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.