Discrete Mathematics with Graph Theory (3rd Edition) 90

# Discrete Mathematics with Graph Theory (3rd Edition) 90 -...

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88 Solutions to Exercises 34. (a) [BB] Write 3n + 2 = P1P2 ... Pm as the product of (necessarily odd) primes. Each of the Pi is either 3 or of the form 3k + 1 or 3k + 2. Now we notice that the product of integers of the form 3k + 1 is of the same form and the product of 3 with such an integer is a multiple of 3. So if none of the Pi is of the form 3k + 2, the product cannot be either. If we drop the word odd, the result is false. For example, 3(2) + 2 = 8 is divisible by only the prime 2, which is not of the form 3n + 2 forn EN. (b) Write 4n + 3 = P1P2··· Pm as the product of (necessarily odd) primes. Now any odd prime is of the form 4k + 1 or 4k + 3. The product of any two integers of the form 4k + 1 is of the same form, so if all the Pi were of this form so would be the product, a contradiction. (c) Write 6n + 5 = P1P2 ... Pm as the product of primes. Each of the Pi is of the form 6k + 1 or 6k + 5. The product of two integers of the form 6k + 1 is also of this form, so at least one of the
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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.

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