Unformatted text preview: Section 4.4 95 The most complex situation occurs when k :2: 3, which we assume henceforth. Since 2 I (x + 1)(x  1), we must have x odd, as before. Since k :2: 3 and 2k I (x + 1)(x  1), some power 2i! of 2, with i :2: 2, must divide either x + 1 or x  1. Suppose 2i! I (x  1) with i :2: 2. Then x + 1 = 2i!a + 2 = 2(1 + 2i!a), and since i :2: 2, this means that 2 I (x + 1), but 4 ~ (x + 1). It follows that i :2: k  1. If i :2: k, we again have the obvious solution x == 1 (mod 2k). However, we have a new possibility; namely, i = k  1; that is, x == 1 (mod 2 k 1 ). In this case, is divisible by 2k, so we have a solution. Arguing similarly if 2i! I (x + 1), we obtain the solutions x = 1, x = 2 k 1  1, x = 2 k 1 + 1, x = 2k  1. 21. (a) Since 7(8) = 56 == 1 (mod 11), the inverse of 7 (mod 11) is 8. The solution to 7x == 3 (mod 11) is x = 7 1 (3) = 8(3) = 24 == 2 (mod 11). (b) Since 15(3) = 45 == 1 (mod 22), the inverse of 15 (mod 22) is 3. The solution to 15x == 9 (mod 22) is x = 15...
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 Summer '10
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 Graph Theory, Prime number

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