Greatest Common Divisor

Definitions

• Let a and b be two non-zero integers. The greatest common divisor of a and b, denoted gcd(a,b) is the largest of all common divisors of a and b.

• When gcd(a,b) = 1, we say that a and b are relatively prime.

Examples

• gcd(10,25) =

• gcd(10,26) =

• gcd(10,27) =

so 10 and 27 are relatively prime!

• gcd(10,30) =

• gcd(-10,-30) =

• gcd(13978,15457) =

52

1

10

10

29

Thm 0.2 GCD is a Linear Combination

• For any nonzero integers a and b, there exist integers s and t such that

gcd(a,b) = as + bt

• Moreover, gcd(a,b) is the smallest postive integer of the form as + bt.

• Note: This is the most important thing to know about greatest common divisors!

Example

• gcd(32,14) = 2• Check that 32(-3) + 14(7) = 2• Also, 32(11) + 14(-25) = 2• In fact, there are infinitely many ways to

write 2 as a linear combination.• 2 is the smallest postive linear combination

of 32 and 14.• (Also, 2 | 32s + 14t for all integers s, t)

Example

• gcd(13978,15457) = 29• Check that 13978(209) + 15457(-189) = 29• Also, 13978(-15248) + 15457(13789) = 29• 29 is the smallest postive linear

combination of 13978 and 15457.

Proof

• The proof consists of three steps.1. Show how to choose an integer d > 0 that

could be the gcd of a and b.

2. Show that d is a common divisor, I.e. show that d | a and d | b

3. Show that any other common divisor c of a and b must also divide d.

1. Choose d

• Given non-zero integers a,b, let S be the set of positive linear combinations of a,b.

S = {as+bt | s,t are integers and as+bt > 0}

• Clearly S is a set of non-empty positive integers.

By the WOP, S has a least element, d.

2: Show d|a and d|b

• Divide a by d, recalling that d = as + bt. a =dq + r = (as + bt)q + r, where 0 ≤ r < d. Then r = a(1-sq)+b(-tq)So r is a linear combination of a and b.

• If r > 0, then r would have to be in S. But this is impossible, since r < d.So r must be 0, a = dq, and d|a.

• Use symmetry (exchange the names of a and b) to see that d|b as well as a.

• So d is a common divisor, as required.

3. Show any other common divisor must divide d

• Let c be any common divisor of a and b. Say a = ch and b = ck.

• Then d = as + bt = chs + ckt = c(hs+kt)

• So c|d.

Corollary

• If a and b are relatively prime, then there are integers s and t with as + bt = 1.

Euclidean Algorithm

15457

+

209

13978

1

–

189

1479

9

+

20

667

2

–

9

145

4

+

2

87

1

–

1

58

1

+

1

29

2

0

0

gcd =

15457(-189)+13978(209)=29

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