Discrete Mathematics University of Kentucky CS 275 Spring ... - MGNet

Theorem: If a and m are relatively prime integers and m>1, then an inverse of a
modulo m exists and is unique modulo m.
Proof: Since gcd(a,m) = 1, )s,t,Z(1 = sa+tb). Hence, sa=tb ' 1 (mod m). Since
tm ' 0 (mod m), it follows that sa ' 1 (mod m). Thus, s is the inverse of a
modulo m. The uniqueness argument is made by assuming there are two
inverses and proving this is a contradiction.
Systems of linear congruences are used in large integer arithmetic. The basis for
the arithmetic goes back to China 1700 years ago.
Puzzle Sun Tzu (or Sun Zi): There are certain things whose number is unknown.
• When divided by 3, the remainder is 2.
• When divided by 5, the remainder is 3, and
• When divided by 7, the remainder is 2.
What will be the number of things? (Answer: 23… stay tuned why).
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Chinese Remander Theorem: Let m 1 , m 2 ,…,m n ,N be pairwise relatively prime.
n
Then the system x ' a i (mod m i ) has a unique solution modulo m = ! .
i=1
m i
Existence Proof: The proof is by construction. Let M k = m / m k , 1;k;n. Then
gcd(M k , m k ) = 1 (from pairwise relatively prime condition). By the previous
theorem we know that there is a y k which is an inverse of M k modulo m k , i.e.,
M k y k ' 1 (mod m k ). To construct the solution, form the sum
x = a 1 M 1 y 1 + a 2 M 2 y 2 + … + a n M n y n .
Note that M j ' 0 (mod m k ) whenever j-k. Hence,
x ' a k M k y k ' a k (mod m k ), 1;k;n.
We have shown that x is simultaneous solution to the n congruences. qed
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