1. **Problem:** Prove that for all integers $n$, the expression $n^2 + n + 3$ is an odd number using direct proof. 2. **Formula and rules:** An integer is odd if it can be written as $2k + 1$ for some integer $k$. 3. **Proof:** - Start with $n^2 + n + 3$. - Factor the first two terms: $n^2 + n = n(n+1)$. - Note that $n$ and $n+1$ are consecutive integers, so one of them is even, making $n(n+1)$ even. - Let $n(n+1) = 2k$ for some integer $k$. - Substitute back: $n^2 + n + 3 = 2k + 3$. - Rewrite $3$ as $2 + 1$: $2k + 2 + 1 = 2(k + 1) + 1$. - Since $k + 1$ is an integer, $n^2 + n + 3$ is of the form $2m + 1$, which is odd. **Final answer:** $n^2 + n + 3$ is always odd for all integers $n$.