Description: Lemma for 2503prm . Calculate a power mod. We calculate 2 ^ 1 9 = 2 ^ 1 8 x. 2 == 1 8 3 2 x. 2 = N + 1 1 6 1 , 2 ^ 3 8 = ( 2 ^ 1 9 ) ^ 2 == 1 1 6 1 ^ 2 = 5 3 8 N + 1 3 0 7 , 2 ^ 3 9 = 2 ^ 3 8 x. 2 == 1 3 0 7 x. 2 = N + 1 1 1 , 2 ^ 7 8 = ( 2 ^ 3 9 ) ^ 2 == 1 1 1 ^ 2 = 5 N - 1 9 4 , 2 ^ 1 5 6 = ( 2 ^ 7 8 ) ^ 2 == 1 9 4 ^ 2 = 1 5 N + 9 1 , 2 ^ 3 1 2 = ( 2 ^ 1 5 6 ) ^ 2 == 9 1 ^ 2 = 3 N + 7 7 2 , 2 ^ 6 2 4 = ( 2 ^ 3 1 2 ) ^ 2 == 7 7 2 ^ 2 = 2 3 8 N + 2 7 0 , 2 ^ 1 2 4 8 = ( 2 ^ 6 2 4 ) ^ 2 == 2 7 0 ^ 2 = 2 9 N + 3 1 3 , 2 ^ 1 2 5 1 = 2 ^ 1 2 4 8 x. 8 == 3 1 3 x. 8 = N + 1 and finally 2 ^ ( N - 1 ) = ( 2 ^ 1 2 5 1 ) ^ 2 == 1 ^ 2 = 1 . (Contributed by Mario Carneiro, 3-Mar-2014) (Revised by Mario Carneiro, 20-Apr-2015) (Proof shortened by AV, 16-Sep-2021)
Ref | Expression | ||
---|---|---|---|
Hypothesis | 2503prm.1 | ||
Assertion | 2503lem2 |