# Prove that if $x$ is odd then the polynomial $x^3-x+7$ is odd.

Prove that if $$x$$ is odd then the polynomial $$x^3-x+7$$ is odd.

I know that for proofs about implications, you assume $$x$$ is odd is true, so I reasoned that $$\text{any odd number is of the form} ~~2k-1$$ and substituted $$x=2k-1$$ into the polynomial and simplified it to $$8k^3-12k^2+4k+7$$. I am not sure if to prove it is odd; do I need it in the form of $$2(\text{polynomial in k}) + 1$$?

Thanks for your help.

• it is to prove x^3-x+7 is odd – user803476 Oct 1 '20 at 20:02
• You already nearly arrived at $2\cdot(4k^3-6k^2+2k+4) -1$ – Hagen von Eitzen Oct 1 '20 at 20:15
• The sum of three odd numbers is odd (to suggest an approach to the question in the title) – Mark Bennet Oct 1 '20 at 20:36

## 4 Answers

In a simpler way, if $$x=2k+1$$ is odd then $$x^3=2h+1$$ is odd and then

$$x^3-x+7=2h+1-2k-1+7=2(h-k+3)+1$$

which is odd.

Or also by modular arithmetic

$$x\equiv 1 \pmod 2 \implies x^3-x+7\equiv 1-1+1\equiv 1 \pmod 2$$

Note that $$x^3-x=(x-1)x(x+1)$$ is the product of $$3$$ consecutive numbers, and at least one of these is even.

Thus $$x^3-x$$ is even and since $$7$$ is odd, so is their sum.

From what you have got $$8k^3-12k^2+4k+7=8k^3-12k^2+4k+8-1=2(4k^3-6k^2+2k+4)-1=2m-1$$, where $$m=4k^3-6k^2+2k+4$$

An odd # times an odd # is odd.
An even # times an even # is even.

Therefore
$$\forall ~x ~\in ~\mathbb{Z}, ~[x ~\text{odd}] ~\iff ~[x^3 ~\text{odd}].$$

Therefore $$\forall ~x \in ~\mathbb{Z}, ~[x^3 - x]~$$ is even.