<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 $n$ 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.
</s>