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Math Help - Divisibility Proofs

  1. #1
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    Divisibility Proofs

    Can anyone help me confirm if I've solved this correctly?

    Many thanks.

    Q.
    7^n+4^n+1 is divisible by 6 for n \in \mathbb{N}

    Attempt: Step 1: For n = 1...
    7^1+4^1+1=12, which can be divided by 6.
    Therefore, n = 1 is true...

    Step 2: For n = k...
    Assume 7^k+4^k+1 can be divided by 6,
    i.e. 7^k+4^k+1=6Z, where Z is an integer...1
    Show that n = k + 1 is true...
    i.e. 7^{k+1}+4^{k+1}+1 can be divided by 4
    7^{k+1}+7\cdot4^k-7\cdot4^k+4^{k+1}+1 => 7(7^k+4^k)+4^k(7-4)+1...from 1 above => 7(6Z-1)+4^k\cdot3+1 => 42Z-7+4^k\cdot3+1 => 42Z+4^k\cdot3-6 => 84Z+4^k\cdot6-12 => 6(14Z+4^k-2)
    Thus, assuming n = k, we can say n = k + 1 is true and true for n = 2, 3,... and all n \in \mathbb{N}_0
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  2. #2
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    Re: Divisibility Proofs

    The idea is correct, but there are several small flaws.

    Quote Originally Posted by GrigOrig99 View Post
    Therefore, n = 1 is true...
    Rather, P(1) is true where P(n) is " 7^n+4^n+1 is divisible by 6."

    Quote Originally Posted by GrigOrig99 View Post
    Show that n = k + 1 is true...
    Should be: "Show that P(k + 1) is true."

    Quote Originally Posted by GrigOrig99 View Post
    i.e. 7^{k+1}+4^{k+1}+1 can be divided by 4
    By 6.

    Quote Originally Posted by GrigOrig99 View Post
    7^{k+1}+7\cdot4^k-7\cdot4^k+4^{k+1}+1 => 7(7^k+4^k)+4^k(7-4)+1
    (7 - 4) should be (4 - 7).

    Quote Originally Posted by GrigOrig99 View Post
    ...from 1 above => 7(6Z-1)+4^k\cdot3+1 => 42Z-7+4^k\cdot3+1 => 42Z+4^k\cdot3-6 => 84Z+4^k\cdot6-12 => 6(14Z+4^k-2)
    What exactly does => mean here? Especially, what is the relationship between 42Z+4^k\cdot3-6 and 84Z+4^k\cdot6-12?

    Quote Originally Posted by GrigOrig99 View Post
    Thus, assuming n = k, we can say n = k + 1 is true
    Assuming n = k, the fact that n = k + 1 is obviously false.

    Edit: Added remark about 4 - 7.
    Last edited by emakarov; July 16th 2012 at 07:33 AM.
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  3. #3
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    Re: Divisibility Proofs

    Quote Originally Posted by GrigOrig99 View Post
    Can anyone help me confirm if I've solved this correctly?

    Many thanks.

    Q.
    7^n+4^n+1 is divisible by 6 for n \in \mathbb{N}

    Attempt: Step 1: For n = 1...
    7^1+4^1+1=12, which can be divided by 6.
    Therefore, n = 1 is true...

    Step 2: For n = k...
    Assume 7^k+4^k+1 can be divided by 6,
    i.e. 7^k+4^k+1=6Z, where Z is an integer...1
    Show that n = k + 1 is true...
    i.e. 7^{k+1}+4^{k+1}+1 can be divided by 4
    7^{k+1}+7\cdot4^k-7\cdot4^k+4^{k+1}+1 => 7(7^k+4^k)+4^k(7-4)+1...from 1 above => 7(6Z-1)+4^k\cdot3+1 => 42Z-7+4^k\cdot3+1 => 42Z+4^k\cdot3-6 => 84Z+4^k\cdot6-12 => 6(14Z+4^k-2)
    Thus, assuming n = k, we can say n = k + 1 is true and true for n = 2, 3,... and all n \in \mathbb{N}_0
    You have a sign error.

    \displaystyle \begin{align*} 7^{k + 1} + 4^{k + 1} + 1 &= 7^{k + 1} + 7 \cdot 4^k - 7 \cdot 4^k + 4^{k + 1} + 1 \\ &= 7\left(7^k + 4k\right) - 4^k\left(7 - 4\right) + 1 \\ &= 7\left(6Z - 1\right) - 3\cdot 4^k + 1 \\ &= 42Z - 7 - 3\cdot 4^k + 1 \\ &= 42Z - 3\cdot 4^k - 6 \\ &= 42Z - 3\cdot 4 \cdot 4^{k - 1} - 6 \\ &= 6\left(7Z - 2\cdot 4^{k - 1} - 1\right) \end{align*}
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    Re: Divisibility Proofs

    Ok, thanks guys,
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  5. #5
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    Re: Divisibility Proofs

    Another way to do it:

    7^n \equiv 1^n \equiv 1 (\mod 6)

    4^n \equiv 4 (\mod 6) (this is a little hard to prove, but checking the first few cases you will see that it always works).

    Therefore, 7^n + 4^n + 1 \equiv 1 + 4 + 1 \equiv 0 (\mod 6), therefore it is always divisible by 6.
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