Adv Alegbra HW Solutions 28

Adv Alegbra HW Solutions 28 - a (or in b ) are even as...

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28 Solution. 1234 = 2 · 617 (the reader is expected to prove that 617 is prime, using 617 < 25) and 5678 = 2 · 17 · 167, so that the gcd = 2. 1.70 (i) Prove that an integer m 2 is a perfect square if and only if each of its prime factors occurs an even number of times. Solution. If m = a 2 and a = p e 1 1 ··· p e n n , then m = p 2 e 1 1 ··· p 2 e n n . Conversely, if m = p 2 e 1 1 ··· p 2 e n n , then m = a 2 , where a = p e 1 1 ··· p e n n . (ii) Prove that if m is a positive integer for which m is rational, then m is a perfect square. Conclude that if m is not a perfect square, then m is irrational. Solution. Let m = p e 1 1 ··· p e n n .I f m is not a perfect square, then at least one of the e i is odd. If m = a / b , then mb 2 = a 2 . The exponent of p i on the left is odd while the exponent of p i on the right is even, and this is a contradiction. 1.71 If a and b are positive integers with ( a , b ) = 1, and if ab is a square, prove that both a and b are squares. Solution. Since a and b are relatively prime, the sets of primes occurring in the factorization of a and of b are disjoint. Hence, if ab is a square, then all the exponents e i in ab = p e 1 1 ··· p e n n are even, and hence all the exponents
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Unformatted text preview: a (or in b ) are even as well. Therefore, both a and b are perfect squares. 1.72 Let n = p r m , where p is a prime not dividing an integer m ≥ 1. Prove that p-³ n p r ´ . Solution. Write a = ³ n p r ´ . By Pascal ’ s formula: a = ³ n p r ´ = n ! ( p r ) ! ( n − p r ) ! . Cancel the factor ( n − p r ) ! and cross-multiply, obtaining: a ( p r ) ! = n ( n − 1 )( n − 2 ) ··· ( n − p r + 1 ). Thus, the factors on the right side, other than n = p r m , have the form n − i = p r m − i , where 1 ≤ i ≤ p r − 1. Similarly, the factors in ( p r ) ! , other than p r itself, have the form p r − i , for i in the same range: 1 ≤ i ≤ p r − 1. If p e | p r m − i , where e ≤ r and i ≥ 1, then p r m − i = bp e ; hence, p e | i ; there is a factorization i = p e j . Therefore, p r − i = p e ( p r − e − j ) ....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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