Unformatted text preview: [(a+b)(ab)]; that is, x 12b. Again, by Proposition 4.2.3, x 1 (2ma+2nb), so x 1 2 and we conclude that x = ±1 or x = ±2. The result follows. 15. [BB] If 91 and 92 are each greatest common divisors of a and b, then 91 ~ 92 (because 92 is greatest) and 92 ~ 91 (because 91 is greatest), so 91 = 92. 16. First note that a = b = 0 ¢:::::} a = a + b = 0, so gcd( a, b) exists if and only if gcd( a, a + b) exists. Suppose this is the case. As in the proof of Lemma 4.2.5, let 91 = gcd(a, a + b) and 92 = gcd(a, b). We prove that 91 ~ 92 and 92 ~ 91. First, since 92"1 a and 92 1 b, then 92 1 (a + b). ThUS,92 divides both a and a + b, so 92 ~ 91, the greatest common divisor of a and a + b. Also, since 91 divides both a and a + b, 91 must divide their difference, (a + b) a = b. Thus, 91 divides both a and b, hence, 91 ~ 92, the greatest common divisor of a and b. 17. (a) [BB] Multiplying 2647(17369) + 8402(5472) = 1 by 4, we have 17369( 10588) + (5472)(33608) = 4....
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 Summer '10
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 Graph Theory, Prime number, Greatest common divisor

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