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Math Help - [SOLVED] Proof of divisibility by Induction

  1. #1
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    [SOLVED] Proof of divisibility by Induction

    I am having difficulty with induction in several areas, however one question at a time.
    Ex. if 6 divides n(n^2 + 5) for all n >=1 prove for n+1.

    What I have so far is (k+1)((k+1)^2 +5)

    = k^3 + 5k + 3k^2 + 3k + 6

    which I have worked out to

    k(k^2 +5) + 3k^2 + 3k + 6

    I know 6| k(k^2 +5) but how to I prove 6 divides the rest?
    Am I able to use
    3k^2 mod 6 + 3k mod 6 + 6 mod 6 = 3+3+0 mod 6 = 0 mod 6 therefore also divisible by 6?
    Last edited by R.Laramee; November 14th 2007 at 07:10 PM. Reason: new idea
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  2. #2
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    Hello, R.Laramee!

    If 6 divides n(n^2 + 5) for all n \geq 1, prove for n+1.


    What I have so far is: . (k+1)[(k+1)^2 +5] \:= \:k^3 + 5k + 3k^2 + 3k + 6
    . . which I have worked out to: . k(k^2 +5) + 3k^2 + 3k + 6

    I know 6\,|\,k(k^2 +5), but how to I prove 6 divides the rest?
    Here one approach . . .

    Actually, we know that: . k(k^2+5) {\color{blue}\,+\, 6} is divisible by 6.
    . . So we are concerned with: . 3k^2 + 3k

    Note that: . 3k^2+3k \:=\:3\!\cdot\!k(k+1)
    . . This is: . 3 \times (\text{product of two consecutive integers})

    With two consecutive integers, one must be even and the other must be odd.

    Hence, k(k+1) is even . . . Therefore, 3k(k+1) is a multiple of 6.

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  3. #3
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    That is a really cool way of looking at it. Thanks
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