In trying to show that $n$ is even, is my final solution correct?
First: If $n$ is even then $n^3+n$ is even.
Since $n$ is even, then: $$n=2\cdot s$$
$$n^3+n = (2\cdot s)\cdot (2\cdot s)\cdot (2\cdot s) + 2\cdot s$$
$$= 6\cdot s^3 + 2\cdot s$$
On
Yeah, it's (in essence) correct, beside the miscalculation with the $6\cdot s^3$ term. I'd write it this way, for presentation purposes:
Suppose $n$ is even. Thus, $\exists k \in {Z}$ such that $n=2k$.
$n=2k \implies n^3+n=(2k)^3+2k = 2(2^2k^3+k),$ where $(2^2k^3+k)=q \in Z.$
Thus, $q\in Z \implies \exists q \in Z$ such that $n^3+n = 2q.$
Therefore, $n^3+n$ is even.
Now, suppose we wanted to show that if $n$ is odd, then $n^3+n$ is even:
Suppose $n$ is odd. Thus, $\exists k \in {Z}$ such that $n=2k+1$.
$n=2k \implies n^3+n=(2k+1)^3+2k+1 $
$$= 8k^3+12k^2+8k+2,$$
$$= 2(4k^3+6k^2+4k+1),$$
where $$(4k^3+6k^2+4k+1)=q \in Z.$$
Thus, $q\in Z \implies \exists q \in Z$ such that $n^3+n = 2q.$
Therefore, $n^3+n$ is even.
Almost. It should be $8s^3$. Close, though! :)