soln.hw1.330

# soln.hw1.330 - 64 , 125 agrees with 0 , 1 , 2 , 3 , 4 , 5.)...

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Abstract Algebra, Homework 1 Solutions Chapter 0 18. Since each p i is a prime we have that p i > 1. Since p i divides the product p 1 p 2 ··· p n , dividing q = p 1 p 2 ··· p n + 1 by any one of the primes p i gives a remainder of 1. This means that p i doesn’t divide q for any of the primes p 1 , p 2 , ··· p n . There are many other possible proofs. This one has the advantage that it is simple and related to the division algorithm. 30. By the division algorithm, for any n we can write n = 6 q + r with r = 0 , 1 , 2 , 3 , 4 or 5. We write n mod 6 = r or we could also note that n mod 6 = r mod 6. Note that n 3 = (6 q + r ) 3 = 6 3 q 3 + 3(6 2 ) q 2 r + (3)(6) qr 2 + r 3 , so n 3 = 6 a + r 3 . In other words, n 3 mod 6 = (6 q + r ) 3 mod 6 = r 3 mod 6. Now we can check that for 0 r 5, r 3 mod 6 = r mod 6. (That is, mod 6 the sequence 0 , 1 , 8 , 27 ,
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Unformatted text preview: 64 , 125 agrees with 0 , 1 , 2 , 3 , 4 , 5.) Putting this all together we have n 3 mod 6 = r 3 mod 6 = r mod 6 = n mod 6. Another proof could be based on the fact that n 3-n = ( n-1) n ( n +1). The right hand side is even since either n or n +1 is even. Since n-1 , n and n +1 are three consecutive numbers one of them is divisible by three. This implies that the right hand side is also divisible by three. Since ( n-1) n ( n + 1) is divisible by 2 and by 3 it is divisible by 6. By Chap. 0, Number 9 (see back of book), since 6 divides n 3-n that means that n 3 mod 6 = n mod 6. 1...
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## This note was uploaded on 10/27/2009 for the course MATH 330 taught by Professor Staff during the Spring '08 term at Ill. Chicago.

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