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Math Help - Proof help

  1. #1
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    Proof help

    Let n be a positive integer:
    1.Factorize n^5-n^3 (which I got to be n^2(n+1)(n-1)) and show that it's divisible by 24.
    2. Prove that 2^2n is divisible by 3.
    3. If n-1 is divisible by three, show that n^3-1 is divisible by nine.

    I can explain using words more or less why, but I struggle to put them into mathematical language.

    Thanks in advance for your help.
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  2. #2
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    Hello, free_to_fly!

    I think I have a proof for #3 . . .


    3. If n - 1 is divisible by three, show that n - 1 is divisible by nine.

    We are told that n - 1 is divisible by 3.
    . . Hence: .n - 1 .= .3a, for some integer a.


    Multiply both sides by (n + n + 1): .(n - 1)(n + n + 1) .= .3a(n + n + 1)

    . . and we have: .n - 1 .= .3a(n + n + 1) .[1]


    The quadratic factor is: .n + n + 1 .= .(n - 1)(n + 2) + 3

    . . Since n - 1 .= .3a, it becomes: .3a(n + 2) + 3 .= .3(an + 2a + 1)

    . . So the quadratic factor is a multiple of 3: .3b, for some integer b.


    Substitute into [1] and we have: .n - 1 .= .3a(3b) .= .9ab, a multiple of 9.

    Therefore: .n - 1 is divisible by 9.

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  3. #3
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    Quote Originally Posted by free_to_fly View Post
    2. Prove that 2^2n is divisible by 3.
    Is there a mistake on this one?
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  4. #4
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    Hello again, free_to_fly!

    1. Factor N .= .n^5 - n^3, and show that it's divisible by 24.

    We have: .N .= .n(n - 1)(n + 1)

    . . . . . . . .N .= .n(n-1)n(n+1)
    . . . . . . . . . . . . . .\__________/

    N contains the product of three consecutive integers.
    . . The product of three consecutive integers is divisible by 3.


    Now we must show that N is divisible by 8.
    There are two cases to consider.

    (1) n is even: .n = 2k for some integer k.
    . . Then: .N .= .(2k)(2k - 1)(2k + 1) .= .8k(2k - 1)(2k + 1)
    Therefore, N is divisible by 8.

    (2) n is odd: .n = 2k + 1 for some integer k.
    . . Then: .N .= .(2k + 1)(2k)(2k + 2) .= .4k(k + 1)(2k + 1)
    Hence, N is divisible by 4.

    Note that k and k + 1 are consecutive integers.
    . . Hence, one of them is even, a multiple of 2.
    Therefore, N is divisible by 8.

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  5. #5
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    Mistakes

    Yep, it was meant to be:
    prove that (2^2n)-1 is divisible by three.

    Thanks fo pointing that out.
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  6. #6
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    Hello, free_to_fly!

    Prove that: .N .= .2^{2n} - 1 is divisible by three.

    We have: .(2)^{2n} .= .(2^2)^n .= .4^n

    . . Then: . N .= .4^n - 1 .= .(3 + 1)^n - 1


    Binomial expansion: . N .= .(3^n + n3^{n-1} + . . . + n3 + 1) - 1

    . . . . . . . . . . . . . . . .N .= .3^n + n3^{n-1} + . . . + n3

    . . . . . . . . . . . . . . . .N .= .3[3^{n-1} + n3^{n-2} . . . + n] ... a multiple of 3


    Therefore, N is divisible by 3.

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