Chapter 2 Lecture Notes_ Divisibility.pdf

SMA3043 ELEMENTARY NUMBER THEORY
SEMESTER 2 2023/2024
DIVISIBILITY THEORY

2
THE DIVISION ALGORITHM
• The division algorithm is one of the first concepts relative to the operation of division.
• It is not actually an algorithm, but this is this theorem's traditional name.
Theorem 1 (Division Algorithm)
For any integers a & b, where bZ+, there exist unique integers q, r such that
a = bq + r 0  r  b .
Where.
a – dividend ; b – divisor
q – quotient ; r – remainder (in the division of a by b)

3
THE DIVISION ALGORITHM (Cont.)
Corollary 1
Let a, b  with b  0, then exist unique integers q and r such that
a = bq + r 0  r  | b | .
Example.
If we divide 26 by 3, then we get a quotient of 8 and remainder of 2.
This can be expressed 26 = 3.8 + 2.
It is a little trickier to see what
q and r should be if a < 0.
Eg:if we divide – 26 is by 3, then the remainder
is not – 2.
however, the equation 26 = 3.8 + 2 can be used to our advantage:
– 26 = 3 (– 8) – 2 = [3 . (– 8) – 3] – 2 + 3 = 3(– 9) + 1
Therefore, q = –9 ; r = 1

4
Proof (Corollary 1):
(To show for a, b , b 0, a = bq + r , 0  r  | b | )
Consider b is negative.
Then,
| b | > 0 .
By Theorem 1,
 unique integers q’ & r such that,
a = | b |q’ + r 0  r  | b |
Note that,
| b | = - b , take q = - q’ in order to get a = bq + r , 0  r  | b | . 

5
Example
Let b < 0, where b = -6 Note: 0  r < |-6| = 6
i. If a = 1 ,
then the division algorithm can be expressed as 1 = 0(-6) + 1 .
ii. If a = -2 ,
then the division algorithm can be expressed as – 2 = 1(-6) + 4 .
iii. If a = 61
then the division algorithm can be expressed as 61= (– 10)(-6) + 1 .
iii. If a = – 59
then the division algorithm can be expressed as – 59 = (10)(-6) + 1 .

6
Application of Division Algorithm
1. The square of an integer leaves remainder 0 or 1 upon division by 4 .
If b = 2,
Then the possible remainder are : r = 0 and r = 1. (by D.A)
When r = 0 ; a = 2q - even
When r = 1 ; a = 2q + 1 - odd
Note: ( DA ) a = bq + r
0  r  2
If a2 ;
a2 = (2q)2 = 4q2 = 4k for some kZ
Or;
a2 = (2q + 1)2 = 4q2 + 4q + 1 = 4(q2 + q) + 1 = 4m + 1 for some mZ

7
Example.
72 = 49 = 4(12) + 1
52 = 25 = 4(6) + 1
62 = 36 = 4(9) + 0
2. The square of any an odd integer is of form (8k + 1).
Note that if b = 4,
Then the possible remainder are : r = 0, 1, 2, 3 (by D.A)
(cont.  )
Example.
Write the square of 12, 15 and 21 in
terms of a = 4q + r where 0  r  2

8
DIVISION ALGORITHM (Cont.)
In the division algorithm, any integers can be represented as one of four form:
• 4q
• 4q + 1
• 4q + 2
• 4q + 3
Odd
Odd
When we square the odd integers;
If (4q + 1)2 ;
(4q + 1)2 = 16q2 + 8q + 1 = 8(2q2 + q) + 1= 8k + 1 for some k= (2q2 + q)  ℤ
Or;
(4q + 3)2 = 16q2 + 24q + 9 = 8(2q2 + 3q + 1) + 1= 8m + 1 for some m = (2q2 + 3q + 1)  ℤ

9
Example.
Write the square of 15, 19 and 21 in terms of (8k + 1).

10
GREATEST INTEGER FUNCTION
Definition 1.
The greatest integer function for a real number x, denote by
the largest integer less than or equal to x, satisfying
1
x x x
  
 
 
x
 
 
Notes:
Note:
The greatest integer
function is very useful in
applications involving
data storage
and data transmission.
• if and only if x is an integer.
• If x is any real numbers, then x can be written as
x x

 
 
where 0 1.
x x t t
   
 
 

11
GREATEST INTEGER FUNCTION (Cont.)
Example.
13
i. 6.5 6
2
 
 
 
 
 
 
13
ii. 6.5 7
2

 
   
 
 
 
 
iii. 2.71828 2
e  
   
   
iv. 11 11
  
 
 
v. 5 5

 
 

12
Proposition 1
For x a real number and n integer:
i. x n x n
  
   
   
iii. if 1
x
x
n
n n
 
 
   
 
 
 
   
ii. −x = ቊ
− x if x ∈ ℤ
− x − 1 if x ∉ ℤ

13
Facts:
Greatest integer function can be used to give explicit formulas for the
quotient and remainder in the division algorithm:
a a
q r a b
b b
   
  
   
   
Example.
Given a = 22, b = 3, by using the greatest integer function give explicit formulas for
the division algorithm.

14
Solution.
Given a = 22 , b = 3.
22
7.33 7
3
q
 
  
 
 
 
 
 
22
22 3 22 3 7.33 22 3 7 1
3
r
 
      
 
 
 
 
Then, the division algorithm:
22 = 3(7) + 1
Exercise: Given a = 46, b = 3, by using the greatest integer function give explicit formulas for the division
algorithm.

15
DIVISORS AND MULTIPLES
Definition 2
Given two integers a and b we say a divides b if there is an integer c
such that b = ac. If a divides b, denoted by a | b . If a does not divide
b, denoted by That is
a | b  b = ac , cZ
We say, a is a divisor of b; a is a factor of b; or b is a multiple of a.
Divisibility

16
DIVISORS AND MULTIPLES
Example
i. Show that 4872 is divisible by 2.
ii. Determine whether 270 is divisible 4.
Divisibility
Solution:
i. Note that, 4872 is divisible by 2, since 4872 = 2 . 2436.
By definition 2,
a = 2; b = 4872 & c = 2436.
Therefore, 2 | 4872.
ii. Since there no xZ such that 270 = x . 4 ,
Then, by definition 2,
4 does not divide 270 @ 4 | 270

18
Divisibility
Example.
Proof.
By Definition 2 (Divisibility)
By Definition 2 (Divisibility)

19
Divisibility
Example (Mathematical Induction Involving divisibility)
Using mathematical induction, show that:
8 | (52n + 7) for n  1 .
Solution:
Let S be a set of positive integers.
To show :
1. 1S
2. Assume kS , to show (k+1) S

20
Divisibility
1. To show 1S
Let kS.
8 | (52(1) + 7)
= 8 | 32 Since 32 = 8(4) (from def. of divisible)
Then, 1S .
2. To show (k+1) S , if kS
For n = 1,
Then, 8 | (52k + 7)
[To show 8 | (52(k+1) + 7) ]

22
Divisibility
Exercise (divisibility)
i. Using mathematical induction, show that:
ii. Using mathematical induction, verify that for n any positive integer, 6n −1 is divisible by 5 .
iii. Prove that if a | b and a | c then a | (bx + cy) for some x, y Z

23
Common Divisors
Definition 3 (Common Divisor)
a is a common divisor of b and c if a | b and a | c for a, b, cZ
Notes:
• 1 is a common divisor of every integer.
• Every integer is a common divisor of b = c = 0.

24
Common Divisors
Example.
Determine the common divisors of 27 and 15. Justify your
answer.
Solution:
The possible divisors of 27: 1 , 3, 9, 27
The possible divisors of 15: 1, 3, 5, 15
Since 1 | 15 and 1 | 27 & 3 | 15 and 3 | 27
Then by definition 3 (definition of common divisor),
1 & 3 are common divisors of 27 and 15.

25
Common Divisors
Exercise.
Determine the common divisors of the following integers. Justify
your answer.
i. 45 and 27.
ii. 55 and 35
iii. 36 and 12
iv. 42 and 14
Answer:
i. 1, 3, 9
ii. 1, 5
iii. 1, 2, 4, 12
iv. 1, 2, 7, 14

27
Example.
Determine the greatest common divisor of
the following:
i. – 12 and 30
ii. – 8 and – 36
Solution.
i. The positive divisors of – 12 are: 1, 2, 3, 4, 6, 12 and
the positive divisors of 30 are: 1, 2, 3, 5, 6, 10, 15, 30.
So, gcd (– 12, 30) = 6.

28
Solution.
ii. The positive divisors of – 8 are 1, 2, 4, 8 and
the positive divisors of – 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36.
So, gcd (– 8, – 36) = 4.
Exercise.
Determine the greatest common divisor of the following:
i. 45 and 60
ii. – 9 and 15
iii. 36 and 54
Answer:
i. Gcd(45, 60) = 15
ii. Gcd(– 9, 15) = 3
iii. Gcd(36, 54) = 18

29
Theorem 3.
Let a, b  Z, where a  0, b  0. Then there exist integers x and
y such that:
gcd (a, b) = ax + by .
Example.
i. gcd (– 12, 30) = 6 = (– 12) 2 + (30) 1
ii. gcd (–8, –36) = 4 = (–8) 4 + (–36) (–1) .

30
Relatively Prime
Definition 5. (Relatively Prime)
Two integers a and b, not both zero, are said to be relatively prime
if gcd (a, b) = 1.
Example.
i. Since gcd(11, 36) = 1, then 11 and 36 are relatively prime.
ii. Since gcd(23, 52) = 1, then 23 and 52 are relatively prime.

31
Relatively Prime
The following theorem characterizes relatively prime integers in terms of
linear combinations.
Theorem 4.
Let a, b  ℤ, not both zero. Then a and b are relatively prime if and only if
there exist integers x and y such that 1 = ax + by.
Proof. (next )

32
Relatively Prime
Proof.
() Let a and b are relatively prime. (To show 1 = ax + by)
Then, gcd(a,b) = 1.
By Theorem 3, exist integers x, y such that
1 = ax + by .
() Let 1 = ax + by for x, yZ and d = gcd(a,b) (To show d = gcd(a,b) = 1)
Then, by definition 4, d | a and d | b.
Hence Theorem 2 yields d | (ax + by) or d | 1.
Since d is a positive integer, by Theorem 2, d = 1.
Then,
gcd(a,b) = 1 @ a and b are relatively prime. 

33
Relatively Prime
Corollary 2.
If gcd (a, b) = d, then gcd , 1.
a b
d d
 

 
 
a/d and b/d are
said to be relatively
prime.
Example.
Let a = – 8 , b = – 36
We have gcd(– 8 , – 36) = 4
Then,
 
8 36
gcd , gcd 2, 9 1
4 4
 
 
   
 
 
Example.
Let a = 4 , b = 6
We have gcd(4 , 6) = 2
Then,
 
4 6
gcd , gcd 2,3 1
2 2
 
 
 
 

35
Relatively Prime
Proof. (Corollary 3)
Let a | c and b | c, with gcd (a, b) = 1 (To show ab | c ).
Then, by definition 2 (divisible),
exist r, s Z such that
c = ar = bs ----------------- (1)
Since gcd (a,b) = 1, then, by Theorem 3;
1 = ax + by for some x, y in Z ------- (2)
Multiplying (2) by c;
c = c. 1 = c(ax + by) = acx + bcy ------- (3)
By substituting (1) into (3);
c = a(bs)x + b(ar)y
= ab(sx + ry) where (sx + ry)Z
Then, by definition 2,
ab | c.

36
Note.
i. Given a, b  Z, not both zero. Then,
gcd (a, b) = gcd (– a, b) = gcd (a, – b) = gcd (–a, –b).
ii. Let a, b, c  Z, no two of which are zero, and d = gcd (a, b, c) .
Then:
d = gcd (gcd (a, b) , c)
= gcd (a, gcd (b, c))
= gcd (gcd (a, c) , b)
iii. For any integer k  0, gcd (ka, kb) = | k | gcd (a, b)

37
Euclid’s Lemma
Theorem 5. (Euclid’s Lemma)
If a | bc, with gcd (a, b) = 1, then a | c.
Proof.
Let a | bc, with gcd (a, b) = 1.
Then,
gcd (a, b) = 1 = ax + by (by Theorem 3)
By multiplying with c,
c = cax + cby ---------- (1)
Since a | bc,
we have, bc = ak. ------ (2)
So, (Cont. )

38
Euclid’s Lemma
Proof. (cont.)
By substituting (2) into (1);
c = cax + cby
= cax + (ak)y
= a (cx + ky )
Therefore, a | c . 

39
Euclid’s Lemma
Example.
Let a = 3, b = 4 & c = 3.
3 | 4(3) , with gcd(3, 4) = 1
Then, by Euclid’s Lemma, clearly 3 | 3.
Note.
If a & b are not relatively prime, the conclusion of Euclid’s Lemma
may fail to hold.
Eg. 12 | 9(8) where gcd(12,9)  1 and gcd(12,8)  1
we have 12 | 9 and 12 | 8

41
Euclid’s Lemma
Proof (Theorem 6).
() Let d = gcd(a, b).
Then, d | a and d | b (by definition 4 –def GCD)
Hence, condition (i) holds.
Next to show condition (ii).
By Theorem 3, d = ax + by for some x, yZ .
Let c | a and c | b.
Then,
c | ax + by or c | d. ( (ii) holds)
( )
Let d be any positive integer which satisfying condition (i)
and (ii).
Given any common divisor c of a and b,

42
Euclid’s Lemma
Proof (Theorem 6).
( )
Let d be any positive integer which satisfying condition (i)
and (ii).
Given any common divisor c of a and b,
we have c | d (from the hypothesis).
This implies c  d,
Consequently d is the greatest common divisor of a and
b. 

43
Common Multiple
Definition 6 (Common Multiple)
Let a, b be two nonzero integers. An integer m is said to be a
common multiple of a and b whenever a | m and b | m.
Note:
• Zero is a common multiple of a and b.
• (ab) and – (ab) are both common multiples
of a and b.

44
Common Multiple
Example
Multiple of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ...
Multiple of 3: 3, 6, 9, 12, 15, 18, 21, 24
We can see that 6 is a common multiple of 2 & 3.
Then,
2 | 6 and 3 | 6.
12 is a common multiple of 2 & 3.
Then,
2 | 12 and 3 | 12.

45
Least Common Multiple
Definition 7 (Least Common Multiple)
Let a and b be two nonzero integers. The least common
multiple of a and b, denoted by lcm (a, b) is the positive integer
m satisfying the following:
i. a | m and b | m
ii. If a | c and b | c with c > 0, then m  c .
Example.
The positive common multiples of –12 and 30 are:
60, 120, 180, ......
Hence,
lcm(-12, 30) = 60.

46
Example.
The positive common multiples of –8 and –36 are:
72, 216, 288, ......
Hence,
lcm(-8, -36) = 72.

47
Note.
i. Given nonzero integers a and b, lcm (a, b) always exists and
lcm (a, b)  | ab |
ii. Let a, b  Z+. Then,
gcd (a, b) lcm (a, b) = ab .
iii. lcm (a, b) = ab if and only if gcd (a, b) = 1.

48
Example (ii).
Given two positive integers 3054 and 12378 where gcd(3054, 12378) = 6.
Determine the lcm(3054, 12378).
Solution.
By (ii), gcd (a, b) . lcm (a, b) = ab
Then,
 
   
3054 12378
lcm 3054,12378 6,300,402.
6

 

49
THE EUCLIDEAN ALGORITHM
• The Euclidean algorithm is one of an efficient methods for
computing the greatest common divisor (GCD).
 Involving repeated application of Division Algorithm.
• Let a and b be two nonzero integers.
Because gcd (| a | , | b |) = gcd (a, b) , we can assume that a  b.
• By applying the Division Algorithm:
Cont. 

50
• If r1 = 0, then we stop. gcd (a, b) = b. If r1  0, we continue the
next step.
• If r2 = 0, then we stop. gcd (a, b) = r1.
• Continue the process until we obtain the remainder zero.
• Then, the last nonzero remainder, rn is the greatest common divisor
of a and b,
i.e., gcd (a, b) = rn (the last nonzero remainder), using the fact that:
gcd (a, b) = gcd (b, r1) = ... = gcd (rn-1, rn) = gcd (rn, 0)

51
Example.
Determine the GCD of 68 and 14 and write your GCD = d in
terms of (ax + by).
Solution.
gcd (68, 14) :
68 = 14 (4) + 12
14 = 12 (1) + 2
12 = 2 (6) + 0
Then,
gcd (68, 14) = 2.
Cont. 

52
Solution. (Cont.)
Next to write d = (ax + by).
To find integers x and y that satisfy 68x + 14y = 2
2 = 14 – 12 (1)
= 14 – [68 – 14 (4)] 1
= – 68 + 14(5)
So, gcd (68, 14) = 2 = 68(–1) + 14(5), where
x = –1, y = 5.

53
Example.
Determine the GCD of 42823 and 6409 and write your GCD
= d in terms of (ax + by).
Solution.
gcd (42823, 6409) :
42823 = 6409 (6) + 4369
6409 = 4369 (1) + 2040
4369 = 2040 (2) + 289
2040 = 289 (7) + 17
289 = 17 (17) + 0
Then,
gcd (42823, 6409) = 17.
Cont. 

54
Solution. (Cont.)
Next to write d = (ax + by).
To find integers x and y that satisfy 42823x + 6409y = 17
17 = 2040 – 289 (7)
= 2040 – [4369 – 2040 (2)] 7
= 2040 (15) – 4369 (7)
= [6409 – 4369] 15 – 4369 (7)
= 6409 (15) – 4369 (22)
= 6409 (15) – [42823 – 6409 (6)] 22
= 6409 (147) – 42823 (22)
So, gcd (42823, 6409) = 17 = 42823 (–22) + 6409 (147) , where
x = –22, y = 147.

55
Exercise.
Using the Euclidean algorithm, calculate the greatest
common divisor of the following numbers and find integers
x, y such that d = (ax + by).
i. 2689 and 4001
ii. 864 and 468
iii. 1819 and 3587
Answer:
i. gcd(2689,4001) = 1.
1 = = – 1117 . 4001 + 1662 . 2689
x = -1117 and y = 1662
ii. gcd(864, 468) = 36.
36 = = 6 . 864 – 11 . 468.
x = 6 and y = -11
iii. gcd(3587, 1819) = 17.
17 = = 71 . 1819 – 36 . 3587
x = 71 and y = - 36

56
LINEAR DIOPHANTINE EQUATION
• Diophantine equations are named after the Greek mathematician Diophantus, c. 250, of Alexandria.
• The great work of Diophantus rests in his Arithmetica, which may be described as the earliest treatise in
Algebra.
• The term Diophantine equation usually refers to any equation in one or more unknowns that is to be
solved in the integers.
• The simplest type of Diophantine equation will be considered:
For a, b, c, d  Z, a, b, c are not both zero. and for which the variables x and y can only have integer
values.
the linear Diophantine equation in two unknowns: ax + by = c;
the linear Diophantine equation in three unknowns: ax + by + cz = d
• A solution of this equation is a pair of integers: x = x0, y = y0 that satisfy the equation.
• A linear Diophantine equation can have either a unique solution, many solutions or no solution.

57
Concerning a Diophantine equation three basic problems arise:
Question 1. Is the equation has solution?
Question 2. If it has solution, is the number of its solutions finite or infinite?
Question 3. If it has solution, determine all of its solutions.
Example.
i. 155x + 45y = 7 has no solution.
ii. 60x + 33y = 9 has infinitely many solutions.
Why?

58
Theorem 7
The linear Diophantine equation ax + by = c has a solution
if and only if
d | c, where d = gcd (a, b) . Furthermore, if x = x0, y = y0 is any particular
solution of the equation, then all other solutions are given by:
0 0
;
b a
x x t y y t
d d
   
   
   
   
The following theorem gives us the solutions for the linear Diophantine equations.
where t is arbitrary integer.

59
Proof. (1st part)
Suppose ax + by = c has a solution x = x0 and y = y0.
Let d = gcd (a, b) .
Then,
d | a  a = dr and d | b  b = ds for r, sZ
Next,
So, c = ax0 + by0
= (dr)x0 + (ds)y0
= d (rx0 + sy0)
Then, by definition 2 (divisible),
d | c.
Suppose d | c.
Then,
Then c = dt for tZ (by definition 2) (Cont. )

60
Proof. (1st part - Cont.)
Let gcd (a, b) = d = ax0 + by0 for some x0, y0  Z.
Thus,
c = (ax0 + by0) t
= ax0t + by0 t
= a(x0t) + b(y0 t)
Hence,
the Diophantine equation has a particular solution: x = x0t and y = y0t.
(2nd part.)
Let x = x0 and y = y0 is the particular solution.

61
(2nd part.) (cont.)

62
(2nd part.) (cont.)

63
(2nd part.) (cont.)


64
Example.
Determine all solutions in the positive integers of:
123x + 360y = 99.
Solution.
Using Euclidean Algorithm to find the gcd (123, 360) :
360 = 123 (2) + 114
123 = 114 (1) + 9
114 = 9 (12) + 6
9 = 6 (1) + 3
6 = 3 (2) + 0
Then, gcd (123, 360) = 3.
Therefore by Theorem 7, equation 123x + 360y = 99 has a solution since 3 | 99.

65
Solution. (Cont.)
Using "backtracking" to find x and y :
3 = 9 – 6
= 9 – [114 – 9 (12)]
= 9 (13) – 114
= [123 –114] 13 – 114
= 123 (13) – 114 (14)
= 123 (13) – [360 – 123 (2)] 14
= 123 (41) + 360 (–14)
Then,
99 = (3) (33) = 123 (1353) + 360 (–462)
Hence the particular solution: x0 = 1353, y0 = – 462.

66
Solution. (Cont.)
Using Theorem 7, the other solutions can be found:
360
1353
3
x t
 
   
 
1353 120t
 
123
462
3
y t
 
    
 
462 41t
   for tℤ
Next to find the positive solutions, i.e., for x > 0 and y > 0 :
1353 + 120t > 0
t > -11
462 41 0
t
  
t < -12
Hence, there is no positive solutions.

67
Example.
Determine all solutions in the positive integers of:
172x + 20y = 1000.
Solution.
Using Euclidean Algorithm to find the gcd (172, 20) :
172 = 20 (8) + 12
20 = 12 (1) + 8
12 = 8 (1) + 4
8 = 4 (2) + 0
Then, gcd (172, 20) = 4.
Therefore by Theorem 7, equation 172x + 20y = 1000 has a solution since 4 | 1000.

68
Solution. (Cont.)
Using "backtracking" to find x and y :
4 = 12 – 8
= 12 – [20 – 12]
= 12 (2) – 20
= 2 [172 – 20(8) ] – 20
= 172 (2) + 20 (–17)
Then,
1000 = (4) (250) = 500 (172) + (–4250) (20)
Hence the particular solution: x0 = 500, y0 = – 4250.

69
Solution. (Cont.)
Using Theorem 7, the other solutions can be found:
20
500
4
x t
 
   
 
500 5t
 
172
4250
4
y t
 
    
 
4250 43t
   for tℤ
Next to find the positive solutions, i.e., for x > 0 and y > 0 :
500 + 5t > 0
t > -100
4250 43 0
36
98
43
t
t
  
 
Hence, t = -99. Therefore the Diophantine
equation has a unique solution x = 5, y =7
corresponding to the value t = -99.

70
Exercise.
1. Determine all solutions in the integers of the following Diophantine equations:
i. 56x + 72y = 40.
ii. 24x + 138y = 18
iii. 221x + 35y = 11
iv. 24x + 138y = 40
2. Find all the solutions (x, y) to the following Diophantine equation 11x + 13y = 369 for
which x and y are both positive.
3. Solve the linear Diophantine Equations: 2x+4y=21, x,y ∈ ℤ.
Answer:
1. i. x = 20+9t ; y = -15-7t
ii. 18+23t ; y = -3-4t
iii. 176+35t ; y = -1111-221t
iv. No solution
2. x = 2214 + 13t, y = −1845 − 11t.
The solutions are (x, y) = (4, 25), (17, 14), and
(20, 3).
3. No solution since gcd(2,4)=2 and 2 does not divide 21.

Chapter 2 Lecture Notes_ Divisibility.pdf

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