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 190: 41

Answer

See below.

Work Step by Step

Based on the given conditions, we have m mod 5= 0, 1, 2, 3, 4 Case1: m mod 5= 0, we can write m=5n, thus $m^2=25n^2=5(5n^2)$; Case2: m mod 5= 1, we can write m=5n+1, thus $m^2=25n^2+10n+1=5(5n^2+2n)+1$; Case3: m mod 5= 2, we can write m=5n+2, thus $m^2=25n^2+20n+4=5(5n^2+4n)+4$; Case4: m mod 5= 3, we can write m=5n+3, thus $m^2=25n^2+30n+9=5(5n^2+6n+1)+4$; Case5: m mod 5= 4, we can write m=5n+4, thus $m^2=25n^2+40n+16=5(5n^2+8n+3)+1$; thus we have covered all the cases for any integer $m$ for the given statement to be true.
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