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. 
 A: You started out exactly backwards when you began ‘I $n^2+n^3$ is an even number’: that’s taking as a hypothesis exactly the statement that you want to prove, so your argument can at best be circular (and hence useless).
There are two reasonably straightforward approaches; I’ll get you started on each of them.


*

*$n^2+n^3=n^2(n+1)$; show that no matter whether $n$ itself is odd or even, $n^2(n+1)$ must be even.

*If $n$ is even, then $n=2m$ for some integer $m$, and $$n^2+n^3=(2m)^2+(2m)^3=4m^2+8m^3=2\left(2m^2+4m^3\right)\;,$$ which is even. If $n$ is odd, then $n=2m+1$ for some integer $m$, and $$n^2+n^3=(2m+1)^2+(2m+1)^3=\;?$$
A: Hint
Pass to $\mod 2$ and take the two cases: $n\equiv 0 \mod 2$ and $n\equiv 1 \mod 2$.
A: Your proof fails when you reuse $n$. To express that $n^2+n^3$ is even you need a new integer variable, $m$ say, and write $n^2+n^3=2m$.
Moreover, You intend to show that if $n^2+n^3$ is even, then $n$ is an integer. You should show that if $n$ is an integer, then $n^2+n^3$ is even.
Incidentally, you conclude that $n\in\{0,1,2\}$, which should have made you suspiciuos. If you simply plug in $n=100$, you obtain $n^2+n^3=1010000$, an even number.
Then again, your idea of factoring is a good step. Just factor $n^2+n^3$ itself.
A: You can't assume that $n^2 + n^3$ is even; it's what you want to prove.
One way to do it is to note that $$n^2 + n^3 = n^2(n+1).  \qquad  (*)$$ If $n$ is even, then $n^2$ is even, so the right-hand side of $(*)$ is even.
If $n$ is odd, then $n+1$ is even, so the right-hand side of $(*)$ is still even.
In both cases, we have shown what we wanted to show.
A: 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.
