Section 4.3 87 (a) [BB] We have a2 = p201.r2 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(a2, b) = p, while if f3 > 1, then gcd(a2, b) = p2. (b) Here, a3 = p301.r3 and b = rf3 s. Now there are three possibilities. If f3 = 1, then gcd(a3, b) = p. If f3 = 2, then gcd(a3, b) = p2, and if f3 ~ 3, then gcd(a3, b) = p3. (c) Here we have a2 = p201.r2 and b3 = p3f3s3. There are two possibilities. If a = 1, then gcd(a2, b3) = p2. If a> 1, then gcd(a2,b3) = 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 = x2, it follows that each ai and each f3i must be even. But this means that a and b are perfect squares. 30. (a) We proceed exactly as in the solution to Problem 17. We write a and b in the form for certain primes Pl, P2, ... , Pr and integers alo a2, ... , ar, f31. f32, ... , f3r.
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