Let's fill in the blanks step by step for Case 1: n is even:

1. Blank 1: $n = 2k$ (By the definition of even, $n$ is a multiple of 2 for some integer $k$.)

2. Blank 2: $k^2 + 8k$ (This is the expression you get when simplifying $n^2 + 15n - 1$ in terms of $k$.)

3. Blank 3: $k^2 + 8k$ (Again, this refers to the expression $k^2 + 8k$ as part of the simplification process.)

4. Blank 4: $k^2 + 8k$ (This is the same expression $k^2 + 8k$ being referred to, which is an integer.)

5. Blank 5: odd (Since $n^2 + 15n - 1 = 2 \times$ an integer $+ 1$, it must be odd.)

So the full proof in this case should look like:

• Let $n$ be an even integer. By the definition of even, $n = 2k$ for some integer $k$.
• Substituting into the expression $n^2 + 15n - 1$, we get $(2k)^2 + 15(2k) - 1 = 4k^2 + 30k - 1 = 2(k^2 + 8k) + 1$.
• Since $k^2 + 8k$ is an integer, the expression $n^2 + 15n - 1$ equals $2 \times$ an integer $+ 1$, so it must be odd.

Would you like more details on this proof?

Here are 5 related questions you might be interested in:

1. How do you prove the second case when $n$ is odd?
2. What general strategy can we use to prove statements by cases?
3. How do odd and even numbers interact under different mathematical operations?
4. Can this method be applied to other types of quadratic expressions?
5. What are some common properties of odd and even functions in mathematics?

Tip: When working with even and odd numbers in algebra, always look for patterns that arise from using $2k$ for even and $2k + 1$ for odd integers.