math336_hw4_solns

# math336_hw4_solns - n 3 7 n â‰ 3(1 n 5 n 3 7 n â‰ 10 n 5 n...

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Homework 4 Partial Solutions Question 9A # 16 Let n be an integer such that 41 - n and n has order 2 mod 41. This last statement implies that [ n ] 2 41 = [1] 41 in Z / 41 Z . Since 41 is prime, Z / 41 Z is a feld, and hence also a commuatative ring with no zero divisors. Since 41 - n , n is a unit in Z / 41 Z . Thus we can apply 8A E18 to get that [ n ] 41 = [1] 41 or [ n ] 41 = [ - 1] 41 . However, iF [ n ] 41 = [1] 41 then n would have order 1 mod 41. But this is a contradiction, so [ n ] 41 = [ - 1] 41 = [40] 41 . This implies that n = 40 + 41 k For some k Z . Question 9B # 12 n 5 5 + n 3 3 + 7 n 15 = 15 15 ± n 5 5 + n 3 3 + 7 n 15 ² = 1 15 ( 3 n 5 + 5 n 3 + 7 n ) . So n 5 5 + n 3 3 + 7 n 15 is an integer iF and only iF 3 n 5 + 5 n 3 + 7 n is divisible by 15 or 3 n 5 + 5 n 3 + 7 n 0 (mod 15). By property (v) on pg. 71, we can show this by showing that 3 n 5 + 5 n 3 + 7 n 0 (mod 3) and 3 n 5 + 5 n 3 + 7 n 0 (mod 5), since 15 = [3 , 5]. IF 3 | n , then 3 n 5 + 5 n 3 + 7 n 0 (mod 3). Otherwise, by ±ermat’s theorem, n 2 1 (mod 3). In this case, 3 n 5 + 5 n 3 + 7 n 3( n 2 ) 2 n + 5 n 2 n + 7 n 3(1) 2 n + 5(1) n + 7 n 15 n 0 (mod 3). IF 5 | n , then 3 n 5 + 5 n 3 + 7 n 0 (mod 5). Otherwise, by ±ermat’s theorem, n 4 1 (mod 5). In this case, 3 n 5 + 5 n 3 + 7 n 3( n 4 ) n + 5
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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‰¤ p-1, ( 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 ) = p-1. (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 p-a . So Ï† ( p n ) counts all number between 1 and p n , inclusive, that are not multiples oF p . Since p n = p n-1 p , then there are p n-1 multiples oF p between 1 and p n inclusive. ThereFore, Ï† ( p n ) = p n-p n-1 iF p is prime. 1...
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