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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
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First factorise it into x³(1-x²).
Then prove that that is divisible by both 3 and 4.
Why did the vector cross the road?
It wanted to be normal.
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Thanks! :-)
Could you please give a few more details....?
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.....don't worry, I think I've got it now! :-) Thanks!
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.....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
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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
Why did the vector cross the road?
It wanted to be normal.
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Cool - thanks mathsyperson! :-)
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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)
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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).
Why did the vector cross the road?
It wanted to be normal.
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