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Section 4.3: Fermat’s Little Theorem
Practice HW (not to hand in)
From Barr Text
p. 284 # 1, 2
• As we will see later, the RSA Cryptosystem will require exponentiation to encrypt and decrypt messages. In this section, we review the basics of exponentiation and demonstrate an efficient method for doing exponentiation in modular arithmetic.
Exponential Notation
• Recall that exponential notation represents an expression of the form
,
where a represents the base of the expression and k represents the exponent. If the exponent k is a positive integer, then
ka
k times multiplied a
k aaaaa
Example 1: Compute and .
Solution:
52 37
In this section, we want to consider the problem
of computing
The next example illustrates a basic computation
with his quantity.
mak MOD
Example 2: Compute .
Solution:
25 MOD 25
The last example illustrates that it is easy to do
modular exponentiation when the exponent k is
small. However, if the exponent becomes larger,
this presents more of a challenge. If we are asked
to compute , for example, we should
note that , which due to size
causes errors in computations due to computer
round off. Our goal next is to present a method
that overcomes this problem.
41 MOD 785
7185 1081.67
• Note: All laws of exponents in the real number system carry over to MOD arithmetic, except for division.
Laws of Exponents
Real Number System Modular Number System
10 a 1 MOD 0 ma
lklk aaa mamaa lklk MOD MOD
kllk aa )( mama kllk MOD MOD )(
11 aa
1),gcd( if exists MOD 1 mama
lkl
k
aa
a 1),gcd( if exists MOD )( 1 mamaa lk
Method of Successive Squaring for
• Idea is to break the exponent k into a sum of powers of 2 (starting with ) and break
in terms of exponential terms as these powers of 2, computing the powers of 2 by “successively squaring” the previous term.
mak MOD
02 ka
Example 3: Compute
Solution:
23MOD 1311
Example 4: Compute
Solution: We first note that the exponent k = 85.
We first determine the powers of 2 that are less
than this exponent. Starting with, we see that
, , , , , ,
We can stop at since
41MOD 785
120 221 422 823 1624 3225 6426
6426 .8512827
We now decompose the exponent k into powers
of 2.
We next write
We next compute the needed powers of 7
needed with respect to the modulus 41. The
ones that we will need are indicated by . Note
that arrows are used to indicate the substitutions
from the previous step.
64164114166451664216485
41MOD )7777(41MOD 741MOD 7 64164164164185
741MOD71
841MOD4941MOD72
2341MOD6441MOD)8(41MOD)7(41MOD7 2224
3741MOD52941MOD)23(41MOD)7(41MOD7 2248
1641MOD136941MOD)37(41MOD)7(41MOD7 22816
1041MOD25641MOD)16(41MOD)7(41MOD7 221632
1841MOD10041MOD)10(41MOD)7(41MOD7 223264
Hence,
3841MOD7Hence,38
41MOD38
141MOD288and3841MOD161Note41MOD)138(
2881816and161237thatNote41MOD)288161(
aboves'fromngSubstituti41MOD)1816237(
41MOD)7777(
41MOD7417
85
641641
64164185
MOD
Note
• In Example 4, to compute by ordinary exponentiation, 84 multiplications are required. Using successive squares requires only 9 multiplications.
41 MOD 785
Fermat’s Little Theorem
Fermat’s Little Theorem in special cases can be
used to simplify the process of modular
exponentiation. We state it now.
Fermat’s Little Theorem: Let p be a prime
number, a an integer where . Then
1. If , then .
2. .
pa 11),gcd( pa 1 MOD 1 pa p
apa p MOD
Example 5: Use Fermat’s Little Theorem to
calculate the remainder when x is divided by the
given divisor m, that is, calculate x MOD m.
a. , m = 31.
Solution:
3021x
b. , m = 941.
Solution:
94015x
c. , m = 941.
Solution:
94115x
Fermat’s Little Theorem can be used to simplify
integers with large exponents if the modulus is
prime. The next example illustrates how this
works.
Example 6: Solve the equation .
Solution:
79 MOD 10782x
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