It is hard to know where to start this, because we don't know much of anything about \(n\text{.}\) We might be able to prove that \(n^3 - n\) is even if we knew that \(n\) was even. In fact, we could probably prove that \(n^3-n\) was even if \(n\) was odd. But since \(n\) must either be even or odd, this will be enough. Here's the proof.

Proof.

We consider two cases: if \(n\) is even or if \(n\) is odd.

Case 1: \(n\) is even. Then \(n = 2k\) for some integer \(k\text{.}\) This give

\begin{align*} n^3 - n \amp = 8k^3 - 2k\\ \amp = 2(4k^2 - k)\text{,} \end{align*}

and since \(4k^2 - k\) is an integer, this says that \(n^3-n\) is even.

Case 2: \(n\) is odd. Then \(n = 2k+1\) for some integer \(k\text{.}\) This gives

\begin{align*} n^3 - n \amp = (2k+1)^3 - (2k+1)\\ \amp = 8k^3 + 6k^2 + 6k + 1 - 2k - 1\\ \amp = 2(4k^3 + 3k^2 + 2k)\text{,} \end{align*}

and since \(4k^3 + 3k^2 + 2k\) is an integer, we see that \(n^3 - n\) is even again.

Since \(n^3 - n\) is even in both exhaustive cases, we see that \(n^3 - n\) is indeed always even.

Solution 3.2.11.1.

Proof.