HW3sol - example: a = 3 , b = 4 , c = 0. (6) Problem 16 ii...

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MATH 3360 – Applicable Algebra. HW 3 Solutions. (1) Problem 5 on page 56. Show that if n is not a prime or a square of a prime, then n has a prime factor smaller than the square root of n . Solution: Let n = pq , where p is the smallest prime dividing n . Assume for contradiction that p > n . Then q > n as well, but this implies that n = pq > n · n = n . Contradiction. (2) Problem 13 on page 58. Solution: In the back of the book, p.572. (3) Problem 6 on page 73. Solution: Many solutions are possible, for example a = 17. (4) Problem 7 on page 74. Solution: Since m is not a prime, m = p · q , where p is the smallest prime dividing m . If p , q , then clearly ( p · q ) | ( m - 1)! and we are done. So, say m = p · p . Note that since m > 4, we have 2 < p = m , so 2 p < ( m - 1). Thus, p · (2 p ) | ( m - 1)! which implies that ( p · p ) = m | ( m - 1)! and we are done. (5) Problem 12 on page 77. Solution: No. There are many counterexamples to this statement. For
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Unformatted text preview: example: a = 3 , b = 4 , c = 0. (6) Problem 16 ii on page 77. Solution: Note that 68 -2 mod 7. So we have 68 105 (-2) 105 ((-2) 3 ) 35 (-8) 35 (-1) 53 -1 mod 7. (7) Problem 18 on page 78. Solution: Note that 6 -5 mod 11. If e is odd, we have 5 e + 6 e 5 e + (-5) e 5 e-5 e 0 mod 11. If e is even, say e = 2 k , we have 5 e + 6 e 5 2 k + (-5) 2 k 5 2 k + 5 2 k 2 5 2 k mod 11. So the question becomes: Does there exist an integer n such that 2 5 2 k = 11 n ? The answer is clearly no, since the right side has 11 in its prime factorization and the left side doesnt. 1...
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).

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