Prove that the set of all odd integers cubed is a proper subset of the odds I'm working through a "transitions to higher math" type of book and want to prove the following. (So no abstract algebra, number theory, etc. as the book only assumes up to some simple calculus.)
Let $A = \{ (2n + 1)^3 | n \in \mathbb{Z} \}$ and $B = \{ 2n + 1 \vert n \in \mathbb{Z} \}$. Prove that $A \subset B$ (i.e., $A$ is a proper subset of $B$).

*

*To show that $A \subseteq B$, let $x \in A$. Then $\exists n \in \mathbb{Z}: x = (2n + 1)^3$. Expand the term and factor to get $x = 2 (4n^3 + 6n^2 + 3n) + 1$. Since $n$ is an integer, the term in parentheses is as well, say $m = 4n^3 + 6n^2 + 3n$. Then $x = 2m + 1$ and so $x \in B$.


*To show that $A$ is a proper subset, I must also show that $A \neq B$. This amounts to showing that $\exists b \in \mathbb{Z}: b \in B \land b \notin A$.
Suppose $n = 1$. Then $b = 2n + 1 = 3$ and $3 \in B$. We claim that $3 \notin A$.
Suppose otherwise. Then $\exists m \in \mathbb{Z}: 3 = (2m + 1)^3$.
$$
\begin{align}
3 & = (2m + 1)^3 \\
& = 8m^3 + 12m^2 + 6m + 1 \\
\iff 2 & = 2(4m^3 + 6m^2 + 3m) \\
\iff 1 & = 4m^3 + 6m^2 + 3m \\
\iff 1 & = m(4m^2 + 6m + 3) \\
\end{align}
$$
For the last equation to be true, we must have both $m = 1$ and $4m^2 + 6m + 3 = 1$. If we set them equal to one another, after some algebra we get $4m^2 + 5m + 3 = 0$. Simple execution of the quadratic formula shows that $m$ (both solutions) is complex, contradicting our assumption that $m \in \mathbb{Z}$.
Is this a correct proof? Is there a simpler or more elegant solution?
 A: Another approach is to note that the function $f(x)=x^3$ has derivative $f'(x)=3x^2 \geq 0$ for all $x$, so $f(x)$ is monotonically increasing on any interval that doesn't include $0$.  And since $x \lt 0 \Rightarrow f(x) \lt 0$ and $x \gt 0 \Rightarrow f(x) \gt 0$, it's also true that $f(x)$ is monotonically increasing on any interval that contains $0$.  Thus, $f(1) \lt 3$ and $f(2) \gt 3$ is enough to demonstrate that $3$ is not the cube of any integer.
A: To show $3$ is not a cube, note that $|n^3| \ge |n|$ for any integer $n$.  So you only have to check that $n^3 \ne 3$ for $n = -3, -2, -1, 0, 1, 2, 3$...
If you want to show $3$ is not the cube of an odd number, you have even fewer to check.
A: Your approach seems logical to me, and the proof does hold water, so well done. That said, if I had to give a critique, you can save quite a bit of trouble for yourself with the second portion, proving that $3$ is not a cube.
For starters, the bit at the end about solving the quadratic equation $4m^2 + 5m + 3 = 0$ is unnecessary, since all that we would need to yield a contradiction would be to say that $m = 1$ is not a solution. From where you were, I might write "Because $m = 1, 4m^2 + 5m + 3 = 4(1)^2 + 5(1) + 3 = 12,$ but because we know that $4m^2 + 5m + 3 = 0,$ this means that we've obtained $12 = 0,$ which is a clear contradiction." I find paths like these which cut out extra calculation to be both more convincing, and quite satisfying.
Also, although like I said I think the way you went about the whole second part made a lot of sense for trying to find a first solution, there are other lighter-weight options which I would recommend. For instance, the bog-standard proof of the irrationality of $\sqrt{2}$ can be adapted to show that $3^{1/3}$ is irrational, and therefore not an integer, and it would require less algebra than this.
Going even further, @RobertIsrael's guess-and-check approach is pretty reasonable in terms of the amount of work required, but it gets a touch ugly when we try to generalize it, so I would recommend a method where we begin by proving that the cube root function is always increasing. Calculus trivializes this, but here's a proof which goes without:

Lemma: For two real numbers $m$ and $n,$ if $m^3 > n^3,$ then $m > n.$
Proof: If $m^3 > n^3,$ then $m^3 - n^3 = (m - n)(m^2 + mn + n^2) > 0,$ and because $m^2 + mn + n^2 = (m + \frac12 n)^2 + \frac34 n^2 > 0,$ we can divide both sides of the inequality by this to obtain $m - n > 0,$ so $m > n.$

This implication ends up going both ways, but this is all we'll need for my recommended method:

Proof that $3$ is not the cube of an odd integer using the fact that the cube root function increases: Suppose that $n^3 = 3$ for some odd integer $n.$ Because $n^3 > 1^3 = 1,$ we must have that $n > 1,$ but because $3^3 = 27 > n^3,$ we must have $3 > n.$ This means $n$ must be between $1$ and $3,$ but there is no odd integer between $1$ and $3,$ so this causes a contradiction.

So, this method required a bit more of a lateral first step, proving the lemma, but after that we get a method which could be used to prove any integer between the cubes of two consecutive integers cannot be the cube of an integer, which only requires calculating two values.
To be clear, it's perfectly reasonable that your first tack had some inefficiencies, there's a very clear difference in mindset between getting a workable proof and going back to try to improve it. I actually had to do that myself several times while writing out the proof for my lemma, since my original ideas were a bit out of order. (and believe it or not, this version is better than what I had to start, even though I do have a tendency to get scatterbrained) It's part of the writing process.
