Introduction to Number Theory

Ian Ludden

University of Illinois at Urbana-Champaign

14 Sep 2017

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 1 / 18

Objectives

After today’s lecture, you will be able to:

Define the divides relation and use it in proofs

State the Fundamental Theorem of Arithmetic

Describe the division algorithm

Apply the Euclidean algorithm (and partially explain why it works)

Understand congruence mod k and prove claims involving it

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 2 / 18

Objectives

After today’s lecture, you will be able to:

Define the divides relation and use it in proofs

State the Fundamental Theorem of Arithmetic

Describe the division algorithm

Apply the Euclidean algorithm (and partially explain why it works)

Understand congruence mod k and prove claims involving it

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 2 / 18

Objectives

After today’s lecture, you will be able to:

Define the divides relation and use it in proofs

State the Fundamental Theorem of Arithmetic

Describe the division algorithm

Apply the Euclidean algorithm (and partially explain why it works)

Understand congruence mod k and prove claims involving it

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 2 / 18

Objectives

After today’s lecture, you will be able to:

Define the divides relation and use it in proofs

State the Fundamental Theorem of Arithmetic

Describe the division algorithm

Apply the Euclidean algorithm (and partially explain why it works)

Understand congruence mod k and prove claims involving it

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 2 / 18

Objectives

After today’s lecture, you will be able to:

Define the divides relation and use it in proofs

State the Fundamental Theorem of Arithmetic

Describe the division algorithm

Apply the Euclidean algorithm (and partially explain why it works)

Understand congruence mod k and prove claims involving it

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 2 / 18

Why Do We Care About Number Theory?

Backbone of modern cryptography, e.g. RSA

Fast, practical algorithms for many problems, e.g. modularexponentiation

Pseudo-random algorithms, e.g. for games

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 3 / 18

Why Do We Care About Number Theory?

Backbone of modern cryptography, e.g. RSA

Fast, practical algorithms for many problems, e.g. modularexponentiation

Pseudo-random algorithms, e.g. for games

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 3 / 18

Why Do We Care About Number Theory?

Backbone of modern cryptography, e.g. RSA

Fast, practical algorithms for many problems, e.g. modularexponentiation

Pseudo-random algorithms, e.g. for games

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 3 / 18

The Divides Relation

Definition

Let a and b be integers. Then a divides b (a|b) if and only if b = an forsome integer n. We say b is a multiple of a.

Examples:

5|55, because 55 = 5 · 11

14 6 | 7 (but 7|14, because 14 = 7 · 2)

6|0, because 0 = 6 · 00 6 | n for any non-zero integer n

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 4 / 18

The Divides Relation

Definition

Let a and b be integers. Then a divides b (a|b) if and only if b = an forsome integer n. We say b is a multiple of a.

Examples:

5|55, because 55 = 5 · 11

14 6 | 7 (but 7|14, because 14 = 7 · 2)

6|0, because 0 = 6 · 00 6 | n for any non-zero integer n

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 4 / 18

The Divides Relation

Definition

Let a and b be integers. Then a divides b (a|b) if and only if b = an forsome integer n. We say b is a multiple of a.

Examples:

5|55, because 55 = 5 · 11

14 6 | 7 (but 7|14, because 14 = 7 · 2)

6|0, because 0 = 6 · 00 6 | n for any non-zero integer n

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 4 / 18

The Divides Relation

Definition

Let a and b be integers. Then a divides b (a|b) if and only if b = an forsome integer n. We say b is a multiple of a.

Examples:

5|55, because 55 = 5 · 11

14 6 | 7 (but 7|14, because 14 = 7 · 2)

6|0, because 0 = 6 · 0

0 6 | n for any non-zero integer n

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 4 / 18

The Divides Relation

Definition

Let a and b be integers. Then a divides b (a|b) if and only if b = an forsome integer n. We say b is a multiple of a.

Examples:

5|55, because 55 = 5 · 11

14 6 | 7 (but 7|14, because 14 = 7 · 2)

6|0, because 0 = 6 · 00 6 | n for any non-zero integer n

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 4 / 18

Proof Using Divides

Claim: For any integers p, q and r , where p 6= 0, if p | 3q and 3q | r , thenp | 3q + r .

Proof.

Let p, q, and r be any integers with p 6= 0. Assume p | 3q and 3q | r .That is, for some integers k and m, 3q = pk and r = (3q)m. Consider:

3q + r = pk + (3q)m

= pk + (pk)m

= pk(1 + m)

= p(k(1 + m)).

We have found an integer n = k(1 + m) such that 3q + r = pn, sop | 3q + r .

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 5 / 18

Proof Using Divides

Claim: For any integers p, q and r , where p 6= 0, if p | 3q and 3q | r , thenp | 3q + r .

Proof.

Let p, q, and r be any integers with p 6= 0. Assume p | 3q and 3q | r .

That is, for some integers k and m, 3q = pk and r = (3q)m. Consider:

3q + r = pk + (3q)m

= pk + (pk)m

= pk(1 + m)

= p(k(1 + m)).

We have found an integer n = k(1 + m) such that 3q + r = pn, sop | 3q + r .

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 5 / 18

Proof Using Divides

Claim: For any integers p, q and r , where p 6= 0, if p | 3q and 3q | r , thenp | 3q + r .

Proof.

Let p, q, and r be any integers with p 6= 0. Assume p | 3q and 3q | r .That is, for some integers k and m, 3q = pk and r = (3q)m.

Consider:

3q + r = pk + (3q)m

= pk + (pk)m

= pk(1 + m)

= p(k(1 + m)).

We have found an integer n = k(1 + m) such that 3q + r = pn, sop | 3q + r .

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 5 / 18

Proof Using Divides

Claim: For any integers p, q and r , where p 6= 0, if p | 3q and 3q | r , thenp | 3q + r .

Proof.

Let p, q, and r be any integers with p 6= 0. Assume p | 3q and 3q | r .That is, for some integers k and m, 3q = pk and r = (3q)m. Consider:

3q + r = pk + (3q)m

= pk + (pk)m

= pk(1 + m)

= p(k(1 + m)).

We have found an integer n = k(1 + m) such that 3q + r = pn, sop | 3q + r .

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 5 / 18

Proof Using Divides

Claim: For any integers p, q and r , where p 6= 0, if p | 3q and 3q | r , thenp | 3q + r .

Proof.

Let p, q, and r be any integers with p 6= 0. Assume p | 3q and 3q | r .That is, for some integers k and m, 3q = pk and r = (3q)m. Consider:

3q + r = pk + (3q)m

= pk + (pk)m

= pk(1 + m)

= p(k(1 + m)).

We have found an integer n = k(1 + m) such that 3q + r = pn, sop | 3q + r .

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 5 / 18

Prime Numbers and the FTA

Definition

An integer q ≥ 2 is prime if the only positive factors of q are q and 1. Aninteger q ≥ 2 is composite if it is not prime.

Theorem

Fundamental Theorem of Arithmetic: Every integer q ≥ 2 can bewritten uniquely as the product of one or more prime factors.

Examples:

260 = 2 · 2 · 5 · 13

180 = 2 · 2 · 3 · 3 · 5

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 6 / 18

Prime Numbers and the FTA

Definition

An integer q ≥ 2 is prime if the only positive factors of q are q and 1. Aninteger q ≥ 2 is composite if it is not prime.

Theorem

Fundamental Theorem of Arithmetic: Every integer q ≥ 2 can bewritten uniquely as the product of one or more prime factors.

Examples:

260 = 2 · 2 · 5 · 13

180 = 2 · 2 · 3 · 3 · 5

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 6 / 18

Prime Numbers and the FTA

Definition

An integer q ≥ 2 is prime if the only positive factors of q are q and 1. Aninteger q ≥ 2 is composite if it is not prime.

Theorem

Fundamental Theorem of Arithmetic: Every integer q ≥ 2 can bewritten uniquely as the product of one or more prime factors.

Examples:

260 = 2 · 2 · 5 · 13

180 = 2 · 2 · 3 · 3 · 5

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 6 / 18

Proof of Infinite Primes

Claim: There are infinitely many prime numbers.

Proof.

(By contradiction) Assume there are finitely many primes. Label them p1,p2, ..., pn, where n is the (finite) number of primes.Let p = p1 · p2 · ... · pn + 1.Note that p is greater than every prime, so p is not prime. Therefore, p iscomposite and must be divisible by some prime pi , 1 ≤ i ≤ n.However, if we divide p by pi , we get a remainder of 1, so no pi divides p.Therefore, p is prime, a contradiction. Our assumption that there arefinitely many primes must be false, so there are infinitely many primes.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 7 / 18

Proof of Infinite Primes

Claim: There are infinitely many prime numbers.

Proof.

(By contradiction) Assume there are finitely many primes. Label them p1,p2, ..., pn, where n is the (finite) number of primes.

Let p = p1 · p2 · ... · pn + 1.Note that p is greater than every prime, so p is not prime. Therefore, p iscomposite and must be divisible by some prime pi , 1 ≤ i ≤ n.However, if we divide p by pi , we get a remainder of 1, so no pi divides p.Therefore, p is prime, a contradiction. Our assumption that there arefinitely many primes must be false, so there are infinitely many primes.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 7 / 18

Proof of Infinite Primes

Claim: There are infinitely many prime numbers.

Proof.

(By contradiction) Assume there are finitely many primes. Label them p1,p2, ..., pn, where n is the (finite) number of primes.Let p = p1 · p2 · ... · pn + 1.

Note that p is greater than every prime, so p is not prime. Therefore, p iscomposite and must be divisible by some prime pi , 1 ≤ i ≤ n.However, if we divide p by pi , we get a remainder of 1, so no pi divides p.Therefore, p is prime, a contradiction. Our assumption that there arefinitely many primes must be false, so there are infinitely many primes.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 7 / 18

Proof of Infinite Primes

Claim: There are infinitely many prime numbers.

Proof.

(By contradiction) Assume there are finitely many primes. Label them p1,p2, ..., pn, where n is the (finite) number of primes.Let p = p1 · p2 · ... · pn + 1.Note that p is greater than every prime, so p is not prime. Therefore, p iscomposite and must be divisible by some prime pi , 1 ≤ i ≤ n.

However, if we divide p by pi , we get a remainder of 1, so no pi divides p.Therefore, p is prime, a contradiction. Our assumption that there arefinitely many primes must be false, so there are infinitely many primes.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 7 / 18

Proof of Infinite Primes

Claim: There are infinitely many prime numbers.

Proof.

(By contradiction) Assume there are finitely many primes. Label them p1,p2, ..., pn, where n is the (finite) number of primes.Let p = p1 · p2 · ... · pn + 1.Note that p is greater than every prime, so p is not prime. Therefore, p iscomposite and must be divisible by some prime pi , 1 ≤ i ≤ n.However, if we divide p by pi , we get a remainder of 1, so no pi divides p.Therefore, p is prime, a contradiction. Our assumption that there arefinitely many primes must be false, so there are infinitely many primes.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 7 / 18

Caution: Some False Things Seem True

There are some conjectures that seem true because it’s hard to find acounterexample.

Claim: ∀n ∈ Z, n2 + n + 41 is prime.

12 + 1 + 41 = 43 is prime

22 + 2 + 41 = 47 is prime

32 + 3 + 41 = 53 is prime

Looks good so far...

412 + 41 + 41 = 41(41) + 41(1) + 41(1) = 41 · 43 is not prime

Takeaway: Don’t believe a mathematical claim, even though it looks true,until you have a proof.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 8 / 18

Caution: Some False Things Seem True

There are some conjectures that seem true because it’s hard to find acounterexample.

Claim: ∀n ∈ Z, n2 + n + 41 is prime.

12 + 1 + 41 = 43 is prime

22 + 2 + 41 = 47 is prime

32 + 3 + 41 = 53 is prime

Looks good so far...

412 + 41 + 41 = 41(41) + 41(1) + 41(1) = 41 · 43 is not prime

Takeaway: Don’t believe a mathematical claim, even though it looks true,until you have a proof.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 8 / 18

Caution: Some False Things Seem True

There are some conjectures that seem true because it’s hard to find acounterexample.

Claim: ∀n ∈ Z, n2 + n + 41 is prime.

12 + 1 + 41 = 43 is prime

22 + 2 + 41 = 47 is prime

32 + 3 + 41 = 53 is prime

Looks good so far...

412 + 41 + 41 = 41(41) + 41(1) + 41(1) = 41 · 43 is not prime

Takeaway: Don’t believe a mathematical claim, even though it looks true,until you have a proof.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 8 / 18

Caution: Some False Things Seem True

There are some conjectures that seem true because it’s hard to find acounterexample.

Claim: ∀n ∈ Z, n2 + n + 41 is prime.

12 + 1 + 41 = 43 is prime

22 + 2 + 41 = 47 is prime

32 + 3 + 41 = 53 is prime

Looks good so far...

412 + 41 + 41 = 41(41) + 41(1) + 41(1) = 41 · 43 is not prime

Takeaway: Don’t believe a mathematical claim, even though it looks true,until you have a proof.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 8 / 18

Caution: Some False Things Seem True

There are some conjectures that seem true because it’s hard to find acounterexample.

Claim: ∀n ∈ Z, n2 + n + 41 is prime.

12 + 1 + 41 = 43 is prime

22 + 2 + 41 = 47 is prime

32 + 3 + 41 = 53 is prime

Looks good so far...

412 + 41 + 41 = 41(41) + 41(1) + 41(1) = 41 · 43 is not prime

Takeaway: Don’t believe a mathematical claim, even though it looks true,until you have a proof.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 8 / 18

Caution: Some False Things Seem True

There are some conjectures that seem true because it’s hard to find acounterexample.

Claim: ∀n ∈ Z, n2 + n + 41 is prime.

12 + 1 + 41 = 43 is prime

22 + 2 + 41 = 47 is prime

32 + 3 + 41 = 53 is prime

Looks good so far...

412 + 41 + 41 = 41(41) + 41(1) + 41(1) = 41 · 43 is not prime

Takeaway: Don’t believe a mathematical claim, even though it looks true,until you have a proof.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 8 / 18

Caution: Some False Things Seem True

There are some conjectures that seem true because it’s hard to find acounterexample.

Claim: ∀n ∈ Z, n2 + n + 41 is prime.

12 + 1 + 41 = 43 is prime

22 + 2 + 41 = 47 is prime

32 + 3 + 41 = 53 is prime

Looks good so far...

412 + 41 + 41 = 41(41) + 41(1) + 41(1) = 41 · 43 is not prime

Takeaway: Don’t believe a mathematical claim, even though it looks true,until you have a proof.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 8 / 18

Caution: Some False Things Seem True

There are some conjectures that seem true because it’s hard to find acounterexample.

Claim: ∀n ∈ Z, n2 + n + 41 is prime.

12 + 1 + 41 = 43 is prime

22 + 2 + 41 = 47 is prime

32 + 3 + 41 = 53 is prime

Looks good so far...

412 + 41 + 41 = 41(41) + 41(1) + 41(1) = 41 · 43 is not prime

Takeaway: Don’t believe a mathematical claim, even though it looks true,until you have a proof.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 8 / 18

GCD and LCM

Definition

if c | a and c | b, then c is a common divisor of a and b. The largestsuch number is the greatest common divisor (GCD) of a and b, writtenas gcd(a, b).

Definition

if a | c and b | c , then c is a common multiple of a and b. The smallestsuch number is the least common multiple (LCM) of a and b, writtenas lcm(a, b).

Fact

lcm(a, b) =ab

gcd(a, b)

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 9 / 18

GCD and LCM

Definition

if c | a and c | b, then c is a common divisor of a and b. The largestsuch number is the greatest common divisor (GCD) of a and b, writtenas gcd(a, b).

Definition

if a | c and b | c , then c is a common multiple of a and b. The smallestsuch number is the least common multiple (LCM) of a and b, writtenas lcm(a, b).

Fact

lcm(a, b) =ab

gcd(a, b)

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 9 / 18

GCD and LCM

Definition

if c | a and c | b, then c is a common divisor of a and b. The largestsuch number is the greatest common divisor (GCD) of a and b, writtenas gcd(a, b).

Definition

if a | c and b | c , then c is a common multiple of a and b. The smallestsuch number is the least common multiple (LCM) of a and b, writtenas lcm(a, b).

Fact

lcm(a, b) =ab

gcd(a, b)

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 9 / 18

The Division Algorithm

Theorem

For any integers a and b, where b > 0, there are unique integers q (thequotient) and r (the remainder) such that a = bq + r and 0 ≤ r < b.

Examples:

Divide 17 by 3: 17 = 5 · 3 + 2

Divide -10 by 7: −10 = 7 · (−2) + 4

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 10 / 18

The Division Algorithm

Theorem

For any integers a and b, where b > 0, there are unique integers q (thequotient) and r (the remainder) such that a = bq + r and 0 ≤ r < b.

Examples:

Divide 17 by 3: 17 = 5 · 3 + 2

Divide -10 by 7: −10 = 7 · (−2) + 4

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 10 / 18

The Division Algorithm

Theorem

For any integers a and b, where b > 0, there are unique integers q (thequotient) and r (the remainder) such that a = bq + r and 0 ≤ r < b.

Examples:

Divide 17 by 3: 17 = 5 · 3 + 2

Divide -10 by 7: −10 = 7 · (−2) + 4

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 10 / 18

The Euclidean Algorithm

A long time ago, Euclid came up with a fast algorithm for computing gcd.Suppose remainder(a, b) returns the remainder when a is divided by b.Then we can compute the gcd as follows:gcd(a,b: positive integers)

x := a

y := b

while (y > 0)

begin

r := remainder(x,y)

x := y

y := r

end

return x

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 11 / 18

The Euclidean Algorithm (cont.)

Example: Compute gcd(255, 483).

x y r Equation

255 483 255 255 = 483 · 0 + 255483 255 228 483 = 255 · 1 + 228255 228 27 255 = 228 · 1 + 27228 27 12 228 = 27 · 8 + 1227 12 15 27 = 12 · 1 + 1512 15 12 12 = 15 · 0 + 1215 12 3 15 = 12 · 1 + 312 3 0 12 = 3 · 4 + 03 0 3 —

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 12 / 18

The Euclidean Algorithm: Why it Works

For the Euclidean algorithm to work, it needs to:

(a) terminate (i.e. y needs to eventually become 0).

(b) not change the output gcd when resetting variables inside the loop.

(c) return a at the end because gcd(a, 0) = a.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 13 / 18

The Euclidean Algorithm: Sketch of Proof

Claim: gcd(a, b) = gcd(b, remainder(a, b)).

Proof.

Suppose a = bq + r .

Lemma 1: Any k that divides a and b also divides r .

Lemma 2: Any k that divides b and r also divides a.

Therefore, the set of common factors of a and b is the same as theset of common factors of b and r .

We conclude that gcd(a, b) = gcd(b, remainder(a, b)), since identicalsets of common factors will have the same maximum.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 14 / 18

The Euclidean Algorithm: Sketch of Proof

Claim: gcd(a, b) = gcd(b, remainder(a, b)).

Proof.

Suppose a = bq + r .

Lemma 1: Any k that divides a and b also divides r .

Lemma 2: Any k that divides b and r also divides a.

Therefore, the set of common factors of a and b is the same as theset of common factors of b and r .

We conclude that gcd(a, b) = gcd(b, remainder(a, b)), since identicalsets of common factors will have the same maximum.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 14 / 18

The Euclidean Algorithm: Sketch of Proof

Claim: gcd(a, b) = gcd(b, remainder(a, b)).

Proof.

Suppose a = bq + r .

Lemma 1: Any k that divides a and b also divides r .

Lemma 2: Any k that divides b and r also divides a.

Therefore, the set of common factors of a and b is the same as theset of common factors of b and r .

We conclude that gcd(a, b) = gcd(b, remainder(a, b)), since identicalsets of common factors will have the same maximum.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 14 / 18

The Euclidean Algorithm: Sketch of Proof

Claim: gcd(a, b) = gcd(b, remainder(a, b)).

Proof.

Suppose a = bq + r .

Lemma 1: Any k that divides a and b also divides r .

Lemma 2: Any k that divides b and r also divides a.

Therefore, the set of common factors of a and b is the same as theset of common factors of b and r .

We conclude that gcd(a, b) = gcd(b, remainder(a, b)), since identicalsets of common factors will have the same maximum.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 14 / 18

The Euclidean Algorithm: Sketch of Proof

Claim: gcd(a, b) = gcd(b, remainder(a, b)).

Proof.

Suppose a = bq + r .

Lemma 1: Any k that divides a and b also divides r .

Lemma 2: Any k that divides b and r also divides a.

Therefore, the set of common factors of a and b is the same as theset of common factors of b and r .

We conclude that gcd(a, b) = gcd(b, remainder(a, b)), since identicalsets of common factors will have the same maximum.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 14 / 18

The Euclidean Algorithm: Sketch of Proof

Claim: gcd(a, b) = gcd(b, remainder(a, b)).

Proof.

Suppose a = bq + r .

Lemma 1: Any k that divides a and b also divides r .

Lemma 2: Any k that divides b and r also divides a.

Therefore, the set of common factors of a and b is the same as theset of common factors of b and r .

We conclude that gcd(a, b) = gcd(b, remainder(a, b)), since identicalsets of common factors will have the same maximum.

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 14 / 18

The Euclidean Algorithm: Sketch of Proof

Lemma 1: Any k that divides a and b also divides r .

Proof.

Let a, b, q, r , and k be integers. Suppose a = bq + r , k | a, and k | b.

Since k | a, a = kn for some integer n. Since k | b, b = km for someinteger m.

Since a = bq + r , r = a− bq.

So r = kn − (km)q = k(n −mq).

We know n −mq is an integer, so k | r .

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 15 / 18

The Euclidean Algorithm: Sketch of Proof

Lemma 1: Any k that divides a and b also divides r .

Proof.

Let a, b, q, r , and k be integers. Suppose a = bq + r , k | a, and k | b.

Since k | a, a = kn for some integer n. Since k | b, b = km for someinteger m.

Since a = bq + r , r = a− bq.

So r = kn − (km)q = k(n −mq).

We know n −mq is an integer, so k | r .

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 15 / 18

The Euclidean Algorithm: Sketch of Proof

Lemma 1: Any k that divides a and b also divides r .

Proof.

Let a, b, q, r , and k be integers. Suppose a = bq + r , k | a, and k | b.

Since k | a, a = kn for some integer n. Since k | b, b = km for someinteger m.

Since a = bq + r , r = a− bq.

So r = kn − (km)q = k(n −mq).

We know n −mq is an integer, so k | r .

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 15 / 18

The Euclidean Algorithm: Sketch of Proof

Lemma 1: Any k that divides a and b also divides r .

Proof.

Let a, b, q, r , and k be integers. Suppose a = bq + r , k | a, and k | b.

Since k | a, a = kn for some integer n. Since k | b, b = km for someinteger m.

Since a = bq + r , r = a− bq.

So r = kn − (km)q = k(n −mq).

We know n −mq is an integer, so k | r .

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 15 / 18

The Euclidean Algorithm: Sketch of Proof

Lemma 1: Any k that divides a and b also divides r .

Proof.

Let a, b, q, r , and k be integers. Suppose a = bq + r , k | a, and k | b.

Since k | a, a = kn for some integer n. Since k | b, b = km for someinteger m.

Since a = bq + r , r = a− bq.

So r = kn − (km)q = k(n −mq).

We know n −mq is an integer, so k | r .

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 15 / 18

Congruence mod k

Definition

If k is any positive integer, two integers a and b are congruent mod k ifk | (a− b).We write this as a ≡ b (mod k).

Examples:

3 ≡ 38 (mod 7)

19 ≡ 7 (mod 12)

−22 ≡ 3 (mod 5)

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 16 / 18

Congruence mod k

Definition

If k is any positive integer, two integers a and b are congruent mod k ifk | (a− b).We write this as a ≡ b (mod k).

Examples:

3 ≡ 38 (mod 7)

19 ≡ 7 (mod 12)

−22 ≡ 3 (mod 5)

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 16 / 18

Congruence mod k

Definition

If k is any positive integer, two integers a and b are congruent mod k ifk | (a− b).We write this as a ≡ b (mod k).

Examples:

3 ≡ 38 (mod 7)

19 ≡ 7 (mod 12)

−22 ≡ 3 (mod 5)

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 16 / 18

Proof with Congruence mod k

Claim: For any integers a, b, c, d , and k , where k > 0, if a ≡ b (mod k)and c ≡ d (mod k), then ac ≡ bd (mod k).

Proof.

Let a, b, c , and d be any integers, and let k be any positive integer.Suppose a ≡ b (mod k) and c ≡ d (mod k).

By definition of congruence mod k, k | (a− b) and k | (c − d). Bydefinition of divides, this means (a− b) = kn and (c − d) = km,where n and m are integers. So a = b + kn and c = d + km.

Multiplying, we have ac = bd + bkm + dkn + k2nm.

So ac = bd + k(bm + dn + knm).

Let q = bm + dn + knm. Since b, d , n,m, and k are integers, q is aninteger. Then, ac = bd + kq, which means (ac − bd) = kq.

So k | (ac − bd). By definition of congruence mod k,ac ≡ bd (mod k).

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 17 / 18

Proof with Congruence mod k

Claim: For any integers a, b, c, d , and k , where k > 0, if a ≡ b (mod k)and c ≡ d (mod k), then ac ≡ bd (mod k).

Proof.

Let a, b, c , and d be any integers, and let k be any positive integer.Suppose a ≡ b (mod k) and c ≡ d (mod k).

By definition of congruence mod k , k | (a− b) and k | (c − d). Bydefinition of divides, this means (a− b) = kn and (c − d) = km,where n and m are integers. So a = b + kn and c = d + km.

Multiplying, we have ac = bd + bkm + dkn + k2nm.

So ac = bd + k(bm + dn + knm).

Let q = bm + dn + knm. Since b, d , n,m, and k are integers, q is aninteger. Then, ac = bd + kq, which means (ac − bd) = kq.

So k | (ac − bd). By definition of congruence mod k ,ac ≡ bd (mod k).

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 17 / 18

Proof with Congruence mod k

Claim: For any integers a, b, c, d , and k , where k > 0, if a ≡ b (mod k)and c ≡ d (mod k), then ac ≡ bd (mod k).

Proof.

Let a, b, c , and d be any integers, and let k be any positive integer.Suppose a ≡ b (mod k) and c ≡ d (mod k).

By definition of congruence mod k , k | (a− b) and k | (c − d). Bydefinition of divides, this means (a− b) = kn and (c − d) = km,where n and m are integers. So a = b + kn and c = d + km.

Multiplying, we have ac = bd + bkm + dkn + k2nm.

So ac = bd + k(bm + dn + knm).

Let q = bm + dn + knm. Since b, d , n,m, and k are integers, q is aninteger. Then, ac = bd + kq, which means (ac − bd) = kq.

So k | (ac − bd). By definition of congruence mod k ,ac ≡ bd (mod k).

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 17 / 18

Proof with Congruence mod k

Claim: For any integers a, b, c, d , and k , where k > 0, if a ≡ b (mod k)and c ≡ d (mod k), then ac ≡ bd (mod k).

Proof.

Let a, b, c , and d be any integers, and let k be any positive integer.Suppose a ≡ b (mod k) and c ≡ d (mod k).

By definition of congruence mod k , k | (a− b) and k | (c − d). Bydefinition of divides, this means (a− b) = kn and (c − d) = km,where n and m are integers. So a = b + kn and c = d + km.

Multiplying, we have ac = bd + bkm + dkn + k2nm.

So ac = bd + k(bm + dn + knm).

Let q = bm + dn + knm. Since b, d , n,m, and k are integers, q is aninteger. Then, ac = bd + kq, which means (ac − bd) = kq.

So k | (ac − bd). By definition of congruence mod k ,ac ≡ bd (mod k).

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 17 / 18

Proof with Congruence mod k

Claim: For any integers a, b, c, d , and k , where k > 0, if a ≡ b (mod k)and c ≡ d (mod k), then ac ≡ bd (mod k).

Proof.

Let a, b, c , and d be any integers, and let k be any positive integer.Suppose a ≡ b (mod k) and c ≡ d (mod k).

By definition of congruence mod k , k | (a− b) and k | (c − d). Bydefinition of divides, this means (a− b) = kn and (c − d) = km,where n and m are integers. So a = b + kn and c = d + km.

Multiplying, we have ac = bd + bkm + dkn + k2nm.

So ac = bd + k(bm + dn + knm).

Let q = bm + dn + knm. Since b, d , n,m, and k are integers, q is aninteger. Then, ac = bd + kq, which means (ac − bd) = kq.

So k | (ac − bd). By definition of congruence mod k ,ac ≡ bd (mod k).

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 17 / 18

Proof with Congruence mod k

Claim: For any integers a, b, c, d , and k , where k > 0, if a ≡ b (mod k)and c ≡ d (mod k), then ac ≡ bd (mod k).

Proof.

Let a, b, c , and d be any integers, and let k be any positive integer.Suppose a ≡ b (mod k) and c ≡ d (mod k).

By definition of congruence mod k , k | (a− b) and k | (c − d). Bydefinition of divides, this means (a− b) = kn and (c − d) = km,where n and m are integers. So a = b + kn and c = d + km.

Multiplying, we have ac = bd + bkm + dkn + k2nm.

So ac = bd + k(bm + dn + knm).

Let q = bm + dn + knm. Since b, d , n,m, and k are integers, q is aninteger. Then, ac = bd + kq, which means (ac − bd) = kq.

So k | (ac − bd). By definition of congruence mod k ,ac ≡ bd (mod k).

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 17 / 18

Proof with Congruence mod k

Claim: For any integers a, b, c, d , and k , where k > 0, if a ≡ b (mod k)and c ≡ d (mod k), then ac ≡ bd (mod k).

Proof.

Let a, b, c , and d be any integers, and let k be any positive integer.Suppose a ≡ b (mod k) and c ≡ d (mod k).

By definition of congruence mod k , k | (a− b) and k | (c − d). Bydefinition of divides, this means (a− b) = kn and (c − d) = km,where n and m are integers. So a = b + kn and c = d + km.

Multiplying, we have ac = bd + bkm + dkn + k2nm.

So ac = bd + k(bm + dn + knm).

Let q = bm + dn + knm. Since b, d , n,m, and k are integers, q is aninteger. Then, ac = bd + kq, which means (ac − bd) = kq.

So k | (ac − bd). By definition of congruence mod k ,ac ≡ bd (mod k).

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 17 / 18

References

Building Blocks for Theoretical Computer Science, Ch. 4 by Margaret M. Fleck

Mathematics for Computer Science, Ch. 4 by Lehman, Leighton, and Meyer

Dr. Madhu Parthasarathy’s old lecture slides

Ian Ludden (UIUC) Intro to Number Theory CS173 Fall 2017 18 / 18