Biconditional proof - Don't know if I'm wrong Let $n$ be a positive integer. Prove that $n^3 + 1$ is odd if and only if $n^2 − 1$ is odd. 
What kind of proofs did you use?
I chose direct proof;
$$n^3 + 1 = (2k+1)^3 +1 = 2(k^3+k^2+k+1) = 2k~$$
$$n^2 - 1 = (2k+1)^2 -1 = 2(2k^2+2k+1)   = 2k~$$
$$LHS = RHS$$
Am I doing something wrong? I felt like this was too straight forward. Am I missing out on including the "$n$ be a positive integer" somewhere in my proof?
Thanks in advance.
 A: You proved that if $n$ is odd the $n^k-1$ is even for $k=2,3$.
Hint for your statement:
$n^3 + 1$ is odd iff $n^3$ is even iff $n$ is even iff $n^2$ is even iff $n^2-1$ is odd.
A: $$n^3+1-(n^2-1)=n^2(n-1)+2$$
Now as $n(n-1)$ is even for all integer $n,$ 
and $n(n-1)$ divides $n^2(n-1),$
$n^2(n-1)+2$ is even
$\implies n^3+1,n^2-1$  must have the same parity.
A: If $n^2-1$ is odd, $n$ is even. Therefore $n^3$ is even, therefore $n^3+1$ is odd.
A: Here is a direct conditional proof that if n denotes a positive integer, then n^3 + 1 is odd if and only if n^2 - 1 is odd:
1)      ASSUMPTION:  n denotes a positive integer.
2)      The number n is odd, or n is even--per Step 1).
3)      If n^3 + 1 is odd, then n^2 - 1 is odd.
    DIRECT CONDITIONAL SUBPROOF

    a)      Assumption:  n^3 + 1 is odd.

    b)      The number n is even.

            DIRECT NON-CONDITIONAL SUBPROOF

            i)      The number n is not odd.

                    SUBPROOF BY CONTRADICTION

                    (1)     ASSUMPTION:  The number n is odd.

                    (2)     n^3 + 1 = (an odd number)^3 + 1--per Step (1)
                                    = (an odd number) + 1
                                    = (an even number).

                    (3)     There exists a contradiction--per Steps (2) and a).  Q.E.D.

            ii)     The number n is even--per the Disjunctive Syllogism rule applied to Steps 2) and i), respectively.  Q.E.D.

    c)      n^2 – 1 = (an even number)^2 – 1--per Step b)
                    = (an even number) - 1
                    = (an odd number).  Q.E.D.

4)      If n^2 - 1 is odd, then n^3 + 1 is odd.
    DIRECT CONDITIONAL SUBPROOF

    a)      ASSUMPTION:  n^2 - 1 is odd.

    b)      The number n is even.

            DIRECT NON-CONDITIONAL SUBPROOF

            i)      The number n is not odd.

                    SUBPROOF BY CONTRADICTION

                    (1)     ASSUMPTION:  The number n is odd.

                    (2)     n^2 – 1 = (an odd number)^2 – 1--per Step (1)
                                    = (an odd number) - 1
                                    = (an even number)

                    (3)     There exists a contradiction--per Steps (2) and a).  Q.E.D.

            ii)     The number n is even--per the Disjunctive Syllogism rule applied to Steps 2) and i), respectively.  Q.E.D.

    c)      n^3 + 1 = (an even number)^3 + 1--per Step b)
                    = (an even number) + 1
                    = (an odd number).  Q.E.D.

5)      The number n^3 + 1 is odd if and only if n^2 - 1 is odd--per Steps 3) and 4).  Q.E.D.
