Direct Proof with Quantifiers

Question

How can i use direct proof to prove that the function $n^2+5n+6$ is even when $n$ is odd and even, for all natural numbers?

Answer 1

If $n$ is even, then $n = 2k$ for some integer $k$. Now, $$\begin{align} n^2 + 5n + 6 &= (2k)^2 + 5(2k) + 6\\ &= 4k^2 + 10k + 6 \\ &= 2(k^2 +5k + 3) \\ &= 2m, \text{ for some integer } m \end{align}$$ where the last equality holds because the set of integers is closed under addition. That is, the sum of integers will always be an integer.

We can work it out similarly if $n$ is odd, i.e. $n=2k+1$ for some integer $k$.

User Chinnapparaj R gives another approach to this problem in their comment. Hopefully you find these both of some use.