[Math] Proof by contrapositive: $x^3 + 1$ is even if and only if $x$ is uneven

Question

$x^3 + 1$ with $x \in \mathbb{Z}$ is even iff $x$ is uneven.

I want to prove this using a proof by contrapositive, so this is my work:

Assume that $x$ is even, so $x = 2k$ with $k \in \mathbb{Z}$. Then $n^3 + 1 = 8k^3 + 1 = 2(4k^3) + 1$. Since $4k^3$ is an integer, we have proven that $x^3 + 1$ is odd if $x$ is even.

I'm pretty sure this is technically correct, but I'm worried that this proof is incomplete because of the "if and only if", I assume that means I have to prove something else to fully prove the statement.

Answer 1

You're right, it is incomplete. In order to prove the other direction of the biconditional:

($x$ is odd) $\rightarrow$ ($x^3 + 1$ with $x\in \mathbb Z$ is even)

You need to prove its contrapositive:

$(x^3 + 1$ with $x\in \mathbb Z$ odd) $\rightarrow$ ($x$ is even.)