Euler's theorem proof of Number Theory & Euler's function formula

Euler's function:
Euler's function is a very important function in number theory. Euler's function refers to the number of positive integers (including 1) with a positive integer N, less than N and interlace with N, it is recorded as PHI (n ).

Complete remainder set:
Defines a set of numbers less than N and with N mutual quality as Zn, and calls this set a complete remainder set of N. Apparently | Zn | = PHI (n ).

Related Nature:
For prime numbers P, Phi (p) = p-1.
For two different prime numbers, p and q, their product N = p * Q satisfies the requirements of PHI (n) = (p-1) * (Q-1 ).
This is because ZN = {1, 2, 3 ,..., n-1}-{P, 2 p ,..., (Q-1) * p}-{q, 2q ,..., (P-1) * Q}, then PHI (n) = (n-1)-(q-1)-(p-1) = (p-1) * (Q-1) = PHI (p) * PHI (q ).

Euler's theorem:
For positive integers A and N of the mutual quality, there is^{Phi (n) limit 1 mod n.}

Proof:
(1) Order ZN = {x1, x2 ,..., x phi (n)}, S = {A * X1 mod N, A * X2 mod N ,..., A * x PHI (n) mod n },
Then ZN = S.
(1) because of the mutual quality of A and N, XI (1 ≤ I ≤ PHI (N) and N, so a * XI and N ing, so a * Xi mod n ε Zn.
② If I =j, then x<sub>I =xJ, and a * X is available from the mutual quality of A and N.I mod n = A * xJ mod n (Elimination Law ).

(2) A Phi (n) * X1 * X2 *... * x PHI (n) mod n
Round (A * X1) * (A * x2) *... * (A * x PHI (N) mod n
Round (A * X1 mod n) * (A * X2 mod n) *... * (A * x PHI (n) mod n
Limit X1 * X2 *... * x PHI (n) mod n
Compare the left and right sides of the equation, because XI (1 ≤ I ≤ PHI (N) and n are of mutual quality, a Phi (n) limit 1 mod n (dealignment Law ).
Note:
Elimination law: If gcd (C, p) = 1, AC hybrid BC mod p contains a Hybrid B mod p.</sub>

Ferma theorem:
If the positive integer a interacts with the prime number P, there is<sup>P-1 limit 1 mod p.
This theorem is very simple. It can be proved simply by using the Euler's theorem, because PHI (p) = p-1.

Reference Source:
Http://zhidao.baidu.com/question/15882452.html? SI = 2</sup>

"

Supplement: Euler's function formula

(1) Euler's function of PK

For a given prime number P, Phi (p) = p-1. For a positive integer n = PK,

 Phi(N) = PK-1

Proof:
Less than PThe number of Positive Integers of K is P.K-1, of which
And PPositive Integers with non-Interconnectivity of K have{P * 1, p * 2,..., p * (pk-1-1 )}TotalPK-1-1Items
So PHI (n) = PK-1-(PK-1-1) = PK-PK-1.

(2) Euler's function of p * q

Assume that p and q are two positive integers of mutual quality, then the Euler's function of p * q is

(P * q) = PHI (p) * PHI (Q), gcd (p, q) = 1.

Proof:
N = p * q, gcd (p, q) = 1
According to the remainder theorem of China
There is a one-to-one ing between Zn and ZP × ZQ.
(My idea is: A in ZP, B in ZQ in B * P + A * q in Zn .)
Therefore, the number of elements in the complete remainder set of N is equal to the number of ZP x ZQ elements in the set.
The number of elements in the latter is PHI (p) * PHI (q ).
? (P * q) =? (p) *? (q ).

(3) Euler's function of any positive integer

Any integer N can be expressed as the product of its prime factor:

I
N = zhangpIKI (number of prime factors whose I is N)
I = 1

According to the previous two conclusions, it is easy to conclude that the Euler's function is:

         I                      I
 Φ(n) = ∏  piki -1(pi -1) = n ∏ (1 - 1 / pi)
        i=1                    i=1

For any n> 2, 2 | PHI (N), because it must existP_I-1Is an even number.

For program code, see: http://blog.csdn.net/Rappy/archive/2007/08/16/1747489.aspx

Reference Source:
Http://blog.csdn.net/ray58750034/archive/2006/03/27/640074.aspx