Two integers have the same parity if and only if their sum is even

(i) Let \(m\) and \(n\) be two integers having the same parity. Then \(m\) and \(n\) are either both even or both odd.

Case 1. (\(m\) and \(n\) both even.) If \(m\) and \(n\) are both even, then \(m=2k\) and \(n=2j\) for some integers \(k,j\). Then \[m+n=2k+2j=2(k+j)\] and hence \(m+n\) is even.

Case 2. (\(m\) and \(n\) both odd). If \(m\) and \(n\) are both odd, then \(m=2k+1\) and \(n=2j+1\) for some integers \(k,j\). Then \[m+n=(2k+1)+(2j+1)=2(k+j+1)\] and hence \(m+n\) is even.

(ii) We will prove the reverse implication, that two integers have the same parity if their sum is even, by contraposition. Hence, suppose that \(n\) and \(m\) have opposite parity. Without loss of generality we may assume that \(m\) is even and \(n\) is odd. Then \(m=2k\) and \(n=2j+1\) for some integers \(k,j\). So \[m+n=2k+(2j+1) = 2(k+j)+1,\] which proves that \(m+n\) is odd. Thus by contraposition we have proved that if their sum is even, two integers have the same parity.