<s>[INST] Prove if $n^3$ is odd, then $n^2 +1$ is even 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.
 [/INST] 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. 
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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.
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