# Is this valid to prove that $2 \ \vert \ x^3 \Rightarrow 2 \ \vert \ x$?

Proof by contrapositive:

Let $$x \in \mathbb{Z}$$. Assume that $$2 \nmid x$$. Thus, $$\forall k \in \mathbb{Z}$$, $$2k \neq x \Rightarrow (2k)^3 \neq x^3 \Rightarrow 8k^3 \neq x^3 \Rightarrow 2(4k^3) \neq x^3$$. B/c $$4k^3 \in \mathbb{Z}$$, $$2 \nmid x^3$$. $$\therefore 2 \nmid x \Rightarrow 2 \nmid x^3$$, so the contrapositive $$2 \ \vert \ x^3 \Rightarrow 2 \ \vert \ x$$ is true.

I didn't realize that $$2 \nmid x \Rightarrow x$$ is odd when I did this proof, but is the above method valid?

• "B/c $4k^3 \in \mathbb{Z}$, $2 \nmid x^3$. " I'm not following the reasoning here. I'm not convinced that $2 \nmid x^3.$ May 15 at 22:00

This method is not valid. You showed that $$2(4k^3)\neq x^3$$ for all $$k\in\mathbb Z$$ correctly, but but to show that $$2\nmid x^3$$, you need to show that $$2n\neq x^3$$ for all integers $$n$$, not just integers of the form $$4k^3$$ for some $$k\in\mathbb Z$$.

To illustrate why this is important, consider the following (false) statement: $$4\mid x^2\Rightarrow 4\mid x$$. Using the proof technique proposed we could proceed to "prove" this statement by contrapositive:

Suppose $$4\nmid x$$. Then $$4k\neq x$$ for any $$k\in \mathbb Z$$, and $$16k^2=4(4k^2)\neq x^2$$ for any $$k\in\mathbb Z$$.

But if we consider $$x=6$$, the flaw in the argument becomes clear: while $$6^2=36=4\cdot 9$$, $$9$$ cannot be written as $$4k^2$$ for any integer $$k$$. Thus, even though $$4\nmid 6$$, $$4\mid 6^2$$.

• You can get the right formatting with \not\mid by the way May 15 at 21:22
• @pancini Thanks! you learn something new every day. May 15 at 21:25

There are a couple issues:

• You didn't really show that $$2$$ does not divide $$x^3$$. Rather, you showed that, for all $$k\in\Bbb Z$$, $$x^3\neq 2(4k^3)$$. It's not clear why $$x^3$$ must be of the form $$2(4k^3)$$ if $$2|x$$.
• The statement $$2k\neq x\implies (2k)^3\neq x^3$$ is true, but only because $$x\mapsto x^3$$ is injective. (Note that $$2k\neq x$$ does not imply $$(2k)^2\neq x^2$$, for example.)

Here's a hint for another approach: a non-unit $$p\in\Bbb Z$$ is prime $$\iff$$ $$p|ab$$ implies $$p|a$$ or $$p|b$$ for all $$a,b\in\Bbb Z$$.

Suppose $$2 \mid x^3$$.

Assume $$2 \nmid x$$,

then $$\exists r \in \mathbb{N}$$ such that $$x = 2r + 1$$.

Then $$x^3 = (2r + 1)^3$$ $$=8r^3 + 12r^2 + 18r + 1$$

Since $$2 \mid x^3$$, then it follows that $$2 \mid (8r^3 + 12r^2 + 18r + 1)$$, But, while $$2 \mid 8r^3$$, $$2 \mid 12r^2$$ and $$2 \mid 18r$$, $$2 \nmid 1$$

Now we arrive at a contradiction such that:

$$x^3 = 8r^3 + 12r^2 + 18r + 1$$ $$2 \mid x^3$$ but $$2 \nmid (8r^3 + 12r^2 + 18r + 1)$$

Therefore, the original assumption is not correct, and that it is indeed true that $$2 \mid x$$.

$$2 \mid x^3 \implies 2 \mid x$$