The Euclidean Algorithm
The Euclidean Algorithm appears in Book VII in Euclid’s The Elements, written around 300 BC. It
is one of the oldest mathematical algorithms.
It is also one of the most applicable. The algorithm provides a systematic way to find the greatest
common divisor GCD of two integers and provide additional important information about the relationship
between the GCD and the two integers involved.
Modern technology uses a variety of algorithms based on modular arithmetic including the public-key
encryption RSA algorithm. Many of these algorithms in turn rely on the Euclidean Algorithm as an
algorithm acting on the ring of integers or as an algorithm acting on a ring of polynomials.
Here we introduce the Euclidean algorithm for the integers. The Euclidean Algorithm on the
set of polynomials is similar. The concepts here may be generalized to any algebraic system which obeys
the division algorithm; such rings are called Euclidean Domains.
1.1
Introduction
We say that the integer d divides the integer a (written d|a) if there is an integer k such that a = dk. For
example, −5|20 since 20 = (−5)(−4). So the divisors of 20 are −20, −10, −5, −4, −2, −1, 1, 2, 4, 5, 10, 20.
Given two integers a and b, we seek divisors d which divide both of these integers. We are in particular
interested in the largest such divisor, the greatest common divisor of both a and b. (The set of all
common divisors of a and b is exactly the set of divisors of the greatest common divisor.)
Hereafter we abbreviate “greatest common divisor” of a and b by GCD or GCD(a, b).
In order to avoid issue about ”size”, we will define the GCD of integers a and b as the positive integer
d that satisfies the following condition:
If c|a and c|b then c|d
(1)
If a and b have common divisors (other than −1, 1) then there is a common prime p dividing both of
them. The GCD of a and b is 1 if and only if the only common divisors of a and b are −1 and 1. In this
case (since −1 and 1 divide every integer) we say that a and b “have no common divisors.” Equivalently
we say a and b are relatively prime.
If an integer d divides both integers a and b then for any integer q, it divides a − qb. In particular, if
d is the greatest divisor of a and b and r is the remainder (guaranteed by the division algorithm) upon
division of a by q, then d also divides r.
Conversely, if d divides b and d divides r = a−qb then d also divides a. Therefore the greatest common
divisor of a and b is also the greatest common divisor of b and r.
This is the essence of the Euclidean algorithm. We replace a pair of integes a and b by a smaller pair
of integers b and r and iterate the process until we reach the smallest possible pair of integers.
Suppose we are given two integers, a and b (for example, a = 843 52256 45419 and b = 105 46961 61403).
Instead of hunting for divisors of a (8435225645419), we may divide the smaller number b (1054696161403)
into the larger, and use the division algorithm to get a remainder r = a − qb, where 0 ≤ r < b. (In this
case q = 7 and the remainder is r = 23436 45805.) Now we want the GCD of b and r, which are smaller
numbers. Since the size of the positive integers is dropping, we may repeat this step, replacing a pair of
integers
ai and bi
by
bi and ri = ai − qbi
1
until we finally get a remainder of zero. Since, in the last step, the GCD of zero and an integer bk is just
bk (every integer divides zero!), then the final integer bk is the GCD of a and b.
This algorithm carries more information than might be obvious at first glance. Suppose we write
a = a(1) + b(0), and b = a(0) + b(1), and, at each stage, write the new number in the form as + bt for
integers s and t. Since each step in our algorithm involves computing ai −qbi , we may think of this process
as an elementary row operation on a matrix of integers with rows of the form bi = as + bt, si , ti .
(See Wikipedia for more on matrices and elementary row operations.)
We will work out our example with a = 8435225645419 and b = 1054696161403 in detail.
Table 1, below, is the algorithm in tabular form. The four columns represent −q (where q is the
quotient in the current step of the division algorithm), bi = as + bt, s, and t. In the last two rows, we
compute the GCD of 12347 and 0, and so the GCD of the original two numbers must be 12347. The
process of computing the GCD of a = 8435225645419 and b = 1054696161403 requires nine rows.
Table 1: Euclidean algorithm in tabular form
−q
-7
-1
-449
-42
-5
-4
-214
as + bt
a = 8435225645419
b = 1054696161403
1052352515598
2343645805
55549153
10581379
2642258
12347
0
s
1
0
1
-1
450
-18901
94955
-398721
85421249
t
0
1
-7
8
-3599
151166
-759429
3188882
-683180177
Notice that this algorithm always allows us to write the GCD of a and b in the form as + bt. Here
12347 = a(−398721) + b(3188882).
The equation
GCD(a, b) = as + bt
(2)
is called Bezout’s Identity. The Euclidean Algorithm not only finds the GCD of a and b but it also
finds the integers s and t which satisfy Bezout’s Identity.
1.2
Modular Arithmetic and the Group of Units of Zn
We say that an integer a is congruent to b modulo n if n divides a − b. We write a ≡ b mod n.
Intuitively, this means that a and b should have the same remainder upon division by n.
For example, 13 and 34 are congruent modulo 7. By our definition, this is because 7 divides 34 − 13,
but we might also notice that if we divided 13 or 34 by 7, we would get a remainder of 6 in each case. So
34 ≡ 13 ≡ 6 mod 7.
We say a and n are relatively prime if GCD(a, n) = 1.
If d > 1 is the GCD of a and n, then there is no solution to the equation ax ≡ 1 mod n, for a solution
to this equation implies an integer k such that 1 = ax + kn, and any integer dividing a and n must also
divide ax + kn.
Conversely, if an integer a is relatively prime to an integer n then we may use the Euclidean algorithm
to compute s and t such that 1 = as + nt. Then, modulo n, we have that 1 ≡ as, and so s is the inverse
2
of a mod n.
We have just proven that a has a multiplicative inverse modulo n if and only if GCD(a, n) = 1, and,
furthermore, when this occurs, the Euclidean algorithm in tabular format will reveal the inverse to us!
We call the set of integers between 0 and n that are relatively prime to n the set of units of n. and
write
U (n) := {k ∈ Z : 0 < k < n, GCD(k, n) = 1}.
(U (n) is also called the group of units of n.) This set, U (n), is a critical set of study in the arithmetic
of computations modulo n.
For example, let us set n = 20 and look at U (20) = {1, 3, 7, 9, 11, 13, 17, 19}.
We can multiply these number modulo 20. Below (Table 2) is the start of a multiplication table for
the units of 20.
Table 2: Multiplication modulo 20
·
1
3
7
9
11
13
17
19
1.3
1
1
3
7
9
11
13
17
19
3
3
9
1
7
13
19
11
17
7
7
1
9
9
9
7
11
11
13
1
19
13
13
19
17
17
11
19
19
17
More on the Group of Units of Zn
We prove here some basic algebra results for the group of units.
First we prove that the product of two units is again (congruent to) a unit. So the group of units is
“closed” under multiplication. For example, in U(20), since 7 and 11 are both relatively prime to 20 then
so is 77. And thus, when we reduce 77 modulo 20, we will obtain a number (17) already in the group of
units.
Lemma 1. If i and j are both relatively prime to n then so is ij.
Proof. By the Euclidean algorithm, write 1 = is + nt for some integers s and t and write 1 = ju + nv
for integers u and v. Then
(1)(1) = (is + nt)(ju + nv).
Rewrite the right hand side in the form (ij)s0 + (n)t0 for integers s0 and t0 and thus GCD(ij, n) = 1.
Exercise for students in an algebra class. Show that U (n) is a group under multiplication.
Lemma 2. Suppose a and b are relatively prime. Then if a divides m and b divides m then ab divides
m.
Proof. Suppose that GCD(a,b)=1 and a|m and b|m. Use the Euclidean algorithm to find integers s, t
such that 1 = as + bt. Multiply this equation by m to get
m = ams + bmt.
3
Since a|m there exists an integer j such that m = aj. Similarly there is an integer k such that m = bk.
This allows us to write the equation for m above in the form
m = a(bk)s + b(aj)t.
So
m = ab(ks + jt).
Since ks + jt is an integer, we have ab dividing m. 2
Lemma 3. Suppose m and n are integers with d := GCD(m, n). Then
m
n
and
are relatively prime.
d
d
m
n
m n
Proof. Write d = ms + nt for integers s and t. Then 1 = s + t. Since , , s and t are all integers,
d
d
d d
m n
then, by the converse of the Euclidean algorithm, GCD( , ) = 1.
d d
1.3.1
The LCM
In order to work further with the set (ring) of integers, we need a few facts about the greatest common
divisor and least common multiples of integers. These facts, naturally enough, come out of the Euclidean
algorithm.
Lemma 4. Suppose m and n are integers with d := GCD(m, n) and l := LCM (m, n). Then l :=
mn
.
d
n
m
Proof. Since d|m and d|n then mn/d = m( ) = n( ) is a multiple of both m and n.
d
d
On the other hand, suppose k is a common multiple of both m and n. Write k = mk1 = nk2 for
k
m
n
integers k1 , k2 . Then = ( )k1 = ( )k2 .
d
d
d
m
n
k
mn
k
Now
and are relatively prime and those both divide . So, by an earlier result,
divides .
d
d
d
d d
d
mn
divides k.
This implies that
d
mn
In other words,
is a common multiple of m and n and it divides any other common multiple. So
d
mn
must be the least common multiple of m and n.
d
Notation. We often write (m, n) for GCD(m, n) and [m, n] for LCD(m, n). Using that notation, the
mn
lemma above proves that [m, n] =
.
(m, n)
4
Euclidean Algorithm, Worksheet 1
On all problems below, the instructions “Use the Euclidean Algorithm...” assume that all computations
done with the Euclidean Algorithm are done in the tabular format.
1. Finish Table 2 (the multiplication table modulo 20.)
2. Finish the following table, to show that 521 and 641 are relatively prime. Then use your work to
find the the integers s and t such that 1 = 641s + 521t.
−q
-1
-4
-2
as + bt
a = 641
b = 521
120
41
s
1
0
1
-4
t
0
1
-1
5
1
0
3. Use the Euclidean algorithm in tabular format to find the GCD of 108 and 605 and compute the
integers s and t such that GCD(108, 605) = 108s + 605t.
4. Compute the inverse of 108 in Z605 .
5. Use the Euclidean algorithm to find the integer x such that 1 = 200x + 641y. (The integer x is “the
inverse of 200 mod 641.”)
6. Find the “inverse” of 521 modulo 625. (That is, find an integer s such that 521s ≡ 1 mod 625.)
7. Use the Euclidean algorithm to find the GCD of a = 232 + 1 and b = 214 · 52 − 1 and compute
the integers s and t such that GCD(a, b) = as + bt. (Fermat, probably around 1640, claimed that
232 + 1 is prime. Was he correct?)
8. Find the smallest pair of integers a and b such that the Euclidean Algorithm, in tabular format, has
nine rows. More generally, what is the “worst case” scenario for the Euclidean Algorithm: which
pair of integers require the most rows?
9. Use Bezout’s Identity to prove Euclid’s Lemma: If a prime p divides ab then either p divides a or
p divides b. (Hint: consider the contrapositive. If p doesn’t divide a and p doesn’t divide b then p
is relatively prime to a and relatively prime to b. Can you get from there, via Bezout’s Identity, to
the claim that p is relatively prime to ab?)
5
Euclidean Algorithm, Worksheet 2
1. Use the Euclidean algorithm (in tabular format) to find the GCD of 11333 and 7213 and compute
the integers s and t such that GCD(11333, 7213) = 11333s + 7213t.
2. Use the Euclidean algorithm to find the GCD of a := (11333)(523) and b := (7213)(523) and
compute the integers s and t such that GCD(a, b) = as + bt.
3. Does 7213 have a multiplicative inverse modulo 11333? (If so, what is it?)
4. Describe the “worse case” scenario for the Euclidean algorithm? In particular what would be the
smallest pair of numbers that would require 20 steps? 1000 steps? n steps?
5. Use the Euclidean algorithm to find the GCD of a = 123456789 and b = 12345000 and compute
the integers s and t such that GCD(a, b) = as + bt.
6. (a) Let a, b ∈ Z. Prove that GCD(a, b) = 1 if and only if there exists integers x, y such that
1 = ax + by.
(b) Prove if GCD(a, x) = 1 and GCD(b, x) = 1 then GCD(ab, x) = 1.
7. Use the Euclidean algorithm (in tabular format) to find integers s and t such that 1 = 5s + 1536t.
Brief solution. s = 1229, t = −4.
8. Find the multiplicative inverse of 5 modulo 1536.
Brief solution. 5−1 = 1229 mod 1536.
9. Use the concept of repeated squaring to compute 6695 modulo 1649.
Brief solution.
6692 ≡ 682
2
682 ≡ 106
mod 1649
mod 1649
Therefore
6695 = (6694 )(669) ≡ (106)(669) ≡ 7
mod 1649.
10. If I square the number 7, and then keep squaring the squares modulo 1649, I get the following list.
72 ≡ 49 mod 1649
74 = 492 ≡ 752 mod 1649
78 = 7522 ≡ 1546 mod 1649
716 = 15462 ≡ 715 mod 1649
732 = 7152 ≡ 35 mod 1649
764 = 352 ≡ 1225 mod 1649
7128 = 12252 ≡ 35 mod 1649
6
7256 = 352 ≡ 1225
..
.
mod 1649
Use this list to compute 71229 mod 1649.
Brief solution. 71229 = (71024 )(7128 )(764 )(78 )(74 )(71 ) ≡ (1225)(35)(1225)(1546)(752)(7) ≡ 669
mod 1649.
11. Factor 1649.
Brief solution. 1649 = 17 · 97.
Comment. Notice that 71229 ≡ 669 mod 1649 and 6695 ≡ 7 mod 1649. More generally, for most
integers M < 1229, if
S ≡ M 1229 mod 1649
then
M ≡ S5
mod 1649.
Given M we can create S; given S we can recover M . The function f (M ) = M 1229 mod 1649 has
an inverse function g(S) = S 5 mod 1649.
Why? This is because 5 and 1229 are inverses modulo 1536. (What is the relationship between 1649
and 1536??)
More later....
7