theorem
Euler's theorem
If $ A $ and $ p $ prime, then $ a ^ {\ psi (n)} \ equiv1 (mod \ p) $, $ \ psi (n) $ is $ 1 {\ sim} n $ and $ n-$ prime number
Set $ x_1, x_2, \ ldots, x _ {\ psi (n)} $ $ and $ n-prime, then consider some of the number of $ ax_1, ax_2, \ ldots, ax _ {\ psi (n)} $
The following two properties:
1, any two $ ax_i, ax_j $ remainder modulo n are not equal.
Proof: Suppose present, i.e. $ ax_i {\ equiv} ax_j (mod \ n) $, there $ n {\ mid} (ax_i-ax_j) $, and because the $ $ $ A $ n-coprime
And $ (x_i-x_j) <n $, so there is no $ n {\ mid} (ax_i-ax_j) $, i.e. absence of $ ax_i {\ equiv} ax_j (mod \ n) $, thus proved.
Summarized: For any $ ax_i $ are true, so there are $ \ psi (n) $ distinct remainder.
2, and any $ ax_i $ $ $ n-prime.
Proof: Since the $ A $ $ $ n-prime, and is $ x_i $ $ $ n-a prime number set, so that the $ ax_i $ $ $ n-prime. I.e., $ gcd (ax_i, n) = 1 $, there Euclidean algorithm, $ gcd (ax_i, n) = gcd (n, ax_i \% n) = 1 $.
According to the above properties, can be drawn $ ax_1 (mod \ n), ax_2 (mod \ n), \ ldots, ax _ {\ psi (n)} (mod \ n) $ after some sort of $ x_1, x_2, \ ldots , x _ {\ psi (n)} $-one correspondence.
You can write $ ax_1 (mod \ n) {\ ast} ax_2 (mod \ n) {\ ast}, \ ldots, {\ ast} ax _ {\ psi (n)} (mod \ n) = x_1 {\ ast } x_2 {\ ast}, \ ldots, {\ ast} x _ {\ psi (n)} $
It is $ ax_1 {\ ast} ax_2 {\ ast}, \ ldots, {\ ast} ax _ {\ psi (n)} {\ equiv} x_1 {\ ast} x_2 {\ ast}, \ ldots, {\ ast } x _ {\ psi (n)} (mod \ n) $
即$(a^{\psi(n)}-1){\ast}x_1{\ast}x_2{\ast},\ldots,{\ast}x_{\psi(n)}{\equiv}0(mod\ n)$
Because $ x_1 {\ ast} x_2 {\ ast}, \ ldots, {\ ast} x _ {\ psi (n)} $ and $ n-$ prime, so $ n {\ mid} (a ^ {\ psi ( n)} - 1) $
I.e., $ a ^ {\ psi (n)} {\ equiv} 1 (mod \ n) $, Euler's theorem proved.
Fermat's Little Theorem
If $ p $ is a prime number, for any $ a $, $ a ^ p {\ equiv} a (mod \ n) $ established.
Proof: The equation can be written as $ a ^ {p-1} {\ ast} a {\ equiv} a (mod \ p) $, because $ a {\ equiv} a (mod \ p) $ permanent establishment, Therefore, only permit $ a ^ {p-1} mod n = 1 $
If $ A $ and $ p $ prime to Euler theorem $ a ^ {p-1} {\ equiv} 1 (mod \ p) $, because $ p $ is a prime number, the $ \ psi (p) = p-1 $.
If $ A $ and $ p $ is not prime, then $ a ^ p $ $ certain multiples of formula A $, so $ a ^ p {\ equiv} a {\ equiv} 0 (mod \ p) $.
Theorem is proved.
Euler's theorem Corollary
If $ A $ and $ p $ prime, then $ a ^ b {\ equiv} a ^ {b \ mod \ \ psi (p)} (mod \ p) $.
Certificate: As above, the equation can be written as $ a ^ {bb \ mod \ \ psi (p)} {\ ast} {a ^ {b \ mod \ \ psi (p)}} {\ equiv} a ^ { b \ mod \ \ psi (p)} (mod \ p) $,
Because $ {a ^ {b \ mod \ \ psi (p)}} {\ equiv} a ^ {b \ mod \ \ psi (p)} (mod \ p) $ constant established,
Therefore, only permit $ a ^ {bb \ mod \ \ psi (p)} mod \ p = 1 $, because $ \ psi (p) {\ mid} (bb \ mod \ \ psi (p)) $,
设 $(b-b\ mod\ \psi(p))=q{\ast}\psi(p)$
And because $ a ^ {q \ psi (p)} = (a ^ {\ psi (p)}) ^ q $, $ a $ p prime and,
Therefore $ (a ^ {\ psi (p)}) ^ q {\ equiv} 1 (mod \ p) $, thereby obtaining certificate $ a ^ {bb \ mod \ \ psi (p)} mod \ p = 1 $ .
Therefore $ a ^ b {\ equiv} a ^ {b \ mod \ \ psi (p)} (mod \ p) $.