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Discrete Mathematics with Graph Theory (3rd Edition) 88

Discrete Mathematics with Graph Theory (3rd Edition) 88 -...

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86 Solutions to Exercises (c) If n is not a power of 2, we can write n = ab with b odd. Then 2 ab + 1 = (2 a + 1)(2 a (b-l) - 2 a (b-2) + ... + 1). Note the last tenn in the second factor is "+"1 because b is odd. 20. (a) If P = 2k + 1 and q = 2l + 1, then P + q = 2(k +.e + 1), so 2 1 (p + q). Also P + q > 2 since P > 2 and q > 2. Thus, P + q > 2 is divisible by 2 and so not prime. (b) No. Consider 2 + 3 = 5. 21. [BB] Suppose, to the contrary, that P and q are consecutive primes (in that order) such that P + q = 2r, where r is a prime. Then P = P + P < P + q = r < q + q = qj 2 2 2 that is, P < r < q. This contradicts the fact that P and q are consecutive primes. 22. Suppose there exist integers x and y such that x 2 - y2 = 2n. Then 21 (x + y)(x - y) so 2 1 (x + y) or 21 (x - y). But x +y = (x - y) + 2y shows that x + y and x - y are both even or both odd. Thus, 2 divides both x + y and x - y so 41 (x 2 - y2), 412n and 21 n, a contradiction. 23. [BB] This is true, and the proof has given Andrew Wiles a place in history. This is Fennat's Last Theorem (4.3.14)! 24. [BB] The condition tells us that in the prime decompositions of a, band e, no prime appears in two of the decompositions. Hence,
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