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译文题目：Euler ’s Theorem and Fermat ’s Theorem 学生姓名： 张云

专 业：数学与统计学院

指导教师： 张吉刚

2009年 12月 30日

Euler’s Theorem and Fermat ’s Theorem

Book: Elementary Methods in number theory

Author :Melvyn B. Nathanson

Page :7167P P

2.5 Euler’s Theorem and Fermat ’s Theorem

Euler ’s theorem and its corollary ,Fermat ’s theorem ,are fundamental results in number theory ,with many applications in mathematics and computer science .In the following sections we shall see how the Euler and Fermat theorems can be used to determine whether an integer is prime or composite ,and how they are applied in cryptography.

Theorem2.12（Euler ）Let m be a positive integer, and let a be an integer relatively prime to m .Then

()()m a m mod 1≡?.

Proof. Let (){}m r r ? ,1be a reduced set of residues modulo m .Since ()1,=m a ，we have ()()()m i m ar i ? ,11,== for 1,,()i m ?= .Consequently, for every (){}m i ? ,1∈there exists ()(){}m i ?σ ,1∈such that

()()m r ar i i mod σ≡.

Moreover ，()m ar ar j i mod ≡ if and only if j i =，and so σ is a permutation of the set (){}m ? ,1 and (){}m ar ar ? ,1 is also a reduced set of residues modulo m .It follows that

()()()()()()()m ar ar ar a m r r r m m mod 2121??? ≡

()()()()m r r r m mod 21σσσ ≡

()()m r r r m m o d 21? ≡ Dividing by ()m r r r ? 21，we obtain

()()m a m mod 1≡?

This completes the proof.

The following corollary is sometimes called Fermat ’s litter theorem.

Theorem 2.13 (Fermat ) Let p be a prime number .If the integer a is not divisible by p ,then

()p a r mod 11≡-

Moreover,

()p a a p mod ≡

for every integer a .

Proof . If p is prime and does not divide a, then ()1,=p a ，()1-=p p ?，and

()()p a a p p mod 11≡≡-?

by Euler’s theorem. Multiplying this congruence by a ，we obtain

()p a a p mod ≡

If p divides a ,then this congruence also holds for a .

Let m be a positive integer and let a be an integer that is relatively prime to m .By Euler ’s theorem,()()m a m mod 1≡?.The order of a with respect to the modulus m is the smallest positive integer d such that ()m a d mod 1≡.Then ()m d ?≤≤1.We shall prove that ()a ord m divides ()m ? for every integer a relatively prime to p

Theorem 2.14 Let m be a positive integer and a an integer relatively prime to m .If d is the order of a modulo m ,then ()m a a l k mod ≡ if and only if ()d l k mod ≡.In particular, ()m a n mod 1≡ if and only if d divides n ,and so d divides ()m ?.

Proof. Since a has order modulo m ,we have ()m a d mod 1≡.If ()d l k mod ≡，then dq l k +=，and so

()()m a a a a a l

q d l dq l k mod ≡==+. Conversely, suppose that ()m a a l k mod ≡.By the division algorithm, there exist integers q and r such that

r dq l k +=-and 10-≤≤d r .

Then

()()m a a a a a a a r k r q d

l r dq l k mod ≡==++ Since ()1,=m a k ，we can divide this congruence by k a and obtain

()m a r

m o d 1≡ Since 10-≤≤d r ， and d is the order of a modulo m, it follows that 0=r ,and so ()d l k mod ≡.

If )(mod 1m a a

n ≡≡，then d divides n .In particular, d divides )(m ?，since )(mod 1)(m a m ≡? by Euler ’s theorem.

For example; let 15=m and a=7.Since 8)15(=?，Euler ’s theorem tells us that

)15(mod 178

≡ Moreover, the order of 7 with respect to 15 is a divisor of 8. We can compute the order as follows:

)15(mod 771≡

)15(mod 44972≡≡

)15(mod 132873

≡≡ )15(mod 19174≡≡

And so the order of 7 is 4.

We shall give a second proof of Euler ’s theorem and its corollaries .we begin with some simple observations about groups. We define the order of a group as the cardinality of the group. Theorem 2.15 (Lagrange ’s theorem) If G is a finite group and H is a subgroup of G , then the order of H divides the order of G

Proof .Let G be a group ,written multiplicatively, and let X be a nonempty subset of G .For every ∈a G ,we define the set

):{X x ax aX ∈=

The map aX X f →: defined by ax x f =)( is a bijection, and so aX X = for all ∈a G .If H is subgroup of G , then aH is called a coset of H . Let aH and bH be cosets of the subgroup H 。If φ≠?bH aH ,then there exist H y x ∈,such that by ax =，or ,since H is a subgroup,az axy b ==-1，where H xy z ∈=-1。

Then ah azh bh ∈= for all H h ∈and so aH bH ?.By symmetry , aH bH ?,and so aH bH =.Therefore , cosets of a subgroup H are either disjoint or equal .Since every element of G belongs to some coset of H （for example , a aH ∈ for all ∈a G ），it follows that the cosets of H partition G .We denote the set of cosets by G H .If G is a finite group, then H and G H are finite ,and

G H G H = In particular, we see that H divides G .

Let G be a group ,written multiplicatively ,and let a G ∈.Let {:}k H a k z =∈.Then

1a H G =∈?。Since k l k l a a a += for all ,k l Z ∈，it follows that H is a subgroup of G .This subgroup is called the cyclic subgroup generated by a ,and written a .Cyclic

subgroups are abelian.

The group G is cyclic if there exists an element G a ∈such that a G =.In this cases ,the element a is called a generator of G 。For example, the group ()?

Z Z 7 is a cyclic group of order 6 generated by Z 73+.The congruence class Z 75+is another generator of this group.

If l k a a ≠for all integers l k ≠，then the cyclic subgroup generated by a is infinite .If there exist integers k and l such that k l < and l k a a =，then 1=-l k a . Let d be the smallest positive integer such that 1=d a .Then the group elements 1，a ，2a ，…，1-d a are distinct .Let Z n ∈.By the division algorithm ,there exist integers q and r such that

r dq n += and 10-≤≤d r 。As ()r r q d r dq n a a a a a ===+，

it follows that

{}{}

10::-≤≤=∈=d r a Z n a a r n ， and the cyclic subgroup generated by a has order d .Moreover ，l k a a = if and only if )(mod d l k ≡。

Let G be a group, and let G a ∈.we define the order of a as the cardinality of the cyclic subgroup generated by a

Theorem 2.16 Let G be a finite group, and G a ∈.Then the order of the element a divides the order of the group G .

Proof. This follows immediately from Theorem 2.15, since the order of a is the order of the cyclic subgroup that a generates.

Let us apply these remarks to the special case when ()?=mZ Z G is the group of units in the ring of congruence classes modulo m .Then G is a finite group of order ()m ?.Let ()1,=m a and let d be the order of mz a + in G ,that is the order of the cyclic subgroup generated by mz a +.By Theorem 2.16,d divides ()m ?，and so

()()()()()()

mZ mZ a mZ a mZ a d m d m m +=+=+=+1???.

Equivalently,

()()m a m mod 1≡? ，

This is Euler ’s theorem.

Theorem 2.17 Let G be a cyclic group of order m ,and let H be a subgroup of G .If a is a generator of G ,then there exists a unique divisor d of m such that H is the cyclic subgroup generated by d a ,and H has order d m .

Proof . Let S be the set of all integers u such that H a u ∈.If u, S v ∈,then u a ，H a v ∈.Since

H is a subgroup ,it follows that H a a a v u v u ∈=+ and ()H a a a v u v u ∈=--1Therefore, S v u ∈±,and S is a subgroup of Z .By Theorem 1.3,there is a unique nonnegative integer d such that dZ S =,and so H is the cyclic subgroup generated by d a .Since H a

m ∈=1,we have S m ∈,and so d is a positive divisor of m .It follows that H has order d m .

Theorem 2.18 Let G be a cyclic group of order m ,and let a be a generator of G .For every integer k ,the cyclic subgroup generated by k a has order d m

,where ()k m d ,=,and

d k a a =.In particular, G has exactly ()m ? generators.

Proof. Since ()m k d ,=,there exist integers x and y such that my kx d +=.Then

()()()x k y m x k my kx d a a a a a ===+

and so k d a a ∈ and d k a a ?.Since d divides k ,there exists an integer z such that dz k =.Then

()z d k a a =，

and so d k a a ∈ and d k a a ?.Therefore, d k a a = and k a has order d m .In

particular k a generates G if and only if 1=d if and only if ()1,=k m ,and so G has

exactly ()m ? generators .This completes the proof.

We can now give a group theoretic proof of Theorem 2.8.Let G be a cyclic group of order m .For every divisor d of m ,the group G has a unique cyclic subgroup of order d ,and this subgroup has exactly )(d ? generators .Since every element of G generates a cyclic

subgroup ,it follows that

()∑=m

d d m ?.

欧拉定理和费马定理

著作:初等数论

作者:Melvyn B. Nathanson

页码:7167P P -

2.5 欧拉定理和费马定理

欧拉定理及其推论，费马定理都是数论中的重要结果，而且在数学和计算机领域中都有很多的应用.在本节中，我们将可以看到，欧拉定理和费马定理是如何用来判断一个正数是素数还是合数以及它们是怎样应用在密码学中的.

定理2.12（欧拉）设m 是一个正整数，d 是一个与m 互质的整数，则

()()m a m mod 1≡?

证明：设(){}m r r ? ,1是模m 的既约剩余类.由于()1,=m a ，我们可得()()()m i m ar i ? ,11,==，因此，对于每个(){}m i ? ,1∈，必存在()(){}m i ?σ ,1∈使

得

()()m r ar i i mod σ≡

而且，当且仅当j i =，()m ar ar j i mod ≡，所以σ是集合(){}m ? ,1的排列.(){}

m ar ar ? ,1也是模m 的既约剩余类，从而有

()()()()()()()1212m od m m r r r m a ar ar ar m ???≡

()()()()m r r r m mod 21σσσ ≡

()()m r r r m mod 21? ≡

两边同除以()m r r r ? 21，我们得

()()m a m mod 1≡?

证明完毕.

下面的推论有时称作费马小定理

定理2.13（费马）设p 是一个素数，p 不整除整数a ，则

()p a r mod 11≡-

而且，对于每个整数a ，都有

()p a a p mod ≡

证明：设p 是素数，p 不整除a ，则()1,=p a ，()1-=p p ?，由欧拉定理知 ()()p a a p p mod 11≡≡-?

这个同余类两边同乘以a ，我们可得

()p a a p mod ≡.

如果p 整除a ，则这个同余式对a 也成立.

设m 是个正整数，a 是一个与m 互质的整数.由欧拉定理知，()()m a

m mod 1≡?，a 关于模m 的阶是使得()m a d mod 1≡的最小正整数.那么()m d ?≤≤1.我们用()a ord m 来表示a

关于模m 的阶.我们将证明，对于每个与p 互质的整数a ，都有()a ord m 整除()m ?.

定理2.14 设m 是一个正整数，a 是一个与m 互质的整数.如果d 是模m 的阶，那么当且仅当()d l k mod ≡有()m a a l k mod ≡.特别地，当且仅当d 整除n ，才有()m a n mod 1≡，所以d 整除()m ?.

证明：由于a 关于模m 的阶为d ，即()m a

d m o d 1≡.如果()d l k mo d ≡，那么

dq l k +=，所以 ()()m a a a a a l

q d l dq l k mod ≡==+.

相反的，假设()m a a l k mod ≡.由除法性质得，必存在整数q 和r 使得

r dq l k +=-及10-≤≤d r .

那么

()()m a a a a a a a r k r q d l r dq l k mod ≡==++

由于()1,=m a k ，我们用k a 来整除这个同余类，从而得到

()m a r mod 1≡

由于0-≤≤d r ，d 是模m 的阶，那么)(mod 1,0d k r ≡=

假设)(mod 1m a a n ≡≡，则d 整除n ，尤其d 整除)(m ?，根据欧拉定理可得

)(mod 1)(m a m ≡?

例如，7,15==a m ,由于8)15(=?，我们由欧拉定理知

)15(mod 178≡

而且，7关于15的阶是8的一个约数.我们这样计算阶：

)15(mod 771

≡ )15(mod 44972≡≡

)15(mod 132873≡≡

)15(mod 19174≡≡

所以7的阶是4.

我们将给出欧拉定理的另一种证明及它的推论.我们将从关于群的一些简单观察开始.我们把群的阶定义为群的基数

定理2.15（拉格朗日定理）假设G 是一个有限群，H 是G 的子群，则H 的阶整除G 的阶

证明：设G 是一个群，运算为乘法，X 是G 的一个非空子集.对于任意∈a G,我们定义这个集合：

):{X x ax aX ∈=

由ax x f =)(定义的映射aX X f →:是一个双射，所以对于所有的∈a G ，aX X =.假设H 是G 的子群，那么aH 被称作H 的一个陪集.设aH 和bH 是子群H 的陪集.φ≠?bH aH ，则存在H y x ∈,，使得by ax =，或者，由于H 是一个子群，az axy b ==-1，这里H xy z ∈=-1.对于所有的H h ∈，

ah azh bh ∈=，所以aH bH ?；同理，?,所以aH bH =因此子群H 的陪集不交或相等.由于G 的每个元素属于

H 的陪集（例如，对于所有的a G ∈都有a aH ∈）

，从而可得出H 划分G 的陪集.我们用G H 表示陪集.假设G 是一个有限集，则H ，G H 是有限的，及

G H G H = 特别地，我们有H 整除G

设G 是一个群，运算为乘法.设a G ∈，H ={∈k a k :Z }，则01a H G =∈?.由于对所有的k ，∈l Z,k l k l a a a +=，从而可得出H 是G 的子群.这个子群被称作由a 生成的循环子群，记作a ，循环子群都是交换的.

如果存在一个元素G a ∈，使得a G =，则群G 是循环群.此时，元素a 称作G 的生成元.

例如，群(Z/7Z ?)是由73+Z 生成的阶为6的循环群.同余类75+Z 是这个群的另一个生成

元.如果对于所有整数l k ≠，都有l k a a ≠，则由a 生成的这个循环群是无限的.如果存在k

和l ，使得k l <，l k a a =，则1=-l k a

.设d 是最小的正数，且使得1=d a ，则这个群的元素1，a ，2a ，…，1-d a

是不同的.设Z n ∈，由除法知，存在整数q 和r ，使得r dq n +=，

10-≤≤d r .由于 ()r

r q d r dq n a a a a a ===+， 从而有

=a {∈n a n :Z }{}

10:-≤≤=d r a r ， 由a 生成的这个循环群的阶为d ，而且，当且仅当)(mod d l k ≡，才有l k a a =. 设G 是一个群，G a ∈.我们可以将a 的阶定义为由a 生成的循环子群的基数. 定理2.16 设G 是一有限群，G a ∈，则元素a 的阶整除群G 的阶.

证明：由定理2.15立即可得知，因为a 的阶是由a 生成的循环子群的阶. 我们可以应用到一种特殊情形：当()?=mZ Z G 是模m 的剩余类环中的，则G 是阶为()m ?的有限群.设()1,=m a ，d 是G 中mz a +的阶，则d 也是由mz a +生成的循环子群的阶.根据定理2.16知，d 整除()m ?，及

m a m +)(?Z (m a +=Z )()m ?=((m a +Z m d m d +=1)))

(?Z

同样地，

()()m a m mod 1≡? ，

这就是欧拉定理.

定理2.17 设G 是阶为m 的一个循环群，H 是G 的子群.如果

是G 的生成元，则存在唯一的m 的约数d ，使得H 是由d a 生成的循环子群，其中H 的阶为d

m . 证明：设S 是所有整数u 的集合，使得H a u ∈，如果u ，S v ∈，则u a ，H a v ∈.

由于H 是一个子群，则H a a a v u v u ∈=+，()H a a a v u v u ∈=--1，因此，S v u ∈±，S 是

Z 的一个子群.由定理1.3知，存在唯一的非负整数d ，使得d S =Z ，所

H 是由d a 生成的一个循环子群.由于H a m ∈=1，我们可得S m ∈，d 是m 的正约数，H 的阶为d m . 定理 2.18 设G 是阶为m 的循环群，a 是G 的一个生成元，对于每个整数k ，由k a 生成

的循环子群的阶为d m ，则()k m d ,=，d k a a =.特别地，G 恰好有()m ?个生成元.

证明：由于()m k d ,=，则存在整数x ，y ，使得my kx d +=.则

()()()x

k y m x k my kx d a a a a a ===+，

所以k d a a ∈，k d a a ?.由于d 整除k ，则存在一个整数z ，使得dz k =.那么

()z d k a a =，

所以，d k a a ∈，d k a a ?.因此，d k a a =，k a 的阶为d

m

.特别地，当且仅当1=d ，()1,=k m ，所以G 恰好有()m ?个生成元.证明完毕. 我们现在可以给出定理2.8的群论证明.设G 是阶m 的循环群.对于m 的每个约数，群G 有唯一的阶为d 的循环群，而且这个子群恰好有)(d ?个生成元.由于G 的每个元素都可以生成一个循环子群，从而有

()∑=m

d d m ?