Unformatted text preview: n 3 + 7 n â‰¡ 3(1) n + 5 n 3 + 7 n â‰¡ 10 n + 5 n 3 â‰¡ 5(2 n + n 3 ) â‰¡ 0 (mod 5). ThereFore, 3 n 5 + 5 n 3 + 7 n â‰¡ 0 (mod 3) and 3 n 5 + 5 n 3 + 7 n â‰¡ 0 (mod 5), so 3 n 5 + 5 n 3 + 7 n â‰¡ (mod 15), and hence n 5 5 + n 3 3 + 7 n 15 is an integer. Question 9C # 7 (i) IF p is prime, then For all 1 â‰¤ a â‰¤ p1, ( a, p ) = 1. This implies that [ a ] is a unit in Z /p Z . Ï† ( p ) is the number oF units in Z /p Z , so Ï† ( p ) = p1. (ii) Ï† ( p n ) is the number oF integers a , where 1 â‰¤ a â‰¤ p n and ( a, p n ) = 1. Since p is prime, ( a, p n ) = 1 iF and only iF pa . So Ï† ( p n ) counts all number between 1 and p n , inclusive, that are not multiples oF p . Since p n = p n1 p , then there are p n1 multiples oF p between 1 and p n inclusive. ThereFore, Ï† ( p n ) = p np n1 iF p is prime. 1...
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This note was uploaded on 06/21/2008 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell.
 Spring '08
 BILLERA
 Math, Algebra

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