Divisibility question....

ASADM · 2009-03-16 10:45:03

Hi guys,

How could I go about proving that for all x>=1, x^3-x^5 is divisible by 12?

Any suggestion would be greatly appreciated! :-)

x

mathsyperson · 2009-03-16 10:56:48

First factorise it into x³(1-x²).

Then prove that that is divisible by both 3 and 4.

ASADM · 2009-03-16 11:49:59

Thanks! :-)

Could you please give a few more details....?

ASADM · 2009-03-16 11:53:39

.....don't worry, I think I've got it now! :-) Thanks!

ASADM · 2009-03-16 12:25:06

.....actually, I can do the divisible by 4 bits fine, but I'm struggling with showing things are divisible by 3. :-s

Eg - how can I show that 8n^3+12n^2+6n+1 is divisible by 3? (this is x^3 for odd x. ie - x = 2n+1 )

Many thanks. x

mathsyperson · 2009-03-16 13:17:28

For divisibility by 3, split it into three cases instead of odds and evens.

Prove it for:
x = 3n
x = 3n+1
x = 3n+2

ASADM · 2009-03-16 14:14:35

Cool - thanks mathsyperson! :-)

JaneFairfax · 2009-03-16 14:28:52

Mathsys method is cumbersome. For divisibility by 3, just factorize.

Any three consecutive integers will contain a multiple of 3 and hence their product is divisible by 3.

Last edited by JaneFairfax (2009-03-16 23:49:48)

mathsyperson · 2009-03-16 23:44:28

Ooh, that's nicer.

Now you've written it like that, it's easier to see that it can actually be divided by 8 as well (and hence by 24).

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