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: 31

Answer

a. Let $m$ and $n$ be integers. Then either both are even, $m$ is even and $n$ is odd, $m$ is odd and $n$ is even, or both are odd. In the first case, we know from 4.2.3 (1) that the sum and difference of even integers is always even, so $m+n$ and $m-n$ are both even. In the second case, we know from 4.2.3 (5) that even plus odd makes odd, and from 4.2.3 (7) that even minus odd makes odd, so $m+n$ and $m-n$ are both odd. In the third case, we know from 4.2.3 (5) that odd plus even makes odd, and from 4.2.3 (6) that odd minus even makes odd, so $m+n$ and $m-n$ are both odd. Finally, in the fourth case, we know from 4.2.3 (2) that the sum and difference of odd integers is always even, so $m+n$ and $m-n$ are both even. In each of the four cases, either $m+n$ and $m-n$ were both even, or $m+n$ and $m-n$ were both odd. Therefore, since these are the only possible cases, it must be that the sum and difference of two integers are always either both even or both odd. b. $m=9$ and $n=5$, or $15$ and $13$ c. $m=13$ and $n=9$, or $m=23$ and $n=21$

Work Step by Step

b. Let $56=m^{2}-n^{2}=(m-n)(m+n)$ for some positive integers $m$ and $n$. By the unique factorization of integers, $56=2^{3} \times 7$ gives the unique prime factorization of $56$. Hence, the only ways to write $56$ as a multiple of two positive integers are as follows: $56=7\times 8=14\times 4=28\times 2=56\times 1$. By part (a) above, $m-n$ and $m+n$ are either both odd or both even, so we limit our options to $14\times 4$, and $28\times 2$. Since $m$ and $n$ are both positive, the larger of each pair must correspond to $m+n$, and the smaller to $m-n$. Therefore, the only solutions are given by the solutions to the following systems of equations: (1) $m+n=14$ and $m-n=4$; and (2) $m+n=28$ and $m-n=2$. Thus, we have either $m=9$ and $n=5$, or $m=15$ and $n=13$. c. Let $88=m^{2}-n^{2}=(m-n)(m+n)$ for some positive integers $m$ and $n$. By the unique factorization of integers, $88=2^{3}11$ gives the unique prime factorization of $88$. Hence, the only ways to write $88$ as a multiple of two positive integers are as follows: $88=11\times 8=22\times 4=44\times 2=88\times 1$. By part (a) $m-n$ and $m+n$ are either both odd or both even, so we limit our options to $22\times 4$ and $44\times 2$. Since $m$ and $n$ are both positive, the larger of each pair must correspond to $m+n$, and the smaller to $m-n$. Therefore, the only solutions are given by the solutions to the following systems of equations: (1) $m+n=22$ and $m-n=4$; and (2) $m+n=44$ and $m-n=2$. Thus, we have either $m=13$ and $n=9$, or $m=23$ and $n=21$.
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