Question

Linear Algebra:

Answer 1

For all integers n,
\[n^3-n\] is divisible by 3.

How to show this?

Answer 2

what did you try so far?

Answer 3

I know that I can use induction method but not sure how to apply it yet so i've been trying to manipulate it to prove it by that method

Answer 4

Hint: factor n out, simplify to get the logic

Answer 5

\[n(n^2-1)\]

Answer 6

This is what i tried also but not sure how that helps me?

Answer 7

to show it is divisible by 3

Answer 8

more

Answer 9

n(n-1)(n+1)

Answer 10

Yup

Answer 11

Would I then try to get a 1/3 out?

Answer 12

Actually this not true for any integer?

Answer 13

you can try out some integers

Answer 14

\[(2)(2-1)(2+1) = 2(1)(3)\]

Answer 15

So i should set the condition: \[n \ge 1\]

Answer 16

you goal is to prove that n(n-1)(n+1)=3k
that is what it means to be divisible by 3

Answer 17

what about negative intergers?
what did the original problem state

Answer 18

for every positive integer is only thing that was stated

Answer 19

okay good
you stated for any integer n, which made think that there must be a problem

Answer 20

if it is positive integers that's good!

Answer 21

But when we use 1 there is a zero ? \[(1)(0)(2) = 0\]

Answer 22

well 0 divisible by any number we get back 0

Answer 23

True so then k is also zero

Answer 24

hey, Positive integer means >0 (strictly )

Answer 25

Non-negative is \(\geq \)0

Answer 26

eh good catch @Loser66 zero is not included

Answer 27

my bad!

Answer 28

so zero if zero is not included there must still be a problem?

Answer 29

what problem?

Answer 30

no you got me wrong! i meant not included in positive integers
but 0 is still divisible by 3

Answer 31

Ohhhh okay I see now so by expanding my original expression and plugging in positive integers I showed that it is divisible by 3 but how would i know to add the "3k" to the right of my expression for next time?

Answer 32

not sure i got you point here hehe

Answer 33

i was replying to @Johnbc

Answer 34

yes those consecutive

Answer 35

No worries

Answer 36

eh i have to go lol

Answer 37

Thank you very much I am still unsure about the right side but I understand the left

Answer 38

good luck with this^_^
@Loser66 gave you a hint to go with this

Answer 39

Plugging in P(n)?

Answer 40

Thank you

Answer 41

you can easily see for n=3 as he showed
but you need to do it for n and n+1

Answer 42

you are welcome