91Ó°ÊÓ

If \(m-n\) is even, then we can say that there exists an integer \(k\) such that \(m-n = 2k\). We have to show that both \(m\) and \(n\) are either even or odd. Consider two cases: Case 1: Assume \(m\) is even. This means there exists some integer \(a\) such that \(m = 2a\). Now substitute this value of \(m\) into the equation \(m-n = 2k\), so \[2a-n = 2k.\] To make the left side of the equation even, \(n\) must also be even. Thus, \(n = 2b\) for some integer \(b\). Now we have \(2a - 2b = 2k\), in which both \(m\) and \(n\) are even. Case 2: Assume \(m\) is odd. This means there exists some integer \(a\) such that \(m = 2a + 1\). Now substitute this value of \(m\) into the equation \(m-n = 2k\), so \[(2a + 1) - n = 2k.\] We need the left side of the equation to be even, so we can assume that \(n\) must also be odd. Thus, let \(n = 2b + 1\) for some integer \(b\). Now we have \[(2a + 1) - (2b + 1) = 2k.\] Expanding and simplifying this equation, we get \(2(a - b) = 2k\), in which both \(m\) and \(n\) are odd.

Now, we need to show that if both \(m\) and \(n\) are even or both odd, then \(m-n\) must be even. Case 1: If both \(m\) and \(n\) are even, then there exist integers \(a\) and \(b\) such that \(m = 2a\) and \(n = 2b\). Then, \(m-n = 2a-2b = 2(a-b)\), which is even. Case 2: If both \(m\) and \(n\) are odd, then there exist integers \(a\) and \(b\) such that \(m = 2a + 1\) and \(n = 2b + 1\). Then, \[m-n = (2a + 1) - (2b + 1) = 2(a-b),\] which is also even. Thus, we have proven that if both \(m\) and \(n\) are even or both odd, then \(m-n\) must be even.