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Math Help - Proof by Induction (n^4 - 4n^2)

  1. #1
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    [SOLVED] Proof by Induction (n^4 - 4n^2)

    Hi.

    I need to prove that n^4 - 4n^2 is divisible by 3.

    The induction hypnosis would be k^4 - 4k^2 is indeed divisible by 3, for k >= 0.

    What I don't understand is after we expand out (k+1) -4(k+1)^2, why do we need to subtract (n^4-4n)?

    I know that n^4 - 4n = 3t for some integer t (divisible by 3).

    http://courses.cs.vt.edu/~cs4104/heath/Spring2005/resources/induction.pdf

    The answer is given somewhere in the middle.

    Hence (n+1)^4 - 4(n+1)2 is divisible by 3 if (n+1)^4 - 4(n+1)^2-(n^4 -4n^2)
    is divisible by 3.
    Thanks!
    Last edited by jwxie; September 18th 2011 at 10:02 PM. Reason: solve
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  2. #2
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    Re: Proof by Induction (n^4 - 4n^2)

    Quote Originally Posted by jwxie View Post
    Hi.

    I need to prove that n^4 - 4n^2 is divisible by 3.

    The induction hypnosis would be k^4 - 4k^2 is indeed divisible by 3, for k >= 0.

    What I don't understand is after we expand out (k+1) -4(k+1)^2, why do we need to subtract (n^4-4n)?
    Here is the trick:
    (n+1)^4-4(n+1)^2=n^4-4n^2+6n^2+4n(n^2-1)-3

    Each of \color{blue}~(n^4-4n^2),~\&~(6n^2-3) is divisible by 3.

    Look at 4n(n^2-1)=4\color{blue}n(n-1)(n+1)

    But n(n-1)(n+1) is the product of three consecutive integers.
    Therefore it must be divisible by 3.
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  3. #3
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    Re: Proof by Induction (n^4 - 4n^2)

    Hi Plato,
    At first I thought you were showing me what they did in the solution. Then I started questioning myself "what trick?" and when I worked on the example I actually worked through the example, and found out that we did the same thing.
    Yes. Indeed. A nice trick Although I still don't get the reason why they would subtract anything in the first place.

    Thanks!
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