Question

Use a direct proof to show that every odd integer is the difference of two squares.

Short answer

A difference of squares exists for every odd integer.

Step-by-step solution

Introduction

Consider the following statement:

Every odd integer is the difference of two squares.

A direct proof shows that a conditional statement \(p \to q\) is true, by showing that, p is true, and then q must also be true.

Assuming an odd integer and solving

Let n be any odd integer, then there is any integer k such that n=2k+1.

Let any integers k, k+1

\({\left( {k + 1} \right)^2} - {k^2} = \left( {{k^2} + 2k + 1} \right) - {k^2}\)

\( = {k^2} + 2k + 1 - {k^2}\)

\(\begin{aligned}&= 2k + 1\\ &= n\end{aligned}\)

Therefore,\(n = {\left( {k + 1} \right)^2} - {k^2}\)

Thus, a difference of squares exists for every odd integer.