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Math Help - Just about done with this problem, but I am stuck on this last part!!

  1. #1
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    Just about done with this problem, but I am stuck on this last part!!

    Q: Prove that 6 divides n^(3) - n whenever n is a nonnegative integer?
    __________________________________________________ _______________
    What I have is this:

    Basis Step:
    If n = 0, then f(n)= n^3 - n = 0^(3) - 0 = 0 - 0 = 0.

    So the basis step is true since its divisible by 6.
    Inductive Hypothesis:
    We assume true, if n = k, then f(k) = k^(3) - k.

    We must prove that if n = k + 1, then f(k+1) = (k+1)^(3) – (k+1)
    Inductive Step:
    f(k+1) = (k+1)^(3) – (k+1)
    = (k + 1)(k + 2)^(2) - (k + 1)
    = (k+1)(k^(2) + 2k + 1) – (k+1)
    = k^(3) + 2k^(2) + k + k^(2) + 2k + 1 – k – 1
    = k^(3) + 3k^(2) + 3k + 1 – k – 1
    = k^(3) + 3k^(2) + 3k – k
    = (k^(3) - k) + (3k^(2) + 3k)
    = (k^(3) - k) + 3(k^(2) + k)
    = f(k) + 3(k^(2) + k)


    Where I am stuck at is I know f(k) is divisible by 6 but the 3(k^(2) + k) is divisible by 3?
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  2. #2
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    Hello, Grillakis!

    Your work is excellent!


    Prove that 6 divides n^3 - n for a nonnegative integer n.


    What I have is this:

    Basis Step:
    If n = 1, then f(1)\:=\:1^3 - 1 = 0
    . .
    So the basis step is true since its divisible by 6.

    Inductive Hypothesis:
    We assume true: if n = k, then: f(k) \:=\:k^3 - k is divisible by 6.

    We must prove that if n \,=\, k + 1,
    . . then: . f(k+1) \:=\:(k+1)^3  (k+1) is divisible by 6.


    Inductive Step:
    f(k+1) \:=\:(k+1)^3  (k+1)
    . . . . . . = \;k^3 + 3k^2 + 2k
    . . . . . . = \;(k^3 - k) + (3k^2 + 3k)
    . . . . . . = \;f(k) + 3(k^2 + k)


    Where I am stuck: is 3(k^2 + k) divisible by 6? . . . . Yes!

    We know that f(k) is divisible by 6: . f(k) = 6a,\:\text{ for some integer }a.


    \text{So we have: }\;f(k+1)\:=\:6a + 3\underbrace{k(k+1)}

    Note that k(k+1) is the product of two consecutive integers.
    Hence, one of them is even and the other is odd.
    And their product is even, say, 2b,\:\text{ for some integer }b.

    So we have: . f(k+1)\:=\:6a + 3(2b) \:=\:6(a + b)\;\hdots\;\text{ which is divisible by 6.}


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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Grillakis!

    Your work is excellent!



    We know that f(k) is divisible by 6: . f(k) = 6a,\:\text{ for some integer }a.


    \text{So we have: }\;f(k+1)\:=\:6a + 3\underbrace{k(k+1)}

    Note that k(k+1) is the product of two consecutive integers.
    Hence, one of them is even and the other is odd.
    And their product is even, say, 2b,\:\text{ for some integer }b.

    So we have: . f(k+1)\:=\:6a + 3(2b) \:=\:6(a + b)\;\hdots\;\text{ which is divisible by 6.}


    Thanks Soroban....I understand that now.
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