Hi guys,
I need to show that for x>=1 x^3-x^5 is divisible by 12.
Not really sure how to start! :-s
Many thanks in advance. x
The trick you often use for these type of questions is proving that (x+1) and (x) give the same remainder.
Easy example: 4^x is always divisible by 4. You take (x+1) which is: 4^(x+1) = 4*4^x. When dividing with 4 you can throw out the 4 and you're left with 4^x. Now if you prove that x=1 is divisible by four, you've also proven it for x+1 and x+2 etc.
Your problem is a little harder ofcourse, and sometimes it's easier to use (x+2) or (x+3) instead of (x+1), but you'll just have to try a little. Hope this helped, and if you still can't solve it I'll try to give you a start.
Hello, AAM!
That's a silly way to write the problem.
Are they deliberately being annoying?
Let $\displaystyle N \:=\:x^3-x^5$Show that $\displaystyle x^3-x^5$ is divisible by 12 for $\displaystyle x \geq 1$
Then: .$\displaystyle N \;=\;-x^3(x^2-1)\;=\;-x^3(x-1)(x+1)$
Let's ignore the leading minus-sign.
$\displaystyle \text{We have: }\:N \;=\;\underbrace{(x-1)\cdot x\cdot(x+1)}_{\text{3 consecutive integers}}\cdot x^2$
With 3 consecutive integers, one of them is a multiple of 3.
. . Hence, $\displaystyle N$ is divisible by 3.
There are two cases: .$\displaystyle \begin{array}{c}\text{(1) }x\text{ is even.} \\ \text{(2) }x\text{ is odd.}\; \end{array}$
$\displaystyle \text{(1) If }x\text{ is even: }\;N \;=\;\underbrace{(x-1)}_{\text{odd}}\cdot\underbrace{x}_{\text{even}}\ cdot\underbrace{(x+1)}_{\text{odd}}\cdot\underbrac e{x^2}_{\text{even}} \quad\hdots\quad \text{Hence, }N\,\text{is divisible by {\color{red}4}.}$
$\displaystyle \text{(2) If }x\text{ is odd: }\:N \;=\;\underbrace{(x-1)}_{\text{even}} \cdot \underbrace{x}_{\text{odd}} \cdot\underbrace{(x+1)}_{\text{even}}\cdot\underbr ace{x^2}_{\text{odd}} \quad\hdots\quad \text{Hence, }N\text{ is divisible by {\color{red}4}.}$
. . Therefore, $\displaystyle N$ is divisible by 12.