Mas4203 Class Notes1

transcript

Bibliography:

1. Elementary Number Theory, James Tattershall, Cambridge 1999.

2. Elements of Number Theory, I.A. Barnett, Prindle, Weber andSchmidt 1969

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1. Introduction

Number Theory: Area of mathematics whose aim is to uncover themany deep subtle relationships between different sorts of numbers.

Some classical unsolved problems in number theory

Are there infinitely many twin primes? (That is pairs of primenumbers (p, p + 2). Examples: 3,5 5,7 11,13 17,19

Are there infinitely many primes of the form N2 + 1?, Examples:5,17,37, ...

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1. Introduction

Number Theory: Area of mathematics whose aim is to uncover themany deep subtle relationships between different sorts of numbers.

Some classical unsolved problems in number theory

Are there infinitely many twin primes? (That is pairs of primenumbers (p, p + 2). Examples: 3,5 5,7 11,13 17,19

Are there infinitely many primes of the form N2 + 1?, Examples:5,17,37, ...

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Assignments

1. Add up the first few odd numbers and see if the numbers you getsatisfy some sort of pattern. Express the pattern as a formula. Provethat the formula is correct.

(a) Do you think that there are infinitely many primes of the formN2 − 1?

(b) Do you think that there are infinitely many primes of the formN2 − 2?

(c) How about the form N2 − 3? How about N2 − 4?

(d) Which values of a do you think give infinitely many primes of theform N2 − a?

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2. Divisibility

Notation: "m divides n": m | n, meaning n = mk for some integer k .

Such an m is called a divisor of n.

m - n: m doesn’t divide n.

The greatest common divisor of a,b is the largest number that dividesboth a and b.

Notation: gcd(a,b) or simply (a,b).

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2. Divisibility

Notation: "m divides n": m | n, meaning n = mk for some integer k .

Such an m is called a divisor of n.

m - n: m doesn’t divide n.

The greatest common divisor of a,b is the largest number that dividesboth a and b.

Notation: gcd(a,b) or simply (a,b).

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Example: (14,35) = 7

Euclidean Algorithm for gcd

Example: Compute (36,132)

132 = 3× 36 + 24

36 = 1× 24 + 12

24 = 2× 12 + 0

(36,132) = 12

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Example: (14,35) = 7

Euclidean Algorithm for gcd

Example: Compute (36,132)

132 = 3× 36 + 24

36 = 1× 24 + 12

24 = 2× 12 + 0

(36,132) = 12

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General Euclidean Algorithm: A = Q × B + R

(a,b) = rn, where

r−1 = a

r0 = b

a = q1 × b + r1

b = q2 × r1 + r2... =

rn−2 = qn × rn−1 + rn

rn−1 = qn+1rn + 0

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Verification: For i = 1,2, ....,n − 1,

ri−1 = qi+1 × ri + ri+1

Why is rn the gcd of a and b?

Suppose d is any divisor of a and b.

a = q1 × b + r1 implies that d | r1

b = q2 × r1 + r2 implies that d | r2. Inductively, d | ri−1 and d | ri impliesthat d | ri+1 (follows from ri−1 = qi+1 × ri + ri+1).

Eventually, one reaches d | rn.

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Verification: For i = 1,2, ....,n − 1,

ri−1 = qi+1 × ri + ri+1

Why is rn the gcd of a and b?

Suppose d is any divisor of a and b.

a = q1 × b + r1 implies that d | r1

b = q2 × r1 + r2 implies that d | r2. Inductively, d | ri−1 and d | ri impliesthat d | ri+1 (follows from ri−1 = qi+1 × ri + ri+1).

Eventually, one reaches d | rn.

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Assignment

Let b = r0, r1, r2, ... be the successive remainders in the Euclideanalgorithm applied to a and b. Show that every two steps reduces theremainder by at least one half. In other words, verify that

ri+2 <12

for every i = 0,1,2, ... Conclude that the Euclidean algorithm willterminate in at most 2 log2 b steps.

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Least Common Multiple

The least common multiple (lcm) of a and b is the smallest number ofwhich a and b are divisors.

Notation lcm(a,b) = [a,b].

Claim: ab = (a,b)[a,b].

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Least Common Multiple

The least common multiple (lcm) of a and b is the smallest number ofwhich a and b are divisors.

Notation lcm(a,b) = [a,b].

Claim: ab = (a,b)[a,b].

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Proof: (Skip!!) [a,b] = ak with b | ak .

If d = (a,b), then a = da′ and b = db′, with (a′,b′) = 1.

Since b | ak = da′k , then db′ | da′k , that is b′ | k .

In other words, k = b′t , and

[a,b] = ab′t = a(bd)t = (

The least common multiple is obtained when t is the smallest naturalnumber, that is t = 1. Hence

[a,b] =ab

(a,b).

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Divisibility by 4 and 8; 5 and 25

In base 10, every integer N has a unique expression

N = an10n + an−110n−1 + ...+ a110 + a0

Claim: N is divisible by 2k if and only if ak−110k−1 + ...+ a0 is divisibleby 2k .

Proof:k−1∑i=0

ai10i = N −∑i≥k

Each term in the summation on the right side is divisible by 2k , so thesummation on the left side is divisible by 2k if and only if N is divisibleby 2k .

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Examples: 2455314 is not divisible by 4 = 22, because 14 is notdivisible by 4.

2477312 is divisible by 8 = 23 because 312 is divisible by 8.

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Examples: 2455314 is not divisible by 4 = 22, because 14 is notdivisible by 4.

2477312 is divisible by 8 = 23 because 312 is divisible by 8.

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It follows easily from 10 = 2× 5 that

N is divisible by 5k if and only if

k−1∑i=0

is divisible by 5k

Assignments

1. The last 3 digits of an integer N are X24. Find the missing integer Xif N is divisible by 8.

2. The last three digits of an integer N are 2X4. Find the missing digitX if N is divisible by 8.

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It follows easily from 10 = 2× 5 that

N is divisible by 5k if and only if

k−1∑i=0

is divisible by 5k

Assignments

1. The last 3 digits of an integer N are X24. Find the missing integer Xif N is divisible by 8.

2. The last three digits of an integer N are 2X4. Find the missing digitX if N is divisible by 8.

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Remainder when dividing by 9

Claim: If N =∑n

i=0 ai10i and

N = 9×Q + R, with R 6= 0

thenn∑

ai = R

We can use this fact to find the least remainder when N is divided by 9.

Add up the digits of N: an + ...+ a1 + a0.

If the sum is > 9, add up the digits of the sum

repeat this process until the sum of digits comes up < 9.

This method is called The method of casting out nines

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Example Find the (least) remainder when 13578 is divided by 9.

1+3+5+7+8=24

The least remainder is equal to 6.

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Proof of the claim behind the method:

LetN = 9×Q + R

ThenN9

= Q +R9

=an10n + ...+ a110 + a0 − (an + ...+ a1 + a0) + (an + ...+ a1 + a0)

=an(10n − 1) + ...+ a1(10− 1)

an + ...+ a0

+an + ...+ a0

= Q +an + ...+ a0

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EquivalentlyN = 9×Q + an + ...+ a0

soR = an + ...+ a1 + a0

As a consequence:

N is divisible by 9 if and only if the sum of its digits is divisible by 9

Another consequence is:

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soR = an + ...+ a1 + a0

As a consequence:

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soR = an + ...+ a1 + a0

As a consequence:

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Proof:N − 9×Q = an + ...+ a0

If N is divisible by 3, then so is an + ...+ a0 since 9×Q is clearlydivisible by 3. Conversely, if an + ...+ a0 is divisible by 3, then so isN = 9×Q + an + ...+ a0.

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Assignments

1. Check 62085 for divisibility by 3 and 9.

2. Check each of the following for divisibility by 6 and 12.

a) 23082

b) 14064

c) 241224

(Divisibility by 6 means by 2 and 3; divisibility by 12 means by 3 and 4)

3. The number X203X is divisible by 24 (= 23.3). Find the missingdigits.

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3. Linear diophantine equations

A diophantine equation is one of the form:

ax + by = c

a, b and c are integers and integer solutions (lattice points) are sought.

Example:2x + 3y = 11

some solutions are (1,3) (7,−1),...

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Here is an example without any solution:

2x + 4y = 3

Any solution (x0, y0) would have to satisfy

2(x0 + 2y0) = 3

but this is impossible because 3 is not divisible by 2.

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Ifax + by = c

has a solution, then gcd(a,b) | c.

Before we prove the converse, let us look at the casec = d = gcd(a,b).

Solvingax + by = gcd(a,b)

Example: Solve22x + 60y = gcd(22,60)

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The Euclidean Algorithm for gcd(22,60) goes:

60 = 2× 22 + 16

22 = 1× 16 + 6

16 = 2× 6 + 4

6 = 1× 4 + 2

4 = 2× 2 + 0

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gcd(22,60) = 2 and going through the algorithm, with a = 60, b = 22,one has:

16 = 60− 2× 22

6 = 22− 1× 16

= 22− 60 + 2× 22

= 3× 22− 60

4 = 16− 2× 6

= 60− 2× 22− 6× 22 + 2× 60

= 3× 60− 8× 22

2 = 6− 1× 4

= 3× 22− 60− 3× 60 + 8× 22

2 = 11× 22− 4× 60

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How to get all solutions

For gcd(a,b) = 1, solveax + by = 1.

Let (x1, y1) be a solution.

Observation: (x1 − kb, y1 + ka), k = 0,±1,±2, .... is also a solution!

Proof:

(x1 − kb)a + b(y1 + ka) = x1a + y1b − k(ba) + (kba) = 1.

Claim: This procedure gives all solutions

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Proof: Suppose (x1, y1) and (x2, y2) are two solutions.

ax1 + by1 = 1

ax2 + by2 = 1

Multiply the first identity by y2, the second by y1 to obtain:

y2ax1 + by2y1 = y2

y1ax2 + y1by2 = y1

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Substract the second from the first identity:

a(y2x1 − y1x2) = y2 − y1

Repeat this with x2 and x1 to get

b(x2y1 − y2x1) = x2 − x1

Let k = (x2y1 − y2x1). Then we see that

x2 = x1 + kb

andy2 = y1 − ka

This settles the case gcd(a,b) = 1. What if gcd(a,b) > 1?

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Let gcd(a,b) = d . We know that

ax + by = d

has at least one solution (x1, y1). But also, it is clear that (x1, y1) is asolution of

y = 1 (∗)

with gcd( ad ,

bd ) = 1

Any other solution of (∗) is given by

x2 = x1 + kbd, y2 = y1 − k

This completes the description of all solutions of ax + by = d .

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Let gcd(a,b) = d . We know that

ax + by = d

has at least one solution (x1, y1). But also, it is clear that (x1, y1) is asolution of

y = 1 (∗)

with gcd( ad ,

bd ) = 1

Any other solution of (∗) is given by

x2 = x1 + kbd, y2 = y1 − k

This completes the description of all solutions of ax + by = d .

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Examples: Find all solutions for 14x + 35y = 7.

35 = 2× 14 + 7

14 = 2× 7 + 0

So7 = 35− 2× 14

(x1, y1) = (−2,1) is a solution. Any other solution is of the form

x2 = −2 + 5k , y2 = 1− 2k , k = ±1,±2, ....

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Examples: Find all solutions for 14x + 35y = 7.

35 = 2× 14 + 7

14 = 2× 7 + 0

So7 = 35− 2× 14

(x1, y1) = (−2,1) is a solution. Any other solution is of the form

x2 = −2 + 5k , y2 = 1− 2k , k = ±1,±2, ....

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We have seen that if ax + by = c has a solution, then gcd(a,b) | c.We will show that

conversely, if gcd(a,b) | c, then ax + by = c has a solution.

Proof:

Let d = (a,b). There are integers x0 and y0 such that

ax0 + by0 = d

These are determined by solving for the remainders and substituting inthe last step of the Euclidean algorithm for gcd .

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Multiply both sides of the equation by c/d

dc) + b(

dc) = c

Let x1 = x0d c, y1 = y0

d c. Then

ax1 + by1 = c

showing that ax + by = c has a solution.

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Example

Find a solution for4x + 6y = 12.

Observe gcd(4,6) = 2 and 2 | 12. So there is a solution. To find one,first solve

4x + 6y = 2

or equivalently2x + 3y = 1

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The Euclidean algorithm goes:

3 = 1× 2 + 1

2 = 2× 1 + 0

Solving for remainders:

gcd(2,3) = 1 = 3− 1× 2

2× (−1) + 3× 1 = 1

So x0 = −1, y0 = 1.

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Multiplying this identity by 12, we get:

2× (−12) + 3× (12) = 12

or equivalently:

4× (−6) + 6× 6 = 12

So a solution of 4x + 6y = 12 is given by x = −6, y = 6

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ax + by = c has a solution if and only if d = gdc(a,b) divides c, inwhich case it is given by x = x0

d c, y = y0d c, where (x0, y0) is a solution

y = 1.

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Assignments: solve

a) 3x+4y=1

b) 6x+9y=2

c) 117x+52y=26

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Direct method of finding solutions

Example: Suppose a man makes a purchase of $1.15 and gives theclerk two $1 bills. In how many ways may he receive his change indimes and quarters?

0 ≤ d = number of dimes

0 ≤ q = number of quarters

10d + 25q = 85

Equivalently2d + 5q = 17

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Direct method of finding solutions

Example: Suppose a man makes a purchase of $1.15 and gives theclerk two $1 bills. In how many ways may he receive his change indimes and quarters?

0 ≤ d = number of dimes

0 ≤ q = number of quarters

10d + 25q = 85

Equivalently2d + 5q = 17

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Solve for the variable with smallest coefficient:

2d = 17− 5q

d = 8 +12− 2q − q

2= 8− 2q +

1− q2

t = 1−q2 must be an integer and, following the rule above

q = 1− 2t

Substituting in the expression of d , we get:

d = 8− 2(1− 2t) + t = 6 + 5t

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All solutions are obtained by letting t = 0,±1,±2, ... but only few ofthese meet the requirement d ≥ 0, q ≥ 0.

d ≥ 0⇒ 6 + 5t ≥ 0⇒ t ≥ −65

q ≥ 0⇒ 12 ≥ t .

so −65 ≤ t ≤ 1

2 , thus t = −1 or t = 0. The solutions are (d ,q) = (1,3)and (d ,q) = (6,1).

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Another example: A man cashes a check at his bank and the tellermistakenly interchanges dollars and cents. After leaving the bank, theman spends 68 cents and then discover that he now has twice theamount of the original check. What was the amount of the check?

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Let 0 ≤ d ≤ 100 denote the number of dollar bills and 0 ≤ c ≤ 100 thatof pennies.

100d + c − 68 = 2(d + 100c)

That is98d − 199c = 68

gcd(98,199)=1 and 1 | 68, so we expect a solution to exist.

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Solve for d which has the smallest coefficient:

d = 2c +3c + 68

z = 3c+6898 must be an integer, and solving for c (which has the smallest

coefficient in this equation)

c = 32z − 22 +2z − 2

Again, u = 2z−23 must be an integer and solving for z gives

z = u + 1 +u2

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t = u2 must be an integer and

u = 2t

Back substituting, one gets:

z = 3t + 1

c = 98t + 10

d = 199t + 21

0 ≤ c ≤ 100 and 0 ≤ d ≤ 100 imply that t = 0 is the only acceptablevalue and the solution is c = 10 and d = 21. The amount of the checkwas $10.21.

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Assignment:

1. Five times a nonnegative integer added to seven times anothernonnegative integer give a sum of 100. Find all possible integers.

5n + 7m = 100

2. Solve:33x + 14y = 113

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Linear diophantine equations in more than 2unknowns

a1x1 + ....+ anxn = b

We will focus on 3 unknowns

ax + by + cz = k

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gcd and lcm when prime factorization is known

a = p1a1 ...pk

b = p1b1 ...pk

gcd(a,b) = p1sm(a1,b1)...pk

sm(ak ,bk )

lcm(a,b) = p1la(a1,b1)...pk

la(ak ,bk )

The definitions can be extended to 3 or more numbers.

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ax + by + cz = k

Necessarily, gcd(a,b, c) divides k for a solution to exist.

Conversely, suppose d = gcd(a,b, c) divides k . Find integersx ′, y ′, z ′ such that

ax ′ + by ′ + cz ′ = d .

(x ′kd

,y ′kd

,z ′kd

is a solution of

au + bv + cw = k

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Example: Find positive solutions to

15x + 12y + 8z = 400

gcd(15,12,8)=1 and 1 | 400 so there is a solution! At each stage, solvefor the unknown with smallest coefficient:

z = 50− x − 7x8− y − 4y

8= 50− x − y − (

7x + 4y8

u = 7x+4y8 must be an integer and from 8u = 7x + 4y , we solve for y :

y = 2u − x − 3x4

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Again, v = 3x4 must be an integer and solving for x gives

x = v +v3

t = v3 must be an integer and

v = 3t

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Recap:v = 3t

x = 3t + t = 4t

y = 2u − 7t

z = 50− 3u + 3t

Solutions are given by two nonindependent integer parameters t andu, each taking values 0,±1,±2, ....

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We need now to determine u and t so that x , y , z are positive integers!

t ≥ 0

u ≥ 7t2

u ≤ 503

and503

+ t − 72

t ≥ 0⇒ t ≤ 203.

So0 ≤ t ≤ 20

t ≤ u ≤ 503

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t=0 u=0,1,...,16t=1 u=4,5,...,17t=2 u=7,8,...,18t=3 u=10,11,...,19t=4 u=14,15,...,20t=5 u=17,18,...,21t=6 u=21,22

Assignment: Find gcd(60,144,280).

Solve60x + 144y + 280z = 4

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Solving systems

To solve the system

15x + 12y + 8z = 400

x + y + z = 40

Solve the first equation:

x = 4t

y = 2u − 7t

z = 50− 3u + 3t

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Then substitute into the second equation:

−u + 50 = 40

Thusu = 10

Sox = 4t

y = 20− 7t

z = 20 + 3t

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The requirement x ≥ 0, y ≥ 0 and z ≥ 0 imply:

t ≥ 0

20 ≥ 7t ⇒ t ≤ 2

The solution set consists of

(x , y , z) = (0,20,20), (4,13,23), (8,6,26)

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A simpler method

Eliminate one of the unknowns, say z, from the system. Obtaining asingle equation in two unknowns. Proceed as before!

Assignment:

1. Solve the system:x + 3y − 4z = 8

2x + y + 3z = 39

2. There are 1000 seats in an auditorium. If men are charged $5 perseat, women $4 and children $0.75, in how many different ways couldthe house be sold to realize $ 3000?

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4. Factorization and the Fundamental Theorem ofArithmetic

Claim: Let p be a prime number. If p | ab, then p | a or p | b.

Proof of claim: Suppose p | ab and p - a, then

ab = kp ⇒ k =abp

→ bp

is an integer

⇒ p | b.

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4. Factorization and the Fundamental Theorem ofArithmetic

Claim: Let p be a prime number. If p | ab, then p | a or p | b.Proof of claim: Suppose p | ab and p - a, then

ab = kp ⇒ k =abp

→ bp

is an integer

⇒ p | b.

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Alternate proof: Suppose p | ab and p - a. Then gcd(p,a) = 1 and sothe diophantine equation

px + ay = 1

has integer solutions. It follows that

bpx + bay = b

has also integer solutions.

Since p | bpx and p | bay , it follows that p | b.

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Prime divisibility property

Let p be a prime number.

If p | a1a2...an, then p | aj for at least one j , 1 ≤ j ≤ n.

The Fundamental Theorem of Arithmetic:

Every integer n ≥ 2 can be factored into a product of primes

n = p1p2...pr

in exactly one way.

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Proof:2 = 2

4 = 2.2

Suppose ∀n ≤ N, n = p1np2n ...prn

If N + 1 is composite, then N + 1 = n1n2 with

2 ≤ n1, n2 ≤ N

Son1 = p1n1

...prn1; n2 = p1n2

...prn2

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Multiplying, we get a prime factorization for N + 1.

N + 1 = n1n2 = p1n1...prn1

p1n2...prn2

Supposen = p1...pr = q1...qs

are two prime factorizations for n. Then p1 | n⇒ p1 | qj for some1 ≤ j ≤ s. So p1 = qj . We may assume after rearranging terms thatp1 = q1.

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Nowp2...pr = q2...qs

repeating the same argument, we reach the fact that

pr = qr

So r = s and qi = pi , for 1 ≤ i ≤ r .

Assignments

1. Suppose that gcd(a,b)=1 and suppose further that a divides theproduct bc. Show that a must divide c.

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Nowp2...pr = q2...qs

repeating the same argument, we reach the fact that

pr = qr

So r = s and qi = pi , for 1 ≤ i ≤ r .

Assignments

1. Suppose that gcd(a,b)=1 and suppose further that a divides theproduct bc. Show that a must divide c.

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5. Congruences

a is congruent to b modulo m is denoted and defined by:

a ≡ b(mod m)⇔ m | a− b

m is called the modulus of the congruence.

Congruences with common modulus can be added, but generally notdivided! This allows us to solve equations like:

Find x ifx + 12 ≡ 5 (mod 8)

x ≡ 5− 12 ≡ −7 (mod 8)

x ≡ 1 (mod 8).

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Modular arithmetic

Theorem: If ai ≡ bi (mod m), for i = 1,2, ...,n, then

a)∑n

i=1 ai ≡∑n

i=1 bi (mod m)

b)∏n

i=1 ai ≡∏n

i=1 bi (mod m)

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If a ≡ b (mod mi), for i = 1,2, ..., k , where m1,m2, ...,mk are pairwiseco-prime, then a ≡ b (mod m) where m =

∏ki=1 mi .

In modular arithmetic, the cancelation law

ac ≡ bc (mod m)⇒ a ≡ b (mod m)

doesn’t necessarily hold.

Theorem:If ac ≡ bc (mod m), then

a ≡ b (modmd)

where d = gcd(c,m).

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Proof:If ac ≡ bc (mod m), then ac − bc = km for some integer k .

Let d = gcd(c,m). Then

(a− b)(cd) = k(

with gcd( cd ,

md ) = 1. Hence, m

d divides a− b or equivalently,

a ≡ b (modmd).

Corollary: If ac ≡ bc (mod m) and gcd(c,m) = 1, then a ≡ b(mod m).

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Exercises:

1. Solve: 4x ≡ 3 (mod 19)

Multiply by 5:

20x ≡ 15 (mod 19)

But 20 ≡ 1 (mod 19), so: 20x ≡ x (mod 19), hence:

x ≡ 15 (mod 19)

2. x2 + 2x − 1 ≡ 0 (mod 7)

Try 1,2, ...,6

x ≡ 2 (mod 7), x ≡ 3 (mod 7).

3. x2 ≡ 3 (mod 10) has no solution.

PR (FIU) MAC 4203 67 / 89

Exercises:

1. Solve: 4x ≡ 3 (mod 19)

Multiply by 5:

20x ≡ 15 (mod 19)

x ≡ 15 (mod 19)

2. x2 + 2x − 1 ≡ 0 (mod 7)

Try 1,2, ...,6

x ≡ 2 (mod 7), x ≡ 3 (mod 7).

3. x2 ≡ 3 (mod 10) has no solution.

PR (FIU) MAC 4203 67 / 89

Exercises:

1. Solve: 4x ≡ 3 (mod 19)

Multiply by 5:

20x ≡ 15 (mod 19)

x ≡ 15 (mod 19)

2. x2 + 2x − 1 ≡ 0 (mod 7)

Try 1,2, ...,6

x ≡ 2 (mod 7), x ≡ 3 (mod 7).

3. x2 ≡ 3 (mod 10) has no solution.PR (FIU) MAC 4203 67 / 89

Solving ax ≡ c (mod m)

Find x so thatax − c = my

Equivalently:ax −my = c

Let d = gcd(a,m). If d - c, then no solution!

If d | c, then find a solution to au + mv = d . Say u = u0, v = v0 hasbeen found. Multiply the equation above by c

So x0 ≡ cu0d (mod m) is a solution to ax ≡ c (mod m).

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To find other solutions, observe that if x1 is one, then m | ax1 − ax0. So

md| a(x1 − x0)

But gcd(m,a) = d , so md | (x1 − x0). In other words,

x1 = x0 + kmd, k = 0,1,2, ...,d − 1.

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Assignments

1. Suppose that ac ≡ bc (mod m) and also assume thatgcd(c,m) = 1. Prove that a ≡ b (mod m).

2. Find all incongruent solutions to each of the following congruences:

(a) 7x ≡ 3 (mod 15)

(b) 6x ≡ 5 (mod 15)

(c) x2 ≡ 1 (mod 8)

(d) x2 ≡ 3 (mod 7)

(e) x2 ≡ 2 (mod 7)

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Properties of divisibility by means of congruences

In the denary scale, any number N admits the expansion:

N = an.10n + ...+ a1.10 + a0

Divisibility by 4: For k ≥ 2,

102 ≡ 0 (mod 4),103 ≡ 0 (mod 4), ...,10k ≡ 0 (mod 4)

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Properties of divisibility by means of congruences

In the denary scale, any number N admits the expansion:

N = an.10n + ...+ a1.10 + a0

Divisibility by 4: For k ≥ 2,

102 ≡ 0 (mod 4),103 ≡ 0 (mod 4), ...,10k ≡ 0 (mod 4)

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Properties of congruences imply then:

an10n + ...+ a2102 ≡ 0 (mod 4)

Thus, if N is divisible by 4, then

a110 + a0 ≡ 0 (mod 4)

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Since 10 ≡ 1 (mod 9), one has

N ≡ an + ...+ a1 + a0 (mod 9).

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Since 10 ≡ −1 (mod 11), then 102 ≡ −10 ≡ 1 (mod 11)

103 ≡ −1 (mod 11), 10k ≡ 1 (mod 11) if k is even, and 10k ≡ −1(mod 11) if k is odd. So

N ≡ a0 − a1 + a2... = (a0 + a2 + ...)− (a1 + a3 + ...) (mod 11).

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Example: Find the remainder if 2,087,190 is divided by 11.

(0 + 1 + 8 + 2)− (9 + 7 + 0) = −5 ≡ 6 (mod 11)

So, 2,087,190 ≡ 6 (mod 11)

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Example: Find the remainder if 2,087,190 is divided by 11.

(0 + 1 + 8 + 2)− (9 + 7 + 0) = −5 ≡ 6 (mod 11)

So, 2,087,190 ≡ 6 (mod 11)

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Application: Checking that ab = c by casting out ninesor 11

ab − c = 0 requires that ab − c be divisible by any integer m.

If m = 9, then ab and c may be replaced each by the sum of its digits!

(a0 + ...+ an)(b0 + ...+ bn) ≡ (c0 + ...+ cn) (mod 9)

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If m = 11, then

(a0 − a1 + ...)(b0 − b1 + ....) ≡ (c0 − c1 + ...) (mod 11)

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Example: Check(3145)(213) = 669,885

by casting out elevens.

(5− 4 + 1− 3)(3− 1 + 2) ≡ 5− 8 + 8− 9 + 6− 6 (mod 11)

(−1)(4) ≡ −4 (mod 11)

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In radix r :

N = anrn + ...+ a1r + a0, 1 ≤ aj ≤ r − 1

r ≡ 1 (mod r − 1)

r2 ≡ 1 (mod r − 1)

r3 ≡ 1, ...., r k ≡ 1 (mod r − 1)

Therefore:

N = anrn + ...+ a1r + a0 ≡ an + an−1 + ...+ a1 + a0 (mod r − 1)

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Assignment: Are these numbers divisible by 11?

a) 2,964,357

b) 225,860,351,672,202

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6. Congruences, Powers and Fermat’s Little Theorem

Fermat’s Little Theorem: For prime p and a 6≡ 0 (mod p), one has:

ap−1 ≡ 1 (mod p).

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Applications:

1. Compute 235 (mod 7).

We use the fact that 26 ≡ 1 (mod 7).

Write 35 = 6× 5 + 5 and use the law of exponents.

235 = 26.5+5 =(

.25 ≡ 1.25 ≡ 32 ≡ 4 (mod 7).

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2. Solvex103 ≡ 4 (mod 11)

x is certainly 6≡ 0 (mod 11)

So,x10 ≡ 1 (mod 11)

(by Fermat’s Little Theorem)

x100 ≡ 1 (mod 11)

x103 ≡ x3 (mod 11)

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We need only solvex3 ≡ 4 (mod 11)

which can be done by trying out x = 1,2, ...,10.

Fermat’s Little Theorem has allowed us to reduce the congruence’spower from 103 to 3.

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x (mod 11) 0 1 2 3 4 5 6 7 8 9 10x3 (mod 11) 0 1 8 5 9 4 7 2 6 3 10

The solution isx ≡ 5 (mod 11).

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Proof of Fermat’s Little Theorem

Lemma: If p is prime and a 6≡ 0 (mod p), then the list

a,2a,3a, ..., (p − 1)a (mod p)

is the same as, possibly in different order,

1,2,3, ...,p − 1 (mod p)

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Proof of the lemma

a,2a, ..., (p − 1)a contains (p-1) numbers, none of which is divisible byp.

Suppose ja and ka are congruent:

ja ≡ ka (mod p)

Then, on one hand, p | (j − k)a and hence p | (j − k).

On the other hand, |j − k | < p − 1. Therefore, j − k = 0 and weconclude that for distinct j and k , ja and ka are distinct (mod p).

But we know there are only p − 1 distinct, nonzero values modulo p,namely 1,2, ...,p − 1

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Proof of Fermat’s Little Theorem (continued)

a.(2a)....(p − 1)a ≡ 1.2.3....(p − 1) (mod p)

ap−1(p − 1)! ≡ (p − 1)! (mod p)

ap−1(p − 1)!− (p − 1)! = kp

So,(ap−1 − 1)(p − 1)! = kp

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Since p - (p − 1)!, it follows that

(ap−1 − 1) = lp

and thus:

ap−1 ≡ 1 (mod p).

Assignments

1. Use Fermat’s Little Theorem to:

(a) Find a number 0 < a < 73 with a ≡ 9794 (mod 73).

(b) Solve x86 ≡ 6 (mod 29).

(c) Solve x39 ≡ 3 (mod 13)

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Since p - (p − 1)!, it follows that

(ap−1 − 1) = lp

and thus:

ap−1 ≡ 1 (mod p).

Assignments

1. Use Fermat’s Little Theorem to:

(a) Find a number 0 < a < 73 with a ≡ 9794 (mod 73).

(b) Solve x86 ≡ 6 (mod 29).

(c) Solve x39 ≡ 3 (mod 13)

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Mas4203 Class Notes1

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