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 4.4 - Page 189: 23

Answer

Let $n$ be some integer such that $n\ mod\ 5=3$. Then by the definition of the $mod$ function, $n=5q+3$ for some integer $q$. Squaring both sides, we get $n^{2}=(5q+3)^{2}=25q^{2}+30q+9$$=5(5q^{2}+6q)+9=$$5(5q^{2}+6q)+5+4=$$5(5q^{2}+6q+1)+4$. Since $0\leq4\lt5$, and since $q^{2}+6q+1$ is an integer, we conclude by the quotient-remainder theorem and the definition of $mod$ that $n^{2}\ mod\ 5=4$. Since $n$ was an arbitrarily chosen integer, it must be that, if $n\ mod\ 5=3$ for any integer $n$, then $n^{2}\ mod\ 5=4$.

Work Step by Step

Page 180 gives the quotient-remainder theorem, and page 181 gives the definition of the $mod$ function.
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