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: 54

Answer

$4(n^2 + n + 1) - 3n^2 = (n + 2)^2 $ $n$ is a integer, so $(n+2)^2$ and $4(n^2 + n + 1) - 3n^2$ are both perfect squares.

Work Step by Step

$4(n^2 + n + 1) - (3n^2)$ is a perfect square if it's equal to one integer squared Let $n$ be any integer $\begin{split} 4(n^2 + n + 1) - 3n^2 & = 4n^2 + 4n + 4 - 3n^2 \\ & = n^2 + 4n + 4 \\ & = (n)^2 + 2(n.2) + (2)^2 \\ \end{split}$ $4(n^2 + n + 1) - (3n^2)$ is a perfect square if it's equal to one integer squared $(n + 2)^2$ is a perfect square, then $4(n^2 + n + 1) - (3n^2)$ is a perfect square too. $4(n^2 + n + 1) - 3n^2 = (n + 2)^2 $
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