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Math Help - Proving a multiple of 8

  1. #1
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    Proving a multiple of 8

    I am stuck on the end part here (or maybe im not) with this proof

    Prove n^2-1 is a multiple of 8 whenever n is an odd integer.

    Heres what I have done

    Let n = even integer
    therefore n = 2K k integer (im still unfamiliar with getting the symbols on the forums)

    (2k)^2 - 1 is a multiple of 8?

    4k^2 - 1 is multiple of 8,
    4k^2 = even. even - 1= odd so that doesnt work

    so
    Let n = odd integer
    Therefor n = 2k + 1 k integer

    (2k+1)^2 - 1 multiple of 8
    4k^2 + 4k +1 - 1 is a multiple of 8

    4k^2 + 4K is a multiple of 8.

    4k(k+1) is a multiple of 8..
    (k+1) is an integer.

    4k(k+1) is a multiple of 8. QED

    Is this correct, or am I missing any parts?
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  2. #2
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    You don't need to do the first bit.

    Quote Originally Posted by probowl8885 View Post
    Let n = odd integer
    Therefor n = 2k + 1 k integer

    (2k+1)^2 - 1

    Just continue on from here.
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  3. #3
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    I figure that, but everything afterwards is good to go? the odd integer part, that is fine the way it is or am I missing something?
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  4. #4
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    I was just noticing, but 4k(k+1), k and k+1 are consecutive integers, should that be noted? k and k+1 are consecutive, so one value is odd and one value is even, therefore odd*even = even. 4 * even is always a multiple of 8. is that correct to say?
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  5. #5
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    You mean n = 2k+1 implies an odd number. Works for me.

    Now you need to say something like as 4k^2 + 4k  = 4(k^2+k) is a multiple of 4 then 4(k^2+k) is also a multiple of 8 as the multiples of 8 are contained in the set that are multiples of 4. i.e. 8M = 4(2M)

    QED..
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