Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.1 - Page 162: 36

Answer

There are no integers in where $6n^2 + 27$ is prime.

Work Step by Step

$6n^2 + 27$ can be factored as $3(2n^2 + 9)$. It is apparent that $2n^2 + 9$ is an integer and greater than $1$, so it will factor into two integers greater than one, one of which will be $3$
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