Parity

Proof. At least one. Let \(n \in \Z \) and let \(q\) be the greatest integer with \(2q \leq n\). Then \(2(q + 1) > n\), so \(r = n - 2q\) satisfies \(0 \leq r < 2\). If \(r = 0\), then \(n = 2q\) is even. If \(r = 1\), then \(n = 2q + 1\) is odd.

Not both. If \(n = 2a = 2b + 1\) with \(a, b \in \Z \), then \(2(a - b) = 1\). But \(2(a - b)\) is \(0\) if \(a = b\), and at least \(2\) in absolute value otherwise. So it is never \(1\).