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Chapter 8 Introduction to Number Theory
2
Contents
Prime Numbers
Fermat’s and Euler’s Theorems
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Prime Numbers
Primes numbers An integer p > 1 is a prime number if and only if it is divisible by only
1 and p.
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Prime Numbers
Integer factorization
Any integer a > 1 can be factored in a unique way as
where p1 < p2 < … < pt are prime numbers and
each ai is a positive integer.
tat
aaa ppppa ...321321
91 = 7 × 13;11101 = 7 × 112 ×13
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Prime Numbers
Another integer factorization If P is the set of all prime numbers, then any positive integer can be w
ritten uniquely in the following form:
The right side is the product over all possible prime numbers p. Most of the exponents ap will be 0.
0each whereP
pp
a apa p
3600 = 24×32×52×70×110×….
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Prime Numbers
Another integer factorization The value of any given positive integer can be specified by listing all
the nonzero exponents.
The integer 12 =22×31 is represented by {a2=2, a3=1}.The integer 18 =21×32 is represented by {a2=1, a3=2}.The integer 91= 72×131 is represented by {a7= 2, a13= 1}.
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Prime Numbers
Multiplication Multiplication of two numbers is adding the corresponding exponents.
k = 12 × 18 = 216
12 = 22 × 31
18 = 21 × 32
------------------216 = 23 × 33
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Prime Numbers
Divisibility
a|b → ap ≤ bp for all p
a = 12; b= 36; 12|36
12 = 22×3;36 = 22×32
a2 = 2 = b2
a3 = 1 ≤ 2 = b3
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Prime Numbers
GCD
k = gcd (a, b) → kp = min(ap, bp) for all p
300 = 22×31×52
18 = 21×32×50
gcd (18, 300) = 21×31×50 = 6
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Fermat’s and Euler’s Theorems
Fermat’s theorem If p is prime and a is a positive integer not divisible by p,
then
ap-1 ≡ 1 (mod p)
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Fermat’s and Euler’s Theorems
Proof of Fermat’s theorem. Outline
Show {1, 2, …, p-1}={a mod p, 2a mod p, …, (p-1)a mod p}
Show .
Since is relatively prime to p, we multiply
to both sides to get .
papp p mod)!1()!1( 1
)!1( p -1)!1( p
pa p mod1 1
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Fermat’s and Euler’s Theorems
Proof of Fermat’s theorem Show {1, 2, …, p-1}={a mod p, 2a mod p, …, (p-1)a mod p}
Show ka mod p for any 1 ≤ k ≤ p-1 is in {1, 2, …, p-1}
by showing that ka mod p ≠ k’a mod p for k≠ k’.
Show ka mod p ≠ k’a mod p for 1 ≤ k≠ k’ ≤ p-1. Proof by contradiction Assume that ka ≡ k’a mod p for some 1 ≤ k≠ k’ ≤ p-1. Since a is relatively prime to p, we multiply a-1 to get k ≡ k’ mod p,
which contradiction the fact that k≠ k’.
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Fermat’s and Euler’s Theorems
Proof of Fermat’s theorem Show .
{1, 2, …, p-1} = {a mod p, 2a mod p, …, (p-1)a mod p}
ppap p mod)!1()!1( 1
pa
papp
papaap...
ppappapap...
p
p
mod1
mod)!1()!1(
mod ])1(...2[)]1(21[
mod )]mod)1((...)mod2()mod[()]1(21[
1
1
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Fermat’s and Euler’s Theorems
An alternative form of Fermat’s Theorem
ap ≡ a mod p
where p is prime and a is any positive integer.
Proof If a and p are relatively prime, we get ap ≡ a mod p by multi
plying a to each side of ap-1 ≡ 1 mod p.
If a and p are not relatively prime, a = cp for some positive integer c. So ap ≡ (cp)p ≡ 0 mod p and a ≡ 0 mod p, which means ap ≡ a mod p.
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Fermat’s and Euler’s Theorems
An alternative form of Fermat’s Theorem
ap ≡ a mod p
where p is prime and a is any positive integer.
p = 5, a = 3 35 = 243 ≡ 3 mod 5
p = 5, a = 10 105 = 100000 ≡ 10 mod 5 ≡ 0 mod 5
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Fermat’s and Euler’s Theorems
Euler’s Totient Function The number of positive integers less than n and relatively prime to n.
)(n
= 3637 is prime, so all the positive number from 1 to 36are relatively prime to 37.
= 2435 = 5×71, 2, 3, 4, 6, 8, 9,11, 12, 13, 16, 17, 18, 19, 22, 23, 24, 26, 27, 29, 31, 32, 33, 34
)37(
)35(
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Fermat’s and Euler’s Theorems
How to compute In general,
For a prime n, (Zn = {1,2,…, n-1})
For n = pq, p and q are prime numbers and p≠ q
)(n
1)( nn
)1()1()( qpn
npp
nn dividing primes theallover runs where,)1
1()(
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Fermat’s and Euler’s Theorems
Proof of
is the number of positive integers less than pq that are relatively prime to pq.
can be computed by subtract from pq – 1 the number of positive integers in {1, …, pq – 1} that are not relatively prime to pq.
The positive integers that are not relatively prime to pq are a multiple of either p or q. { p, 2p,…,(q – 1)p}, {q, 2q, …,(p – 1)q} There is no same elements in the two sets. So, there are p + q – 2 elements that are not relatively prime to pq.
Hence, = pq – 1– (p + q – 2) = pq – p – q +1 = (p – 1)(q – 1)
)(n
)1()1()( qpn
)(n
)(n
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Fermat’s and Euler’s Theorems
Φ(21) = Φ(3)×Φ(7) = (3-1)×(7-1) = 2 ×6 = 12
Z21={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}Φ(3)={3,6,9,12,15,18}Φ(7)={7,14}
where the 12 integers are {1,2,4,5,8,10,11,13,16,17,19,20}
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Fermat’s and Euler’s Theorems
Euler’s theorem For every a and n that are relatively prime:
na n mod1)(
a = 3; n = 10; Φ(10) = 4; 34 = 81 ≡ 1 mod 10a = 2; n = 11; Φ(11) = 10; 210 = 1024 ≡ 1 mod 11
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Fermat’s and Euler’s Theorems
Proof of Euler’s theorem
If n is prime, it holds due to Fermat’s theorem.
Otherwise (If n is not prime), define two sets R and S. show the sets R and S are the same. then, show
na n mod1)(
nan mod11
na n mod1)(
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Fermat’s and Euler’s Theorems
Proof of Euler’s theorem
Set R The elements are positive integers less than n and relatively prime to n. The number of elements is
R={x1, x2,…, xΦ(n)} where x1< x2<…< xΦ(n)
Set S Multiplying each element of R by a∈R modulo n S ={(ax1 mod n), (ax2 mod n),…(axΦ(n) mod n)}.
)(n
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Fermat’s and Euler’s Theorems
Proof of Euler’s theorem
The sets R and S are the same. We show S has all integers less than n and relatively prime to n.
S ={(ax1 mod n), (ax2 mod n),…(axΦ(n) mod n)}
1. All the elements of S are integers less than n that are relatively prime to n because a is relatively prime to n and xi is relatively prime to n, axi must also be relatively prime to n.
2. There are no duplicates in S.If axi mod n = axj mod n, then xi = xj. by cancellation law.
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Fermat’s and Euler’s Theorems
Proof of Euler’s theorem Since R and S are the same sets,
)(mod1
)(mod
)(mod
)mod(
)(
)(
1
)(
1
)(
)(
1
)(
1
)(
1
)(
1
na
nxxa
nxax
xnax
n
n
ii
n
ii
n
n
i
n
iii
n
i
n
iii
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Fermat’s and Euler’s Theorems
Alternative form of the theorem
If a and n are relatively prime, it is true due to Euler’s theorem.
Otherwise, ….
)(mod1)( naa n
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Fermat’s and Euler’s Theorem
The validity of RSA algorithm Given 2 prime numbers p and q, and integers n = pq and m,
with 0<m<n, the following relationship holds.
If m and n are re relatively prime, it holds by Euler’s theorem. If m and n are not relatively prime, m is a multiple of either p or q.
nmmm qpn mod1)1)(1(1)(
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Fermat’s and Euler’s Theorem
Case 1: m is a multiple of p m=cp for some positive integer c. gcd(m, q)=1, otherwise, m is a multiple of p and q and yet m<pq because gcd(m, q)=1, Euler’s theorem holds
by the rules of modular arithmetic,
Multiplying each side by m=cp
qmq mod11
kkqm
qm
qm
n
n
pq
integer somefor ,1
mod1
mod1][
)(
)(
11
nmm
kcnmkcpqmmn
n
mod1)(
1)(
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Fermat’s and Euler’s Theorem
Case 2: m is a multiple of q prove similarly.
Thus, the following equation is proved.
nmmm qpn mod1)1)(1(1)(
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Fermat’s and Euler’s Theorem
An alternative form of this corollary is directly relevant to RSA.
nm
nm
nmm
m
k
kn
nk
mod
theoremsEuler'by ,mod])1[(
mod][( 1)(
1)(
