Abstract Algebra Theory and Applications - Computer Science ...

248 CHAPTER 14 RINGShas a solution. Furthermore, any two solutions of the system are congruentmodulo n 1 n 2 · · · n k .Proof. We will use mathematical induction on the number of equationsin the system. If there are k = 2 equations, then the theorem is true byLemma 14.18. Now suppose that the result is true for a system of k equationsor less and that we wish to find a solution ofx ≡ a 1 (mod n 1 )x ≡ a 2 (mod n 2 ).x ≡ a k+1 (mod n k+1 ).Considering the first k equations, there exists a solution that is unique modulon 1 · · · n k , say a. Since n 1 · · · n k and n k+1 are relatively prime, the systemx ≡ a (mod n 1 · · · n k )x ≡ a k+1 (mod n k+1 )has a solution that is unique modulo n 1 . . . n k+1 by the lemma.□Example 22. Let us solve the systemx ≡ 3 (mod 4)x ≡ 4 (mod 5)x ≡ 1 (mod 9)x ≡ 5 (mod 7).From Example 21 we know that 19 is a solution of the first two congruencesand any other solution of the system is congruent to 19 (mod 20). Hence,we can reduce the system to a system of three congruences:x ≡ 19 (mod 20)x ≡ 1 (mod 9)x ≡ 5 (mod 7).Solving the next two equations, we can reduce the system tox ≡ 19 (mod 180)x ≡ 5 (mod 7).