Number theory

Number theory
• 1.Divisors
• 2.Primer Factorization
• 3.Congruence
• 4.Quadratic Residues
2

1.Divisors
Theorem 1.1. Division Algorithm. Let n and d ≥ 1 be integers. There
exist uniquely determined integers q and r such that n = qd + r and
0 ≤ r < d.
Proof. Let X = {n − td|t ∈ , n − td ≥ 0}. Then X is nonempty (if
n≥0,then n X; if n < 0, then n(1 − d) X). Hence let r be the∈ ∈
smallest member of X . Then r = n − qd for some q ∈ , and it
remains to show that r < d. But if r ≥d, then 0 ≤ r − d = n − (q + 1)d,
so r − d is in X contrary to the minimality of r.
As to uniqueness, suppose that n = q’d + r’, where 0≤ r’< d. We
may assume that r ≤ r’(a similar argument works if r’ ≤ r). Then
0 ≤ r’ − r = (q − q’)d, so (q − q’)d is a nonnegative multiple of d that
is less than d (because r’ − r ≤ r ’< d). The only possibility is
(q − q’)d = 0, so q’ = q, and hence r’ = r.
3

1.Divisors
• Given n and d ≥ 1, the integers q and r in Theorem 1.1 are called,
respectively, the quotient and remainder when n is divided by d.
For example, if we divide n = −29 by d = 7, we find that −29 =
(−5) · 7 + 6, so the quotient is −5 and remainder is 6.
The usual process of long division is a procedure for finding the
quotient and remainder for a given n and d ≥ 1. However, they can
easily be found with a calculator. For example, if n = 3196 and d =
271 then n/d = 11,79 approximately, so q = 11. Then r = n − qd =
215, so 3196 = 11 · 271 + 215, as desired.
If d and n are integers, we say that d divides n, or that d is a
divisor of n, if n = qd for some integer q. We write d|n when this
is the case. Thus, a positive integer p >1is prime if and only if p has
no positive divisors except 1 and p. The following properties of the
divisibility relation | are easily verified:
4

1.Divisors
• Theorem 1.2. Let m and n be integers, not both zero.
Then d = gcd(m, n) exists,and d = xm + yn for some
integers x and y.
Proof. Let X = {sm + tn | s, t ∈ ; sm + tn ≥1}. Then X is
not empty since m2
+ n2
is in X, so let d be the smallest
member of X. Since d X we have d∈ ≥ 1 and
d = xm + yn for integers x and y, proving conditions (i)
and (iii) in the definition of the gcd.
Hence it remains to show that d|m and d|n.We show
that d|n; the other is similar. By the division algorithm
6

1.Divisors
write n = qd + r, where 0 ≤ r < d. Then
r = n − q(xm + yn) = (−qx)m + (1 − qy)n. Hence, if r ≥ 1, then
r X, contrary to the minimality of d. So r = 0 and we have∈
d|n.
When gcd(m, n) = xm + yn where x and y are integers, we say
that gcd(m, n) is a linear combination of m and n. There is an
efficient way of computing x and y using the division
algorithm.
The following example illustrates the method.
7

1.Divisors
• Example . Find gcd(37, 8) and express it as a linear combination of 37
and 8.
Proof. It is clear that gcd(37, 8) = 1 because 37 is a prime; however, no
linear combination is apparent. Dividing 37 by 8, and then dividing each
successive divisor by the preceding remainder, gives the first set of
equations.
37 = 4 · 8 +5 1= 3 − 1 · 2 = 3 − 1(5 − 1 · 3)
8 = 1 · 5 + 3 = 2 · 3 − 5 = 2(8 − 1 · 5) − 5
5 = 1 · 3 + 2 = 2 · 8 − 3 · 5 = 2 · 8 − 3(37 − 4 · 8)
3 = 1 · 2 + 1 = 14 · 8 − 3 · 37
2 = 2 · 1
The last nonzero remainder is 1, the greatest common divisor, and this
turns out always to be the case. Eliminating remainders from the bottom up
(as in the second set of equations) gives 1 = 14 · 8 − 3 · 37.
8

1.Divisors
• Theorem 1.3. Euclidean Algorithm. Given integers m and
n ≥ 1, use the division algorithm repeatedly:
m = q1n + r1 0 ≤ r1 < n
n = q2r1 + r2 0 ≤ r2 < r1
r1 = q3r2 + r3 0 ≤ r3 < r2
...
...
r k-2= qkrk−1 + rk 0 ≤ rk < rk-1
rk−1= qk+1rk
where in each equation the divisor at the preceding stage is
divided by the remainder. These remainders decrease
r1 > r2 > · · · ≥ 0
9

1.Divisors
so the process eventually stops when the remainder
becomes zero. If r1 = 0, then gcd(m, n) = n. Otherwise,
rk = gcd(m, n), where rk is the last nonzero remainder
and can be expressed as a linear combination of m and
n by eliminating remainders.
Proof. Express rk as a linear combination of m and n by
eliminating remainders in the equations from the
second last equation up. Hence every common
divisor of m and n divides rk. But rk is itself a common
divisor of m and n (it divides every ri—work up through
the equations). Hence rk = gcd(m, n).
10

1.Divisors
Two integers m and n are called relatively prime if gcd(m, n) = 1.
Hence 12 and 35 are relatively prime, but this is not true for 12 and 15
Because gcd(12, 15) = 3. Note that 1 is relatively prime to every
integer m. The following theorem collects three basic properties of
relatively prime integers.
Theorem 1.4. If m and n are integers, not both zero:
(i) m and n are relatively prime if and only if 1 = xm + yn for some
integers x and y.
(ii) If d = gcd(m, n), then m/d and n/d are relatively prime.
(iii) Suppose that m and n are relatively prime.
(a) If m|k and n|k, where k ∈ , then mn|k.
(b) If m|kn for some k ∈ , then m|k
11

1.Divisors
• Proof. (i) If 1 = xm + yn with x, y ∈ , then every
divisor of both m and n divides 1, so must be 1 or −1. It
follows that gcd(m, n) = 1. The converse is by the
euclidean algorithm.
(ii). By Theorem 1.2, write d = xm + yn,
where x, y ∈ . Then
1 = x(m/d)+y(n/d) and (ii) follows from (i).
(iii). Write 1 = xm + yn, where x, y ∈ . If k = am and k
= bn, a, b ∈ then k = kxm + kyn = (xb + ya)mn, and
(a) follows. As to (b), suppose that
kn = qm, q ∈ . Then k = kxm + kyn = (kx + qn)m, so
m|k.
12

2.Prime Factorization
Recall that an integer p is called a prime if:
• (i) p ≥ 2.
• (ii) The only positive divisors of p are 1 and p.
The reason for not regarding 1 as a prime is that
we want the factorization of every integer into
primes to be unique. The following result is needed.
13

• Theorem 2.2. Every integer n >1 is a product
of primes.
• Proof. Let pn denote the statement of the theorem. Then p2
is clearly true.
If p2, p3, . . . , pk are all true, consider the integer k + 1. If
k + 1 is a prime, there is nothing to prove. Otherwise,
k + 1 = ab, where 2 ≤ a, b ≤ k. But then each of a and b are
products of primes because pa and pb are both true by the
(strong) induction assumption. Hence ab = k + 1 is also a
product of primes, as required.
15

• Theorem 2.3. Prime Factorization Theorem. Every
integer n ≥ 2 can be written as a product of (one or
more) primes. Moreover, this factorization is unique
except for the order of the factors. That is,
if n = p1p2 · · · pr and n = q1q2 · · · qs ,
where the pi and qj are primes, then r = s and the qj
can be relabeled so that pi = qi for each i.
16

• Proof. The existence of such a factorization was
shown in Theorem 2.2. To prove uniqueness, we
induction the minimum of r and s. If this is 1, then n
is a prime and the uniqueness follows from Euclid’s
lemma. Otherwise, r ≥ 2 and s ≥ 2. Since
p1|n = q1q2 · · · qs Euclid’s lemma shows that p1 divides
some qj , say p1|q1 (after possible relabeling of the qj ).
But then p1 = q1 because q1 is a prime. Hence n/p1 =
p2p3· · · pr = q2q3 · · · qs , so, by induction,
r − 1 = s − 1 and q2, q3, . . . , qs can be relabeled such
that pi = qi for all i = 2, 3, . . . , r. The theorem follows.
17

• It follows that every integer n ≥ 2 can be written in
the form n = p1
n
1 p2
n
2 · · · pr
n
r ,where p1, p2, . . . , pr are
distinct primes, ni ≥ 1 for each i, and the pi and ni are
determined uniquely by n. If every ni = 1, we say that
n is square-free, while if n has only one prime
divisor, we call n a prime power.If the prime
factorization
n = p1
n
1 p2
n
2 · · · pr
n
r of an integer n is given, and if d
is a positive divisor of n, then these pi are the only
possible prime divisors of d (by Euclid’s lemma). It
follows that
18

Prime Factorization
19
Collorary 2.4

Prime Factorization
20
Theorem 2.5

3.Congruences
• Definition 3.1.. If m ≥ 0 is fixed, then integers a and b
are congruent modulo m,denoted by a ≡ b (mod m)
if m | (a – b ). Usually, one assumes that the modulus
m >1 because the cases m = 0 and m = 1 are not very
interesting: if a and b are integers, then a ≡ b (mod 0) if
and only if 0 | (a –b), that is, a = b, and so congruence
mod 0 is ordinary equality.
The congruence a ≡ b (mod 1) is true for every pair of
integers a and b because 1 | (a – b) always. Hence,
every two integers are congruent mod 1.
22

3.Congruences
• If a and b are positive integers, then a ≡ b (mod 10) if
and only if they have the same last digit; more
generally, a ≡ b (mod 10n
)if and only if they have same
last n digits. For example, 526 ≡ 1926 (mod 100).
• London time is 6 hours later than Chicago time. What
time is it in London if it is 10:00 A.M. in Chicago?
Since clocks are set up with 12 hour cycles, this is
really a problem about congruence mod 12. To solve it,
note that 10 + 6 = 16 ≡ 4(mod 12); and so it is 4:00 P.M.
in London.
23

3.Congruences
• Proposition 3.1. If m > 0 is a fixed integer, then for all
integers a, b, c,
(i) a ≡ a (mod m);
(ii) if a ≡ b (mod m), then b ≡ a (mod m);
(iii) if a ≡ b (mod m) and b ≡ c (mod m), then a ≡ c (mod
m).
• Proposition 3.2. Let m > 0 be a fixed integer.
(i) If a = qm + r , then a ≡ r (mod m).
(ii) If 0 ≤ r ’< r < m, then r and r ’are not congruent mod
m; in symbols, r r ’ (mod m).
(iii) a ≡ b (mod m) if and only if a and b leave the same
remainder after dividing by m.
24

3.Congruences
• Proposition 3.3. Let m> 0 be a fixed integer.
(i) If ai ≡ a’i (mod m) for i = 1; 2; … ; n, then
a1 +... + an ≡ a’1 +...+ a’n (mod m):
In particular, if a ≡ a’ (mod m) and b ≡ b’ (mod m),
then
a + b ≡ a’ + b’ (mod m):
(ii) If ai ≡ a’i (mod m) for i = 1; 2; … ; n, then
a1 ... an ≡ a’1 ... a'n (mod m)
In particular, if a ≡ a’ (mod m) and b ≡ b’ (mod m),
then ab ≡ a’b’ (mod m)
(iii) If a ≡ b (mod m), then an
≡ bn
(mod m) for all n >0. 25

3.Congruences
26
Theorem 3.4 (Fermat).

3.Congruences
• Theorem 3.5. If (a;m)= 1, then, for every integer b, the
congruence ax ≡ b (mod m) can be solved for x; in fact,
x = sb, where sa ≡ 1 (mod m). Moreover, any two solutions
are congruent mod m.
Proof. Since (a;m)= 1, there is an integer s with
as ≡ 1( mod m) (because there is a linear combination
1 = sa + tm). It follows that b = sab + tmb and
asb ≡ b (mod m), so that x = sb is a solution. (Note that
Proposition 3.2(i) allows us to take s with 1≤ s < m.)
If y is another solution, then ax ≡ ay mod m, and so
m | a(x - y). Since (a;m)= 1, Theorem 1.4 gives m |(x – y);
that is, x ≡ y (mod m).
27

4.Quadratic Residues
• Definition 4.1. If m is a positive integer ,we say that the
integer a is a quadratic residue of m if (a,m) = 1 and the
congruence x2
≡ a (mod m) has a solution.
If the congruence x2
≡ a (mod m) has a no solution, we say a
is quadratic nonresidue of m.
Example. To detemine which integer are quadratic residues of
11, we compute the squares of the integer 1, 2, 3, …, 10.We
find that 12
≡ 102
≡ 1(mod 11), 22
≡ 92
≡ 4 (mod 11), 32
≡ 82
≡ 9
(mod 11), 42
≡ 72
≡ 5 (mod 11), and 52
≡ 62
≡ 3(mod11).
Hence , the quadratic residues of 11 are 1, 3, 4, 5, 9 ;the integer
2,6,7,8,10 are quadratic nonresidues of 11.
28

• Lemma 4.1. Let p be odd prime and a an integer not divisible
by p. Then the congruence x2
≡ a (mod p) has either no
solutions or exactly two incongruent solutions modulo p.
• Proof. If x2
≡ a (mod p ) has a solution, say x = x0, then we can
easily demonstrate that x = - x0 is second incongruent solution.
Since (-x0)2
= x0
2
≡ a (mod p ) we see that – x0 is solution. We
note that x0 –x0 (mod p), for if x0 ≡ - x0 (mod p), then we have
2x0 ≡ 0 (mod p). This is impossible since p is odd and p x0
(since x0
2
≡ a (mod p ) and p a ).
To show that there are no more than two incogruent solutions,
assume that x ≡ x0 and x ≡ x1 are both solutions of x2
≡ a (mod
p). Then we have x0
2
≡ x1
2
≡ a (mod p) , so that
29
≡
|/
|/

x0
2
- x1
2
= (x0 + x1)(x0- x1) ≡ 0 (mod p).
Hence , p| (x0 +x1) or p | (x0- x1), so that x1 ≡ - x0 (mod p) or
x0 ≡ x1 (mod p). Therefore if there is a solution of x2
≡ a (mod
p), there are exactly two incongruent solution.
Theorem 4.2. If p is an odd prime , then there are exactly
(p-1 )/2 quadratic residues of p and ( p – 1 )/2 quadratic
nonresidues of p among the integer 1, 2, …, p – 1 .
Proof. To find all the quadratic residues of p among the
integers 1, 2, …, p – 1 we compute the least positive residues
modulo p of the squares of the integers 1, 2, p – 1 .
30

• Since there are p – 1 squares to consider and since each
congruence x2
≡ a (mod p) has either zero or two solotions ,
there must be exactly ( p – 1 )/2 quadratic residues of p among
the integer 1, 2, …, p – 1 . The remaining p – 1 – ( p – 1 )/2
= ( p – 1 )/2 positive integers less than p – 1 are quadratic
nonresidues of p .
The special notation associaed with quadratic residues is
described in the following definition.
31
W

• Definition 4.2. Let p be an odd prime and a an integer not
divisible by p . The Legendre symbol is defined by
• The symbol iz named after the French mathematician
Andrien – Marie Legendre who introduced the use of this
notation
32
1
1
if a iz quadraticresidueof pa
if aiz aquadratic nonresidueof pp
  
=  ÷
−  

• Example . The previous example shows that the Legendre
symbol
have the followings values :
33
, 1,2,...,10
11
a
a
 
= ÷
 
1 3 4 5 9
1
11 11 11 11 11
2 6 7 8 10
1
11 11 11 11 11
         
= = = = = ÷  ÷  ÷  ÷  ÷
         
         
= = = = = − ÷  ÷  ÷  ÷  ÷
         

We now present a criterion for deciding whether an integer is
a quadratic residue of prime. This criterion is useful in
demonstraing propeties of the Legendre symbol.
Theorem 4.3. Euler’s Criterion.
Let p be an odd prime and let a be positive integer not
divisible by p. Then
34
( 1)/ 2
(mod ).pa
a p
p
− 
≡ ÷
 

Proof.
Firt ,assume that .Then , the congruence x2
≡ a (mod p)
has a solution, say x = x0. Using Fermat’s little theorem , we see
that
Hence,if ,we know that
35
1
a
p
 
= ÷
 
( 1)/2 2 ( 1)/2 1
0 0( ) 1(mod )p p p
a x x p− − −
= = ≡
1
a
p
 
= ÷
 
( 1)/ 2
(mod )pa
a p
p
− 
≡ ÷
 

• Now cosider the case where .Then , the congruence
x2
≡ a (mod p) has no solutions. For each integer i such that
1≤i≤p-1, there is a unique integer j with 1≤j≤p-1, such that
ij ≡ a (mod p ). Furthermore , since the congruence x2
≡ a
(mod p) has no solutions, we know that i≠j. Thus, we can
group the integer 1, 2,…, p - 1 into (p – 1 )/2 pairs each with
product a . Multiplying these pairs together, we find that
(p – 1 )! ≡ a (p–1)/2
(mod p).Since Wilson’s theorem tell us that
(p – 1 )!≡ - 1 (mod p), we see that – 1 ≡ a (p-1 )/2
(mod p) .In this
case, we also have
36
1
a
p
 
= − ÷
 
( 1)/ 2
(mod ).pa
a p
p
− 
≡ ÷
 

Theorem 4.4. Let p be an odd prime and a and b integers not
divisible by p.Then
37
2
) (mod ), .
) .
) 1.
a b
i if a b p then
p p
a b ab
ii
p p p
a
iii
p
   
≡ = ÷  ÷
   
    
= ÷ ÷  ÷
    
 
= ÷
 

Proof.
i) If a ≡ b (mod p ) then x2
≡ a (mod p) has s solution if an
only if x2
≡ b (mod p) has solution.Hence .
ii) By Euler’s criterion we know that
38
a b
p p
   
= ÷  ÷
   
( 1)/ 2 ( 1)/ 2
( 1)/2
(mod ) , (mod )
( ) (mod )
p p
p
a b
a p b p
p p
ab
ab p
p
− −
−
   
≡ ≡ ÷  ÷
   
 
≡ ÷
 

Hence ,
Since the only posible values of a Lagendre symbol are ±1, we
conclude that
39
( 1)/ 2 ( 1)/2 ( 1)/2
( ) (mod )p p pa b ab
a b ab p
p p p
− − −    
≡ = ≡ ÷ ÷  ÷
    
a b ab
p p p
    
= ÷ ÷  ÷
    

• iii) Since , from part (ii) it folows that
40
1
a
p
 
= ± ÷
 
2
1
a a a
p p p
    
= = ÷  ÷ ÷
   

How the Algorithm Works
Start with two integers for which you want to
find the GCD. Apply the division algorithm,
dividing the smaller number into the larger.
Example: a = 320, b = 296.
320 = 296 · 1 + 24
The first quotient is q0 and the first remainder
is r0.

How the Algorithm Works (cont.)
If you get a remainder of 0, stop. If not, the
divisor from the previous step becomes the
dividend of the next step. The remainder
from the previous step becomes the divisor of
the previous step.
320 = 296 · 1 + 24
296 = 24 · 12 + 8
Continue until you get a remainder of 0.

The Completed Algorithm
• 320 = 296 · 1 + 24
• 296 = 24 · 12 + 8
• 24 = 8 · 3 + 0
• We get a remainder of 0, so we stop. The last
nonzero remainder is the GCD, so (320, 296) is
equal to 8.

Another Example
Compute (346, 592).
592 = 346 · 1 + 246
346 = 246 · 1 + 100
246 = 100 · 2 + 46
100 = 46 · 2 + 8
46 = 8 · 5 + 6
8 = 6 · 1 + 2
6 = 2 · 3 + 0
So (346, 592) = 2.

The Euclidean Algorithm and Bézout’s
Theorem
We can use the Euclidean Algorithm to find
the integers U and V from Bézout’s Theorem.
As an example, let’s use the Euclidean
Algorithm to show that (324, 148) = 4.
324 = 148 · 2 + 28
148 = 28 · 5 + 8
28 = 8 · 3 + 4
8 = 4 · 2 + 0

Finding U and V
• We want to find integers U and V such that
4 = 324U + 148V.
• Take all of the equations (except the last one)
and solve for the remainder.
• 28 = 324 – 148 · 2
• 8 = 148 – 28 · 5
• 4 = 28 – 8 · 3

Back-Substitution
• Notice that the last equation expresses 4 as a
linear combination of 28 and 8.
• 4 = 28 · 1 + 8 · (-3)
• This is not what we want, however. So we use
the previous equation (which has been solved
for 8) to substitute.

Back-Substitution (cont.)
• 4 = 28 · 1 + (148 – 28 · 5) · (-3)
• Now we want to rearrange this so that 4 is
expressed as a linear combination of 28 and
148 (still not quite what we want, but getting
closer)
• We get 4 = 28 · 16 + 148 · (-3)

Back-Substitution (cont.)
Now use the previous equation (which has
been solved for 28) to substitute.
We get 4 = (324 – 148 · 2) · 16 + 148 · (-3)
Once again, multiply out and rearrange until
we get 4 expressed as a linear combination of
324 and 148.
4 = 324 · 16 + 148 · (-35)

Another Example
• Use the Euclidean Algorithm to show that
(15, 36) = 3.
• Use back-substitution to find integers U and V
so that 3 = 15U + 36V.

53
Introduction to Number Theory
•Number theory is about integers and their properties.
•We will start with the basic principles of
• divisibility,
• greatest common divisors,
• least common multiples, and
• modular arithmetic
•and look at some relevant algorithms.

54
Division
•If a and b are integers with a ≠ 0, we say that
a divides b if there is an integer c so that b = ac.
•When a divides b we say that a is a factor of b and that b
is a multiple of a.
•The notation a | b means that a divides b.
•We write a X b when a does not divide b
•(see book for correct symbol).

55
Divisibility Theorems
•For integers a, b, and c it is true that
• if a | b and a | c, then a | (b + c)
• Example: 3 | 6 and 3 | 9, so 3 | 15.
• if a | b, then a | bc for all integers c
• Example: 5 | 10, so 5 | 20, 5 | 30, 5 | 40, …
• if a | b and b | c, then a | c
• Example: 4 | 8 and 8 | 24, so 4 | 24.
•

56
Primes
•A positive integer p greater than 1 is called prime if the
only positive factors of p are 1 and p.
•A positive integer that is greater than 1 and is not prime
is called composite.
•The fundamental theorem of arithmetic:
•Every positive integer can be written uniquely as the
product of primes, where the prime factors are written in
order of increasing size.

57
Primes
•Examples:
3·53·5
48 =48 =
17 =17 =
100 =100 =
512 =512 =
515 =515 =
28 =28 =
15 =15 =
2·2·2·2·3 = 22·2·2·2·3 = 244
·3·3
1717
2·2·5·5 = 22·2·5·5 = 222
·5·522
2·2·2·2·2·2·2·2·2 = 22·2·2·2·2·2·2·2·2 = 299
5·1035·103
2·2·72·2·7

58
Primes
•If n is a composite integer, then n has a prime divisor
less than or equal .
•This is easy to see: if n is a composite integer, it must
have two prime divisors p1 and p2 such that p1⋅p2 = n.
•p1 and p2 cannot both be greater than
• , because then p1⋅p2 > n.
n
n

59
The Division Algorithm
•Let a be an integer and d a positive integer.
•Then there are unique integers q and r, with
0 ≤ r < d, such that a = dq + r.
•In the above equation,
• d is called the divisor,
• a is called the dividend,
• q is called the quotient, and
• r is called the remainder.

60
•Example:
•When we divide 17 by 5, we have
•17 = 5⋅3 + 2.
• 17 is the dividend,
• 5 is the divisor,
• 3 is called the quotient, and
• 2 is called the remainder.

61
•Another example:
•What happens when we divide -11 by 3 ?
•Note that the remainder cannot be negative.
•-11 = 3⋅(-4) + 1.
• -11 is the dividend,
• 3 is the divisor,
• -4 is called the quotient, and
• 1 is called the remainder.

62
Greatest Common Divisors
•Let a and b be integers, not both zero.
•The largest integer d such that d | a and d | b is called
the greatest common divisor of a and b.
•The greatest common divisor of a and b is denoted by
gcd(a, b).
•Example 1: What is gcd(48, 72) ?
•The positive common divisors of 48 and 72 are
1, 2, 3, 4, 6, 8, 12, 16, and 24, so gcd(48, 72) = 24.
•Example 2: What is gcd(19, 72) ?
•The only positive common divisor of 19 and 72 is
1, so gcd(19, 72) = 1.

63
Greatest Common Divisors
•Using prime factorizations:
•a = p1
a
1 p2
a
2 … pn
a
n , b = p1
b
1 p2
b
2 … pn
b
n ,
•where p1 < p2 < … < pn and ai, bi ∈ N for 1 ≤ i ≤ n
•gcd(a, b) = p1
min(a
1
,b
1
)
p2
min(a
2
,b
2
)
… pn
min(a
n
,b
n
)
•Example:
a = 60 =a = 60 = 2222
3311
5511
b = 54 =b = 54 = 2211
3333
5500
gcd(a, b) =gcd(a, b) = 2211
3311
5500
= 6= 6

64
Relatively Prime Integers
•Definition:
•Two integers a and b are relatively prime if
gcd(a, b) = 1.
•Examples:
•Are 15 and 28 relatively prime?
•Yes, gcd(15, 28) = 1.
•Yes, gcd(55, 28) = 1.
•No, gcd(35, 28) = 7.

Fall 2002 CMSC 203 - Discrete Structures 65
Relatively Prime Integers
•Definition:
•The integers a1, a2, …, an are pairwise relatively prime if
gcd(ai, aj) = 1 whenever 1 ≤ i < j ≤ n.
•Examples:
•Are 15, 17, and 27 pairwise relatively prime?
•No, because gcd(15, 27) = 3.
•Are 15, 17, and 28 pairwise relatively prime?
•Yes, because gcd(15, 17) = 1, gcd(15, 28) = 1 and gcd(17,
28) = 1.

Least Common Multiples
•Definition:
•The least common multiple of the positive integers a and
b is the smallest positive integer that is divisible by both a
and b.
•We denote the least common multiple of a and b by
lcm(a, b).
•Examples:
lcm(3, 7) =lcm(3, 7) = 2121
lcm(4, 6) =lcm(4, 6) = 1212
lcm(5, 10) =lcm(5, 10) = 1010

Least Common Multiples
•Using prime factorizations:
•a = p1
a
1 p2
a
2 … pn
a
n , b = p1
b
1 p2
b
2 … pn
b
n ,
•where p1 < p2 < … < pn and ai, bi ∈ N for 1 ≤ i ≤ n
•lcm(a, b) = p1
max(a
1
,b
1
)
p2
max(a
2
,b
2
)
… pn
max(a
n
,b
n
)
•Example:
a = 60 =a = 60 = 2222
3311
5511
b = 54 =b = 54 = 2211
3333
5500
lcm(a, b) =lcm(a, b) = 2222
3333
5511
= 4275 = 540= 4275 = 540

GCD and LCM
a = 60 =a = 60 = 2222
3311
5511
b = 54 =b = 54 = 2211
3333
5500
lcm(a, b) =lcm(a, b) = 2222
3333
5511
= 540= 540
gcd(a, b) =gcd(a, b) = 2211
3311
5500
= 6= 6
Theorem: ab =Theorem: ab = gcd(a,b)lcm(a,b)gcd(a,b)lcm(a,b)

Modular Arithmetic
•Let a be an integer and m be a positive integer.
We denote by a mod m the remainder when a is divided
by m.
•Examples:
9 mod 4 =9 mod 4 = 11
9 mod 3 =9 mod 3 = 00
9 mod 10 =9 mod 10 = 99
-13 mod 4 =-13 mod 4 = 33

Congruences
•Let a and b be integers and m be a positive integer. We
say that a is congruent to b modulo m if
m divides a – b.
•We use the notation a ≡ b (mod m) to indicate that a is
congruent to b modulo m.
•In other words:
a ≡ b (mod m) if and only if a mod m = b mod m.

Congruences
•Examples:
•Is it true that 46 ≡ 68 (mod 11) ?
•Yes, because 11 | (46 – 68).
•Is it true that 46 ≡ 68 (mod 22)?
•Yes, because 22 | (46 – 68).
•For which integers z is it true that z ≡ 12 (mod 10)?
•It is true for any z∈{…,-28, -18, -8, 2, 12, 22, 32, …}
•Theorem: Let m be a positive integer. The integers a and b
are congruent modulo m if and only if there is an integer k
such that a = b + km.

Congruences
•Theorem: Let m be a positive integer.
If a ≡ b (mod m) and c ≡ d (mod m), then
a + c ≡ b + d (mod m) and ac ≡ bd (mod m).
•Proof:
•We know that a ≡ b (mod m) and c ≡ d (mod m) implies
that there are integers s and t with
b = a + sm and d = c + tm.
•Therefore,
•b + d = (a + sm) + (c + tm) = (a + c) + m(s + t) and
•bd = (a + sm)(c + tm) = ac + m(at + cs + stm).
•Hence, a + c ≡ b + d (mod m) and ac ≡ bd (mod m).

The Euclidean Algorithm
•The Euclidean Algorithm finds the greatest common
divisor of two integers a and b.
•For example, if we want to find gcd(287, 91), we divide
287 by 91:
•287 = 91⋅3 + 14
•We know that for integers a, b and c,
if a | b and a | c, then a | (b + c).
•Therefore, any divisor of 287 and 91 must also be a
divisor of 287 - 91⋅3 = 14.
•Consequently, gcd(287, 91) = gcd(14, 91).

•In the next step, we divide 91 by 14:
•91 = 14⋅6 + 7
•This means that gcd(14, 91) = gcd(14, 7).
•So we divide 14 by 7:
•14 = 7⋅2 + 0
•We find that 7 | 14, and thus gcd(14, 7) = 7.
•Therefore, gcd(287, 91) = 7.

75
•In pseudocode, the algorithm can be implemented as
follows:
•procedure gcd(a, b: positive integers)
•x := a
•y := b
•while y ≠ 0
•begin
• r := x mod y
• x := y
• y := r
•end {x is gcd(a, b)}

76
Representations of Integers
•Let b be a positive integer greater than 1.
Then if n is a positive integer, it can be expressed uniquely
in the form:
•n = akbk
+ ak-1bk-1
+ … + a1b + a0,
•where k is a nonnegative integer,
•a0, a1, …, ak are nonnegative integers less than b,
•and ak ≠ 0.
•Example for b=10:
•859 = 8⋅102
+ 5⋅101
+ 9⋅100

77
•Example for b=2 (binary expansion):
•(10110)2 = 1⋅24
+ 1⋅22
+ 1⋅21
= (22)10
•Example for b=16 (hexadecimal expansion):
•(we use letters A to F to indicate numbers 10 to 15)
•(3A0F)16 = 3⋅163
+ 10⋅162
+ 15⋅160
= (14863)10
•

78
•How can we construct the base b expansion of an integer
n?
•First, divide n by b to obtain a quotient q0 and remainder a0,
that is,
•n = bq0 + a0, where 0 ≤ a0 < b.
•The remainder a0 is the rightmost digit in the base b
expansion of n.
•Next, divide q0 by b to obtain:
•q0 = bq1 + a1, where 0 ≤ a1 < b.
•a1 is the second digit from the right in the base b expansion
of n. Continue this process until you obtain a quotient equal
to zero.

79
•Example:
What is the base 8 expansion of (12345)10 ?
•First, divide 12345 by 8:
•12345 = 8⋅1543 + 1
•1543 = 8⋅192 + 7
•192 = 8⋅24 + 0
•24 = 8⋅3 + 0
•3 = 8⋅0 + 3
•The result is: (12345)10 = (30071)8.

80
•procedure base_b_expansion(n, b: positive integers)
•q := n
•k := 0
•while q ≠ 0
•begin
• ak := q mod b
• q := q/b
• k := k + 1
•end
•{the base b expansion of n is (ak-1 … a1a0)b}

81
Addition of Integers
•Let a = (an-1an-2…a1a0)2, b = (bn-1bn-2…b1b0)2.
•How can we add these two binary numbers?
•First, add their rightmost bits:
•a0 + b0 = c0⋅2 + s0,
•where s0 is the rightmost bit in the binary expansion of a
+ b, and c0 is the carry.
•Then, add the next pair of bits and the carry:
•a1 + b1+ c0 = c1⋅2 + s1,
•where s1 is the next bit in the binary expansion of a + b,
and c1 is the carry.

82
•Continue this process until you obtain cn-1.
•The leading bit of the sum is sn = cn-1.
•The result is:
•a + b = (snsn-1…s1s0)2

83
•Example:
•Add a = (1110)2 and b = (1011)2.
•a0 + b0 = 0 + 1 = 0⋅2 + 1, so that c0 = 0 and s0 = 1.
•a1 + b1+ c0 = 1 + 1 + 0 = 1⋅2 + 0, so c1 = 1 and s1 = 0.
•a2 + b2+ c1 = 1 + 0 + 1 = 1⋅2 + 0, so c2 = 1 and s2 = 0.
•a3 + b3+ c2 = 1 + 1 + 1 = 1⋅2 + 1, so c3 = 1 and s3 = 1.
•s4 = c3 = 1.
•Therefore, s = a + b = (11001)2.

84
•How do we (humans) add two integers?
•Example: 7583
+ 4932
5511552211
111111 carrycarry
Binary expansions:Binary expansions: (1011)(1011)22
++ (1010)(1010)22
1100
carrycarry11
1100
11
11(( ))22

85
•Let a = (an-1an-2…a1a0)2, b = (bn-1bn-2…b1b0)2.
•How can we algorithmically add these two binary
numbers?
•First, add their rightmost bits:
•a0 + b0 = c0⋅2 + s0,
•where s0 is the rightmost bit in the binary expansion of a
+ b, and c0 is the carry.
•Then, add the next pair of bits and the carry:
•a1 + b1+ c0 = c1⋅2 + s1,
•where s1 is the next bit in the binary expansion of a + b,
and c1 is the carry.

86
•Continue this process until you obtain cn-1.
•The leading bit of the sum is sn = cn-1.
•The result is:
•a + b = (snsn-1…s1s0)2

87
•Example:
•Add a = (1110)2 and b = (1011)2.
•a0 + b0 = 0 + 1 = 0⋅2 + 1, so that c0 = 0 and s0 = 1.
•a1 + b1+ c0 = 1 + 1 + 0 = 1⋅2 + 0, so c1 = 1 and s1 = 0.
•a2 + b2+ c1 = 1 + 0 + 1 = 1⋅2 + 0, so c2 = 1 and s2 = 0.
•a3 + b3+ c2 = 1 + 1 + 1 = 1⋅2 + 1, so c3 = 1 and s3 = 1.
•s4 = c3 = 1.
•Therefore, s = a + b = (11001)2.

88
•procedure add(a, b: positive integers)
•c := 0
•for j := 0 to n-1
•begin
• d := (aj + bj + c)/2
• sj := aj + bj + c – 2d
• c := d
•end
•sn := c
•{the binary expansion of the sum is (snsn-1…s1s0)2}

89
Mathematical InductionMathematical Induction

90
Mathematical Induction: ExampleMathematical Induction: Example
Show that any postage of ≥ 8¢ can be
obtained using 3¢ and 5¢ stamps.
First check for a few particular values:
8¢ = 3¢ + 5¢
9¢ = 3¢ + 3¢ + 3¢
10¢ = 5¢ + 5¢
11¢ = 5¢ + 3¢ + 3¢
12¢ = 3¢ + 3¢ + 3¢ + 3¢
How to generalize this?

91
• Let P(n) be the sentence “n cents postage can be
obtained using 3¢ and 5¢ stamps”.
• Want to show that
“P(k) is true” implies “P(k+1) is true”
for any k ≥ 8¢.
• 2 cases: 1) P(k) is true and
the k cents contain at least one 5¢.
2) P(k) is true and
the k cents do not contain any 5¢.

92
• Case 1: k cents contain at least one 5¢ coin.
• Case 2: k cents do not contain any 5¢ coin.
Then there are at least three 3¢ coins.
5¢ 3¢ 3¢
Replace 5¢ coin
by two 3¢ coinsk cents k+1 cents
3¢
3¢
3¢ 5¢ 5¢
Replace three
3¢ coins by two
5¢ coins
k cents k+1 cents

93
Domino EffectDomino Effect
• Mathematical induction works like domino
effect:
• Let P(n) be “The nth domino falls backward”.
• If (a) “P(1) is true”;
(b) “P(k) is true” implies “P(k+1) is true”
Then P(n) is true for every n

94
Principle of MathematicalPrinciple of Mathematical
InductionInduction
Let P(n) be a predicate defined for
integers n.
Suppose the following statements are true:
1. Basis step:
P(a) is true for some fixed a∈Z .
2. Inductive step: For all integers k ≥ a,
if P(k) is true then P(k+1) is true.
Then for all integers n ≥ a, P(n) is true.

95
Example: Sum of Odd IntegersExample: Sum of Odd Integers
 Proposition: 1 + 3 + … + (2n-1) = n2
for all integers n≥1.
 Proof (by induction):
1) Basis step:
The statement is true for n=1: 1=12
.
2) Inductive step:
Assume the statement is true for some k≥1
(inductive hypothesis) ,
show that it is true for k+1 .

Example: Sum of Odd IntegersExample: Sum of Odd Integers
 Proof (cont.):
The statement is true for k:
1+3+…+(2k-1) = k2
(1)
We need to show it for k+1:
1+3+…+(2(k+1)-1) = (k+1)2
(2)
Showing (2):
1+3+…+(2(k+1)-1) = 1+3+…+(2k+1) =
1+3+…+(2k-1)+(2k+1) =
k2
+(2k+1) = (k+1)2
.
We proved the basis and inductive steps,
so we conclude that the given statement true. ■
by (1)

97
Important theorems proved byImportant theorems proved by
mathematical inductionmathematical induction
 Theorem 1 (Sum of the first n integers):
For all integers n≥1,
 Theorem 2 (Sum of a geometric sequence):
For any real number r except 1, and any
integer n≥0,
2
)1(
...21
+
=+++
nn
n
1
11
0 −
−
=
+
=
∑ r
r
r
nn
i
i

98
Example
(of sum of the first n integers)
In a round-robin tournament each of the n
teams plays every other team exactly once.
What is the total number of games played?
Solution on the board.

99
Proving a divisibility property byProving a divisibility property by
• Proposition: For any integer n≥1,
7n
- 2n
is divisible by 5. (P(n))
• Proof (by induction):
1) Basis step:
The statement is true for n=1: (P(1))
71
– 21
= 7 - 2 = 5 is divisible by 5.
2) Inductive step:
Assume the statement is true for some k≥1 (P(k))
(inductive hypothesis) ;
show that it is true for k+1 . (P(k+1))

100
Proving a divisibility property byProving a divisibility property by
Proof (cont.): We are given that
P(k): 7k
- 2k
is divisible by 5. (1)
Then 7k
- 2k
= 5a for some a∈Z . (by definition) (2)
We need to show:
P(k+1): 7k+1
- 2k+1
is divisible by 5. (3)
7k+1
- 2k+1
= 7·7k
- 2·2k
= 5·7k
+ 2·7k
- 2·2k
= 5·7k
+ 2·(7k
- 2k
) = 5·7k
+ 2·5a (by (2))
= 5·(7k
+ 2a) which is divisible by 5. (by def.)
Thus, P(n) is true by induction. ■

Number theory

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