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Unformatted text preview: Math 135 Assignment 6 Solutions Winter 2009 1. a) Prove that 5 n 7 + 14 n 4 19 n ≡ 0(mod 7) for all integers n . Solution Since 7 is prime, by Fermat’s Little Theorem, n 7 ≡ n (mod 7). Thus 5 n 7 19 n ≡ 5 n 19 n ≡  14 n ≡ 0(mod 7). Therefore 5 n 7 + 14 n 4 19 n ≡ 14 n 4 + 0 ≡ 0(mod 7), as desired. b) Prove that 14 divides 5 n 7 + 14 n 4 19 n for all integers n . Solution For a number to be divisible by 14, it must be divisible by both 2 and 7. 5 n 7 + 14 n 4 19 n is divisible by 7 from part(a). It is left to show that 5 n 7 + 14 n 4 19 n is divisible by 2. Since 2 is prime, by Fermat’s Little Theorem, n 2 ≡ n (mod 2). Thus n 7 ≡ n 6 ≡ n 5 ≡ n 4 ≡ n 3 ≡ n 2 ≡ n (mod 2). Therefore 5 n 7 + 14 n 4 19 n ≡ 5 n + 14 n 19 n ≡ 0(mod 2). Since 5 n 7 + 14 n 4 19 n is divisible by both 2 and 7 for all integers n , then 5 n 7 + 14 n 4 19 n is divisible by 14 for all integers n . 2. a) Prove by induction that is a ≡ b (mod c ) then a n ≡ b n (mod c ) for all positive integers n . Solution Let P ( n ) be the statement ”If a ≡ b (mod c ), then a n ≡ b n (mod c )”. Clearly P (1) is true. Assume that P ( k ) is true. If a ≡ b (mod c ), then a k ≡ b k (mod c ), for some k ∈ P , k ≥ 1....
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This note was uploaded on 06/03/2010 for the course MATH 135 taught by Professor Peter during the Spring '07 term at University of the West.
 Spring '07
 peter
 Math, Integers

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