2
$\begingroup$

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.

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

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

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $m=4k^3-6k^2+2k+4$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy