781hw2sol

781hw2sol - 18.781, Fall 2007 Problem Set 2 Solutions to...

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Unformatted text preview: 18.781, Fall 2007 Problem Set 2 Solutions to Selected Problems Problem 2.1.17 By Wilsons theorem, we have 70! - 1 (mod 71) , where 70! 63! 64 70 63! (- 7) (- 6) (- 1) (- 1) 7 63! (7!) (mod 71) . Notice that 7! (7 5 2 )(6 4 3) 70 72 - 1 (mod 71) . Therefore, we have (- 1) 70! (- 1) (63!) (7!) (- 1) (63!) (- 1) 63! (mod 71) . That is, 63! + 1 0 (mod 71) . Also, 62 63 (- 9) (- 8) 72 1 (mod 71) , hence 61! + 1 61! (1) + 1 61! (62 63) + 1 63! + 1 0 (mod 71) , as desired. 2 Problem 2.1.25 91 = 7 13 and 7 , 13 are prime numbers. By given condition, a, n are both prime to 7 , 13. Then, by Fermats theorem, we have n 6 1 (mod 7) and a 6 1 (mod 7) . By squaring both sides of each equation, we get n 12 1 (mod 7) and a 12 1 (mod 7) . Hence 7 | n 12- a 12 . Again by Fermats theorem, n 12 1 (mod 13) and a 12 1 (mod 13) . Hence 13 | n 12- a 12 ....
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781hw2sol - 18.781, Fall 2007 Problem Set 2 Solutions to...

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