<s>[INST] Techniques of Proof Prove If n is an integer, then $n^2+n^3$ is an even number. 
I don't know if I'm just reiterating what I'm asked to prove or if my ideas are actually proving the statement. 
If $n^2+n^3$ is an even number then $n^2+n^3=2n$ because $2n$ is an even number. By solving the equation by algebraic operations we get $n^2+n^3-2n=0$ and it follows that $n(n^2+n-2)=0$. By factoring we see that $n(n+2)(n-1)=0$ so $n= 0, 1, 2$ which are all integers so therefore $n$ must be an integer. 
 [/INST] We have two cases, because $n \in \mathbb{Z}$ is even or odd.
If $n$ is even, then there is a $k \in \mathbb{Z}$ such that $n=2k$, so
$$n^2+n^3=(2k)^2+(2k)^3=4k^2+8k^3=2(2k^2+4k^3)$$
with $q \in \mathbb{Z}$ such that $q=2k^2+4k^3$ we have $n^2+n^3=2q$, i.e. $n^2+n^3$ is even.
If $n$ is odd, then there is a $m \in \mathbb{Z}$ such that $n=2m+1$, so
$$\begin{align}n^2+n^3&=(2m+1)^2+(2m+1)^3\\&=(4m^2+4m+1)+(8m^3+12m^2+6m+1)\\
&=8m^3+16m^2+10m+2\\
&=2(4m^3+8m^2+5m+1)
\end{align}$$
with $p \in \mathbb{Z}$ such that $p=4m^3+8m^2+5m+1$ we have $n^2+n^3=2p$, i.e. $n^2+n^3$ is even.
In both cases $n^2+n^3$ is even. Therefore for all $n \in \mathbb{Z}$, $n^2+n^3$ is even.
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