I'm studying for finals and reviewing this question on my midterm. My question is stated above and I can't quite figure out the proof.
On my midterm I used proof by contraposition by stating: If $n^2 +1$ is odd then $n^3$ is even.
I let $n^2+1 = (2m+1)^2 + 1$
$= (4m^2 + 4m + 1) + 1$
$= 2(2m^2 + 2m + .5) + 1$
Let $2m^2 + 2m + .5 = k$
$n^2 + 1 => 2k + 1$
Therefore proving that $n^2 + 1$ was odd making $n^3$ even.
I know my logic was messed up somewhere..some guidance would be nice.
If $n^2+1$ is odd then $n^2+1 = 2m+1 \longrightarrow n^2 = 2m \longrightarrow n^3 = 2mn \longrightarrow n^3=2(mn)$ so that $n^3$ is even.
You assumed $n$ is odd and proved $n^2+1$ is odd (which is false). You need to assume $n^2+1$ is odd and show that $n^3$ is even.
An alternate root might be easier: If $n^3$ is odd, then $n$ is odd. Use that to show that $n^2+1$ is even.
If $n$ is even, then $n=2k$ and $n^3=8k$, and we have that $n^3$ is even.So, if $n^3$ is odd then $n$ is odd, otherwise if $n$ is even, by the previous statement we would have $n^3$ even given a contradiction. So, $n$ is odd, and then $n=2t+1$ so $n^2+1=(2t+1)^2+1= 4t^2+2t+1+1=2(2t^2+t+1)$ then, $n^2+1$ is even.
As I mentioned in my comment, the original assumption you made in your proof was incorrect. Rather than letting $n^2 + 1 = (2m + 1)^2 + 1$ (i.e., $n = 2m + 1$), you should let $n^2 + 1 = 2m + 1$. Rustyn shows how to continue with this implication. However, I'll give a different method of proof, using modular arithmetic.
Suppose $n^3$ is odd. Then $n^3\equiv 1\pmod 2$. However, if we compute the two cubes mod $2$, we have $0^3\equiv 0$ and $1^3\equiv 1$, so that $n\equiv 1\pmod 2$. Hence, $n$ is odd.
Hint $\,\ n^2\!+\!1\,$ odd $\,\Rightarrow\, n^2$ even $\,\Rightarrow\, n^3$ even, $ $ i.e. $\,2\mid \color{#c00}{n^2}\,\Rightarrow\, 2\mid n \color{#c00}{n^2}.$
This can easily be solved with the knowledge that an odd number times an odd number is still and odd number? From here, it's easy to get $n$ is odd, $n^2$ is odd, $n^2 + 1$ is eve.
If $n^3$ is odd, then $n^3-1$ is even. Now, $n^3-1=(n-1)(n^2+n+1)$, since that one of $(n-1)$, $(n^2+n+1)$ (or both) must be even. ' Suppose that $n-1$ is even, then $n$ is odd, $n^2$ is odd and $n^2+1$ is even.
If $n-1$ is odd, then $n$ is even and $n^2+1$ is also odd. Therefore, $n-1$ must be even.
$ \newcommand{even}[1]{#1\text{ is even}} $Many answers already, but I would write down the solution for this yet differently, using the rules \begin{eqnarray} \tag{0} \even{k + m} \;\equiv\; \even k \equiv \even m \\ \tag{1} \even{k \times m} \;\equiv\; \even k \lor \even m \\ \end{eqnarray}
Obviously I'm writing this down in a lot of detail for this question.
We start with the most complex side, and calculate \begin{align} & \even{n^2 + 1} \\ \equiv & \qquad \text{"using $(0)$ and the fact that 1 is not even"} \\ & \lnot(\even{n^2}) \\ \equiv & \qquad \text{"write $\;n^2\;$ as $\;n \times n\;$; rule $(1)$"} \\ & \lnot(\even n \lor \even n) \\ \equiv & \qquad \text{"logic: add one more $\;\even n\;$ -- so that we can introduce $\;n^3\;$"} \\ & \lnot(\even n \lor \even n \lor \even n) \\ \equiv & \qquad \text{"rule $(1)$, twice; write $\;n \times n \times n\;$ as $\;n^3\;$"} \\ & \lnot(\even{n^3}) \\ \end{align}
This proves the required statement. (It actually also proves the other direction.)
