Proving statements by its contrapositive Prove the following statement by proving its contrapositive:
“If $n^3 + 2n + 1$ is odd then n is even” 
Therefore: $\lnot q \rightarrow  \lnot p =$ "if $n^3 + 2n + 1$ is even then $n$ is odd.
So for this I began assuming that: $n=2k+1$
$(2k+1)^3 +2(2k+1)+1 = 8k^3+12k^2 +10k+4 = 2k(4k^2 +6k+5)+4$
The last statement: $2k$ is even, therefore $2k(4k^2 -6k+5)$ is also even and 4 is $2\cdot 2$ which is also even.
Now, my question is, when proving the contrapositive, what's your final conclusion? If it works for the contrapositive, then your theorem holds? Or is there something else?

$\displaystyle n^3+2n+1=n^3-n^2+n^2+n+n+1$
$=n^2(n-1)+n(n+1)+n+1\equiv n+1\pmod2$ as the product of any two consecutive integer is divisible by $2$
$\implies n^3+2n+1$ and $n$ has opposite parity 
