Therefore, $a = 2m$ is even, $b = 2n$ is even, but $m$ and $n$ have opposite parity — so $a + b = 2(m + n)$, $b - a = 2(n - m)$ — wait, no: - AMAZONAWS

📅 March 11, 2026 👤 scraface

Mar 11, 2026

$Therefore, $a = 2m$ is even, $b = 2n$ is even, but $m$ and $n$ have opposite parity — so $a + b = 2(m + n)$, $b - a = 2(n - m)$ — wait, no:$