To prove that if $n$ is an integer and $n^3 + 5$ is odd, then $n$ is even, we will use proof by contradiction.

Proof by Contradiction

1. Assume the opposite of what we want to prove: Suppose $n$ is odd.

2. Express $n$ as an odd integer: If $n$ is odd, then we can write $n = 2k + 1$ for some integer $k$.

3. Calculate $n^3$ based on this assumption: $n^3 = (2k + 1)^3 = 8k^3 + 12k^2 + 6k + 1 = 2(4k^3 + 6k^2 + 3k) + 1$ Since $4k^3 + 6k^2 + 3k$ is an integer, we can conclude: $n^3 = 2 \times (\text{integer}) + 1$ Therefore, $n^3$ is odd.

4. Consider $n^3 + 5$: Since $n^3$ is odd, adding 5 (another odd number) results in an even number: $n^3 + 5 = (\text{odd}) + (\text{odd}) = \text{even}$

5. Contradiction: This contradicts the given information that $n^3 + 5$ is odd.

6. Conclusion: Since our assumption that $n$ is odd leads to a contradiction, $n$ must be even.

Thus, we have proven that if $n^3 + 5$ is odd, then $n$ must be even.

Would you like more details on any part of this proof or have any questions?

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1. How does proof by contradiction work in other types of problems?
2. Why does adding two odd numbers result in an even number?
3. Can you show an example of how to find specific values of $n$ where $n^3 + 5$ is odd?
4. Is there a direct proof method to solve this problem without contradiction?
5. How would the proof change if the requirement was that $n^3 + 5$ is even?

Tip: In proofs by contradiction, always carefully check each logical step, as one incorrect assumption could invalidate the proof.