I really sucks, so wrong eventually.
This is very simple, so read instead.
Xiaoman talked about this brother always felt that, although as a player aged forgetfulness is probably not what I remember. .
Euler's theorem: \ (A ^ {\ varphi (n-)} \ equiv. 1 \ (MOD \ n-) \) , where \ ((a, n) =
1 \) Fermat's little theorem: \ (A ^ {P } -1 \ equiv. 1 \ (MOD \ P) \) , where \ ((a, P) =. 1 \) , is easy to find a special case of Euler's theorem.
Euler's theorem proving :( congruence of the default mode \ (n-\) )
provided \ (X_1, X_2, ..., X _ {\ varphi (n)} \) is \ (1 \) to the \ (n-\) in the \ (n \) are relatively prime numbers, they are easy to find mold \ (n \) pairwise disjoint, and the remainder are with the \ (n \) are relatively prime (nonsense, because after the mold or the original number of Well)
Then we find that \ (aX_1, aX_2, ..., aX _ {\ varphi (n)} \) seems like the two properties. .
Mode \ (n-\) pairwise disjoint: evidence to the contrary, when the \ (aX_i \ equiv aX_j \ (MOD \ n-) \) , then \ (aX_i-aX_j \ equiv 0 \) , then \ (a (X_i-X_j) \ 0 equiv \) , since \ (a \) and \ (n-\) prime, \ (x_j-X_i \) can not be \ (n-\) a multiple mold so \ (n-\) must not \ ( 0 \)
Remainder are associated with \ (n-\) coprime: \ (A \) and \ (n-\) prime, \ (X_i \) and \ (n-\) prime, so \ (aX_i \) also \ (n- \) are relatively prime (which is very emotional understanding orz)
With these two properties, we can find \ (aX_1, aX_2, ..., aX _ {\ varphi (n)} \) mode \ (n-\) after a certain amount is \ (\ varphi (n) \ ) th Unlike \ (n-\) prime number, then it is certainly not \ (X_1, X_2, ..., X _ {\ varphi (n)} \) this set.
Therefore, to obtain \ [X_1 \ cdot X_2 ... X _ {\ varphi (n)} \ equiv aX_1 \ cdot aX_2 ... aX _ {\ varphi (n)} \ (mod \ n) \]
\[\Rightarrow 1 \equiv a^{\varphi(n)}\ (mod\ n)\]
\ (QED. \)