Fermat's Little theory proof 3.42

Fermat's Little theory proof 3.42 - a ] } In other words,...

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MATH 135 Fall 2008 Fermat’s Little Theorem 3.42 If p is prime and p 6 | a , then a p - 1 1 (mod p ). Proof: Suppose p is prime and p 6 | a . Step 1: Show 0 a , 1 a , 2 a , . . . , ( p - 1) a are all distinct mod p Suppose ra sa (mod p ) with 0 s r p - 1. Then p | ( r - s ) a so p | r - s or p | a by Theorem 2.53. Since p 6 | a , then p | r - s . But 0 s r p - 1, so 0 r - s p - 1. Thus, r - s = 0 (since the only multiple of p in the range 0 to p - 1 is 0) so r = s . Therefore, if r 6 = s , then ra 6≡ sa (mod p ), ie. 0 a , 1 a , 2 a , . . . , ( p - 1) a are all distinct mod p . Step 2: Show 1 a · 2 a · 3 a · · · · · ( p - 1) a 1 · 2 · 3 · · · · · ( p - 1) (mod p ) Since 0 a , 1 a , 2 a , . . . , ( p - 1) a are all distinct mod p , then [0 a ], [1 a ], . . . , [( p - 1) a ] are distinct congruence classes in Z p . Since there are exactly p of these classes, then Z p = { [0 a ] , [1 a ] , . . . , [( p - 1)
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Unformatted text preview: a ] } In other words, since [0 a ] = [0], then { [1 a ] , [2 a ] , . . . , [( p-1) a ] } is a rearrangement of { [1] , [2] , . . . , [ p-1] } . So [1 a ] [2 a ] [( p-1) a ] = [1] [2] [ p-1] or 1 a 2 a 3 a ( p-1) a 1 2 3 ( p-1) (mod p ). Step 3: Show a p-1 1 (mod p ) From Step 2, ( p-1)! a p-1 ( p-1)! (mod p ). But p 6 | ( p-1)! so gcd( p, ( p-1)!) = 1. Therefore, we can divide out the common factor of ( p-1)! to get a p-1 1 (mod p )....
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