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Unformatted text preview: Section 4.3 87 (a) [BB] We have a 2 = p201. r 2 with gcd(p, r2) = 1 and b = rf3 s. Also gcd(r2, s) = 1 since gcd(r, s) = 1. We conclude that there are two possibilities: If f3 = 1, then gcd(a 2 , b) = p, while if f3 > 1, then gcd(a 2 , b) = p2. (b) Here, a 3 = p301.r 3 and b = rf3 s. Now there are three possibilities. If f3 = 1, then gcd(a 3 , b) = p. If f3 = 2, then gcd(a 3 , b) = p2, and if f3 ~ 3, then gcd(a 3 , b) = p3. (c) Here we have a 2 = p201. r 2 and b 3 = p3 f3 s3. There are two possibilities. If a = 1, then gcd(a 2 , b 3 ) = p2. If a> 1, then gcd(a 2 ,b 3 ) = p3. 29. [BB] Let the prime decompositions of a and b be a = pflp~2 ... p~r and b = qflqg2 ... q~ . Since gcd(a, b) = 1, we know that Pi =1= qj for any i and j. Thus, ab = pfl ... p~rqfl ... q~. with no simplification possible. Since ab = x 2 , it follows that each ai and each f3i must be even. But this means that a and b are perfect squares....
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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