Prove $w \in \mathbb{z} : k \mod 2 \neq 0 \rightarrow k \in \mathbb{z} : k^3 \mod 2 \neq 0$ directly

Question

Prove $$ k \in \mathbb{Z}: {k}\mod{2} \neq 0 \rightarrow k \in \mathbb{Z} : k^3 \mod 2 \neq 0$$ using direct proof.

I would like to prove that if integer $k$ is not divisible by 2 it implies that $k^3$ is not divisible by 2 either. Could someone provide hint on what should i start with ?

If $k$ is not divisible by $2$, then it must be $k=2n+1$ for some integer $n$ — Jakobian, Sep 13 '18 at 19:47

score 1 · Accepted Answer · answered Sep 13 '18 at 19:51

1

Let $k\in \Bbb Z \;\; : k \mod 2\ne 0.$

$$k \mod 2\ne 0\implies (\exists p\in \Bbb Z)\;\; :$$ $$ \; k=2p+1$$

$$\implies k^3=8p^3+12p^2+6p+1$$

$$\implies \exists q=(4p^3+6p^2+3p)\in \Bbb Z \; : $$ $$k^3=2q+1$$

$$\implies k^3 \mod 2\ne 0$$

answered Sep 13 '18 at 19:51

hamam_Abdallah

62,951

Prove $w \in \mathbb{z} : k \mod 2 \neq 0 \rightarrow k \in \mathbb{z} : k^3 \mod 2 \neq 0$ directly

1 Answers1