Properties of the gcd

1
Properties of the gcd

• Theorem For any two integers a,b there exist
  integers x,y such that xa yb gcd(a,b).
•  
• A proof will not be given at this point.
  However, x,y can be found by applying the
  Euclidean algorithm to a and b, and then, working
  from the end and for every division of u by v
  with quotient q and positive reminder r such
  that u qv r, replacing r by u - qv.

2
The gcd

• Example Applying the Euclidean algorithm
  to find the gcd of 78 and 216 
• (78) (216)
• (216) 2?(78) (60) I (60)
  (78) (78) 1?(60) (18) II (18)
  (60) (60) 3?(18) (6) III (6)
  (18) (18) 3?(6) (0) IV 

3
The gcd

• gcd(78,216)6
• The application of the Euclidean algorithm is
  completed.
•  

4
The gcd

• Finding x, y such that xa yb gcd(a,b)
• By III (6) (60) - 3?(18)
• By II (18) (78) - 1?(60), so
• (60) - 3?(18) (60) - 3?(78) - 1?(60)
  4?(60) - 3?(78)
• By I (60) (216) - 2?(78), so
  
• 4?(60) - 3?(78) 4?(216) - 2?(78) - 3?(78)
  4?(216) - 11?(78)
•  Conclusion (6) 4?(216) - 11?(78)

5
The gcd

• Theorem Let a, b be relatively prime integers.
• Then there exists an integer x such that xa ?
  1 (mod b).
•   Proof By the previous theorem, there exist
  integers x,y such that xa yb gcd(a,b).
  Since a,b are relatively prime, gcd(a,b)1. So
  xa yb 1, hence xa -yb 1, so xa ? 1 (mod
  b).
•  

6
The gcd

• Theorem
• Let a,b,c be integers such that a,b are
  relatively prime and
• a bc. Then a c.
• Proof By a previous theorem there exist integers
  x,y such
• that xa yb gcd(a,b). Since gcd(x,y)1,
• xa yb 1.

7
The gcd

• Since a bc, there exists an integer m such
  that bc ma. So
• c c(xa yb) xac ybc xac yma
• a(xc ym).
• So a divides c.
•