2 implies that [a 1,,a n,a n+1 ] [[a 1,,a n ],a n+1 ] Since I have two positive numbers which divide one another, they re equal: [[a 1,,a n ],a n+1 ] = [a 1,,a n,a n+1 ] This proves the claim Returning to the proof of the induction step, I have [a 1,,a n,a n+1 ] = [[a 1,,a n ],a n+1 ] = [a 1 a n,a n+1 ] = a 1 a n a n+1 The second equality follows by the induction hypothesis (the statement for n) The third equality follows from Lemma 1 and the result for n = 2 Example 6, 25, and7arerelativelyprime(inpairs) Theleastcommonmultiple is[6,25,7] = 1050 = 6 25 7 Theorem (The Chinese Remainder Theorem) Suppose m 1,, m n are pairwise relatively prime (that is, (m i,m j ) = 1 for i j) Then the following system of congruences has a unique solution mod m 1 m 2 m n : x = a 1 (mod m 1 ) x = a 2 (mod m 2 ) x = a n (mod m n ) 2

Notation For example, x 1 x 2 x i x n means x 1 x 2 x i x n omitting x i x 1 x 2 x 4 x 6 means x 1 x 2 x 3 x 5 x 6 This is a convenient (and standard) notation for omitting a single variable term in a product of things Proof Define p k = m 1 m k m n That is, p k is the product of the m s with m k omitted By Lemma 1, (p k,m k ) = 1 Hence, there are numbers s k, t k such that s k p k +t k m k = 1 In terms of congruences, Now let s k p k = 1 (mod m k ) x = a 1 p 1 s 1 +a 2 s 2 p 2 + +a n p n s n If j k, then m k p j, so mod m k all the terms but the k-th term are 0 mod m k : x = a k p k s k = a k 1 = a k (mod m k ) This proves that x is a solution to the system of congruences (and incidentally, gives a formula for x) Now suppose that x and y are two solutions to the system of congruences x = a 1 (mod m 1 ) and y = a 1 (mod m 1 ) x = a 2 (mod m 2 ) and y = a 2 (mod m 2 ) x = a n (mod m n ) and y = a n (mod m n ) Then x = a k = y (mod m k ) so x y = 0 (mod m k ) or m k x y Thus, x y is a multiple of all the m s, so [m 1,,m n ] x y But the m s are pairwise relatively prime, so by Lemma 3, m 1 m n x y, or x = y (mod m 1 m n ) That is, the solution to the congruences is unique mod m 1 m n Example Solve x = 2 (mod 4) x = 7 (mod 9) (4,9) = 1, so there is a unique solution mod 36 Following the construction of x in the proof, p 1 = 9, 9 1 = 1 (mod 4), so s 1 = 1 p 2 = 4, 4 7 = 1 (mod 9), so s 2 = 7 3

Solution: x = a 1 p 1 s 1 +a 2 p 2 s 2 = 18+196 = 214 = 34 (mod 36) Example Solve x = 3 (mod 4) x = 1 (mod 5) x = 2 (mod 3) The moduli are pairwise relatively prime, so there is a unique solution mod 60 This time, I ll solve the system using an iterative method But x = 1 (mod 5), so x = 3 (mod 4), so x = 3+4s 3+4s = 1 (mod 5), 4s = 3 (mod 5), 4 4s = 4 3 (mod 5), s = 2 (mod 5), s = 2+5t Hence, x = 3+4s = 3+4(2+5t) = 11+20t Finally, x = 2 (mod 3), so 11+20t = 2 (mod 3), 20t = 9 = 0 (mod 3), 2t = 0 (mod 3), 2 2t = 2 2 (mod 3), t = 0 (mod 3) Hence, t = 3u Now put everything back: x = 11+20t = 11+20(3u) = 11+60u, or x = 11 (mod 60) Example Calvin Butterball keeps pet meerkats in his backyard If he divides them into 5 equal groups, 4 are left over If he divides them into 8 equal groups, 6 are left over If he divides them into 9 equal groups, 8 are left over What is the smallest number of meerkats that Calvin could have? Let x be the number of meerkats Then x = 4 (mod 5) x = 6 (mod 8) x = 8 (mod 9) From x = 4 (mod 5), I get x = 4+5a Plugging this into the second congruence, I get 4+5a = 6 (mod 8) 5a = 2 (mod 8) 5 5a = 5 2 (mod 8) 25a = 10 (mod 8) a = 2 (mod 8) Hence, a = 2+8b Plugging this into x = 4+5a gives x = 4+5(2+8b) = 14+40b 4

Plugging this into the third congruence, I get 14+40b = 8 (mod 9) 40b = 6 (mod 9) 4b = 3 (mod 9) 7 4b = 7 3 (mod 9) 28b = 21 (mod 9) b = 3 (mod 9) Hence, b = 3+9c Plugging this into x = 14+40b gives x = 14+40(3+9c) = 134+360c The smallest positive value of x is obtained by setting c = 0, which gives x = 134 You can sometimes solve a system even if the moduli aren t relatively prime; the criteria are similar to those for solving system of linear Diophantine equations I ll state the result, but omit the proof Theorem Consider the system x = a 1 (mod m 1 ) x = a 2 (mod m 2 ) (a) If (m 1,m 2 ) a 1 a 2, there are no solutions (b) If (m 1,m 2 ) a 1 a 2, there is a unique solution mod [m 1,m 2 ] Note that if (m 1,m 2 ) = 1, case (b) automatically holds, and [m 1,m 2 ] = m 1 m 2 ie I get the Chinese Remainder Theorem for n = 2 Example Solve x = 5 (mod 12) x = 11 (mod 18) Since (12,18) = 6 11 5, there is a unique solution mod [12,18] = 36 I ll use the iterative method to find the solution x = 5 (mod 12), so x = 5+12s Since x = 11 (mod 18), 5+12s = 11 (mod 18), 12s = 6 (mod 18) Now I use my rule for dividing congruences: 6 divides both 12 and 6, and (6,18) = 6, so I can divide through by 6: 2s = 1 (mod 3) Multiply by 2, and convert the congruence to an equation: Plug back in: s = 2 (mod 3), s = 2+3t x = 5+12s = 5+12(2+3t) = 29+36t, x = 29 (mod 36) c 2014 by Bruce Ikenaga 5