Numbers, Groups and Cryptography

187 Pages·2018·1.89 MB·English

by  Gordan Savin

Numbers, Groups
and
Cryptography
Gordan Savin

Contents
Chapter 1. Euclidean Algorithm 5
1. Euclidean Algorithm 5
2. Fundamental Theorem of Arithmetic 10
3. Uniqueness of Factorization 14
4. Efficiency of the Euclidean Algorithm 16
Chapter 2. Groups and Arithmetic 21
1. Groups 21
2. Congruences 25
3. Modular arithmetic 28
4. Theorem of Lagrange 31
5. Chinese Remainder Theorem 35
Chapter 3. Rings and Fields 41
1. Fields and Wilson’s Theorem 41
2. Field Characteristic and Frobenius 45
3. Quadratic Numbers 48
Chapter 4. Primes 53
1. Infinitude of primes 53
2. Primes in progression 56
3. Perfect Numbers and Mersenne Primes 59
4. The Lucas Lehmer test 63
Chapter 5. Roots 67
1. Roots 67
2. Property of the Euler Function 70
3. Primitive roots 72
4. Discrete logarithm 75
5. Cyclotomic Polynomials 77
Chapter 6. Quadratic Reciprocity 81
1. Squares modulo p 81
2. Fields of order p2 85
3. When is 2 a square modulo p? 88
4. Quadratic Reciprocity 91
Chapter 7. Applications of Quadratic Reciprocity 95
3
4 CONTENTS
1. Fermat primes 95
2. Quadratic Fields and the Circle Group 98
3. Lucas Lehmer test revisited 100
Chapter 8. Sums of two squares 107
1. Sums of two squares 107
2. Gaussian integers 111
3. Method of descent revisited 116
Chapter 9. Pell’s Equations 119
1. Shape numbers and Induction 119
2. Square-Triangular Numbers and Pell Equation 122
3. Dirichlet’s approximation 125
4. Pell Equation: classifying solutions 128
5. Continued Fractions 131
Chapter 10. Cryptography 135
1. Diffie-Hellman key exchange 135
2. RSA Code 138
3. ElGamal Code 142
Chapter 11. Primality testing 147
1. Miller Rabin test 147
2. p−1 method 151
3. p+1 method 154
4. Quadratic sieve 157
Chapter 12. Elliptic curves 161
1. Cubic curves 161
2. Degenerate curves 165
3. Curves modulo p 170
4. The curve y2 = x3+bx 174
Chapter 13. Factoring and testing using elliptic curves 177
1. Lenstra’s factoring method 177
2. Degenerate curves modulo p 181
3. Elliptic curve test for Mersenne primes 184
CHAPTER 1
Euclidean Algorithm
1. Euclidean Algorithm
Euclid was a Greek mathematician who lived in Alexandria around 300
B.C. He devised a clever and fast algorithm to find the greatest common
divisor of two integers. In modern language, the Euclidean Algorithm is
the division with remainder. Recall that dividing two positive integers a,b
means finding a positive integer q, called a quotient, such that
a = qb+r
with 0 ≤ r < b. The number r is a remainder of the division. If r = 0 then
we say that b divides a, and write b|a. For example,
3 | 6 whereas 4 (cid:45) 6.
Dividing two integers is probably the most difficult of the four standard
binary operations. However, some of the most fundamental properties, such
as the uniqueness of factorization, are based on the Euclidean Algorithm.
But how is this related to finding the Greatest Common Divisor (or
simplygcd)oftwointegersmandn? Iffactorizationsofmandnintoprime
factors are known, then it is easy to figure out what the gcd is. Consider,
for example, m = 756 and n = 360. Then
756 = 22·33·7 and 360 = 23·32·5.
We see that 2 and 3 are the only primes appearing in factorizations of
both numbers. Moreover, 22 and 32 are the greatest powers of 2 and 3,
respectively, which divide both 756 and 360. It follows that
gcd(756,360) = 22·32 = 36.
There are two issues here, however. Firstly, we have secretly assumed the
uniqueness of factorization. The second issue, also very important, is that
factoring into primes is a very difficult process, in general. A much better
way was discovered by Euclid. Assume, for example, that we want to find
the greatest common divisor of 60 and 22. Subtract 60−22 = 38. Notice
that any number dividing 60 and 22 also divides 22 and 38. Conversely,
any number that divides 22 and 38 also divides 22+38 = 60 and 22. Thus,
instead of looking for common divisors of 60 and 22, we can look at common
divisors of 22 and 38 instead. Since the later pair of numbers is smaller, the
5
6 1. EUCLIDEAN ALGORITHM
problemoffindingthegreatestcommondivisorof60and22hasjustbecome
easier, and we have accomplished that without ever dividing or multiplying
two numbers. In fact, we can do even better. Instead of subtracting 22 from
60 once, we can subtract it twice, to get 60−2×22 = 16. Thus we get that
gcd(60,22) = gcd(22,16).
Of course, we do not stop here. Since 22−16 = 6, it follows that
gcd(22,16) = gcd(16,6)
and so on. At every step we replace a pair (a,b) by a pair (b,r) where r is
the remainder of the division of a by b. The whole division process in this
case is given here:
60 = 2·22+16
22 = 1·16+6
16 = 2·6+4
6 = 1·4+2
4 = 2· 2 +0
This shows that
gcd(60,22) = gcd(22,16) = ··· = gcd(4,2) = 2.
The last statement is obvious since 2 divides 4. In general, starting with a
pairofnumbersaandbweusethedivisionalgorithmtogenerateasequence
of numbers (b > r > r > ...) as follows. First, we divide a by b:
1 2
a = q b+r ,
1 1
then divide b by r , and so on...
1
b = q r +r
2 1 2
r = q r +r
1 3 2 3
.
.
.
r = q r +r
n−2 n n−1 n
r = q r +0.
n−1 n+1 n
Thisprocessstopswhentheremainderis0. Sinceb > r > r ...itstops
1 2
in less than b steps. We claim that the last non-zero remainder r is equal
n
to the greatest common divisor of a and b. In order to verify this, notice
that the first equation a = qb+r implies that any common divisor of b and
1
r also divides a. Likewise, if we rewrite the first equation as a−qb = r ,
1 1
it is clear that any common divisor of a and b also divides r . This shows
1
that gcd(a,b) = gcd(b,r ). Arguing in this fashion, we see that
1
gcd(a,b) = gcd(b,r ) = ... = gcd(r ,r ) = r
1 n−1 n n
1. EUCLIDEAN ALGORITHM 7
where, of course, the last statement is obvious since r divides r . We
n n−1
can summarize what we have discovered:
Euclidean Algorithm gives an effective way to compute the greatest com-
mon divisor of two integers. Moreover, the algorithm does not rely on the
uniqueness of factorization.
The Euclidean Algorithm can be viewed as a special case of the Contin-
ued Fraction Algorithm. We need some notation before proceeding to the
definition of the algorithm. If α is a real number then let [α] denote the
√
greatest integer less than or equal to α. For example, since 2 = 1.4... the
√
greatest integer less than or equal to 2 is 1:
√
[ 2] = 1.
Note that [α] = α if α is an integer. The Continued Fraction Algorithm is
defined as follows:
(1) Let α > 1, and put β = α−[α].
(2) If β = 0 stop, else put α = 1/β and go to (1).
1
In order to illustrate the relation between the two algorithms, let us
workoutthecasewhenα = 60/22. ThentheContinuedFractionAlgorithm
generates the following numerical data.
i [α ] β α
i i i+1
0 2 16/22 22/16
1 1 6/16 16/6
2 2 4/6 6/4
3 1 2/4 4/2
4 2 0 STOP
Here, of course, we put α = α and β = β. As it can be seen from the
0 0
table, the Continued Fraction Algorithm generates the same numbers as
the Euclidean Algorithm starting with the pair (60,22). More precisely, if
α = a/b and b > r > r ... are non-negative integers generated by the
1 2
Euclidean Algorithm applied to the pair (a,b), then
r r r
1 2 3
β = , β = , β = ...
1 2
b r r
1 2
and
[α] = q , [α ] = q , [α ] = q ...
1 1 2 2 3
Sincer = 0forsomen, β = 0andthealgorithmstopsforeveryrational
n+1 n
number (fraction). The name (continued fraction) comes from the fact the
process can be recorded as
60 16 1
= 2+ = 2+ = ...
22 22 1+ 6
16
8 1. EUCLIDEAN ALGORITHM
1
... = 2+ .
1+ 1
2+ 1
1+12
The continued fraction, in turn, gives a series of - so called - partial conver-
gents:
1 1 1 1
2, 2+ , 2+ ,2+ and 2+ .
1 1+ 1 1+ 1 1+ 1
2 2+1 2+ 1
1 1+12
An easy calculation gives that these five partial convergents are equal to,
respectively,
8 11 30 60
2,3, , and = .
3 4 11 22
A rather different phenomenon occurs when we apply the algorithm to
√
α = 2. Then going through the loop for the first time yields
√ √ √ √
(1) α = 2, β = 2−[ 2] = 2−1.
(2)
1 1 √
α = = √ = 2+1.
1
β 2−1
Going through the loop for the second time yields
√ √ √ √
(1) α = 2+1, β = 2+1−[ 2+1] = 2−1.
1 1
(2)
1 1 √
α = = √ = 2+1.
2
β1 2−1
√
Thus the second output α = 2+1 is the same as the first output of the
2
algorithm. Of course, the third output will be the same, and so on:
√
α = α = α = ... = 2+1.
1 2 3
We see that the Continued Fraction Algorithm is stuck in the loop in this
√ √
case. In particular, this shows that 2 is not a fraction. We can write 2
as a continued fraction
√ 1
2 = 1+ .
2+ 1
2+ 1
2+ 1
...
This is a rather confusing expression, however. A proper way to view this
expressionisintermsofpartialconvergents. Apartialconvergentisobtained
by stopping the continued fraction expression at a point. For example, the
√
third convergent for 2 would be
√ 1 7
2 = 1+ = .
2+ 1 5
2
1. EUCLIDEAN ALGORITHM 9
We mention, without a proof, that the partial convergents form a sequence
√
of rational approximations of 2. For example, the first four convergents
√
for 2 are
3 7 17
1, = 1.5, = 1.4, ≈ 1.41...
2 5 12
Exercises
1) Use the Euclidean Algorithm to find the greatest common divisor of
a) 1084 and 412.
b) 1979 and 531.
c) 305 and 185.
2) Use calculations from the previous exercise to express in a continued
fraction form:
a)
1084
.
412
b)
1979
.
531
c)
305
.
185
3) If a | b and b | c show that a | c.
4) If a | b and a | c show that a | b+c.
√
5) Use the Continued Fraction Algorithm to show that 3 is not rational.
6) Compute the fourth convergent of the continued fraction expansion of
√ √
3. Note how well it approximates 3.
√
7) Use the Continued Fraction Algorithm to show that 7 is not rational.
8) Compute the fourth convergent of the continued fraction expansion of
√ √
7. Note how well it approximates 7.
9) Find the number α whose continued fraction expansion is
1
α = 1+ .
1+ 1
1+ 1
1+ 1
...
10 1. EUCLIDEAN ALGORITHM
2. Fundamental Theorem of Arithmetic
Pick three non-zero integers a,b and c. A purpose of this section is to
find integer solutions of the following equation:
ax+by = c.
This is crucial to proving uniqueness of factorization, which will be done in
the next section.
We shall first derive necessary conditions. Consider, for example, the
equation
9x+12y = 5.
It does not have any integer solutions because 3 divides the left hand side
(since 3 divides 9 and 12) but 3 does not divide the right hand side, which
is 5. Thus, the equation ax + by = c does not have solutions unless the
greatest common divisor of a and b divides c. This is a necessary condition.
We shall see that this is also a sufficient solution. For example, if we replace
5 by 3 in the above equation, then
9x+12y = 3
does have a solution, since 9(−1) + 12(1) = 3. Before we state the main
result of this section we make the following reduction. Let d denote the
greatest common divisor of a and b. We assume now that d divides c, so we
can write c = kd for some integer k. If (x ,y ) is a solution of ax+by = d,
0 0
hence
ax +by = d
0 0
then, after multiplying this equation by k, we have
a(kx )+b(ky ) = kd = c,
0 0
making (kx ,ky ) a solution of the original equation ax+by = c. Thus we
0 0
have reduced to solving:
ax+by = gcd(a,b).
Theorem 1. (Fundamental Theorem of Arithmetic) If a,b are two pos-
itive integers, then there exist integers x,y such that ax+by = gcd(a,b).
Proof. The proof uses the idea behind the Euclidean Algorithm. The
greatest common divisor does not change if we replace the pair (a,b) by the
pair of smaller numbers (a−b,a). If we can solve the equaltion
(a−b)x+by = gcd(a−b,b)
then, by rewriting this equation and using gcd(a−b,b) = gcd(a,b), we get
a solution to the original equation
ax+b(y−x) = gcd(a,b).