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    Prove that 7^n + 4^n + 1 is divisible by 6 for all positive integers n.

    I showed it was true for n=1
    Assumed true for n=k
    7^k + 4^k + 1 = 6m where m is a positive integer
    Now to prove that the statement is true for n=k+1
    7^(k+1) + 4^(k+1) + 1
    7x7^k + 4x4^k + 1
    7x7^k +4(6m - 7^k -1) + 1
    7x7^k + 24m -4x7^k -4 +1
    3x7^k + 24m -3
    Now this is the part I always get stuck on. How do I proceed to take a factor of 6 out now?
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    (Original post by thers)
    Prove that 7^n + 4^n + 1 is divisible by 6 for all positive integers n.

    I showed it was true for n=1
    Assumed true for n=k
    7^k + 4^k + 1 = 6m where m is a positive integer
    Now to prove that the statement is true for n=k+1
    7^(k+1) + 4^(k+1) + 1
    7x7^k + 4x4^k + 1
    7x7^k +4(6m - 7^k -1) + 1
    7x7^k + 24m -4x7^k -4 +1
    3x7^k + 24m -3
    Now this is the part I always get stuck on. How do I proceed to take a factor of 6 out now?
    Consider the first and third terms together and notice that x^n - 1 = (x-1)(1+x+x^2+…+x^{n-1}).

    Alternatively, start again by considering the expression [7^{k+1} + 4^{k+1} + 1] - [7^k +4^k +1]. If you can show this is divisible by 6, then you're done (why?).
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    I tried that f(k+1)-f(k) method beforehand but it comes out as 6x7^k + 3x4^k which I can't take a factor of 6 out.
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    (Original post by thers)
    I tried that f(k+1)-f(k) method beforehand but it comes out as 6x7^k + 3x4^k which I can't take a factor of 6 out.
    Can you see a factor of 2 in that second term? Does this solve your problem?
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    (Original post by Farhan.Hanif93)
    Can you see a factor of 2 in that second term? Does this solve your problem?
    Unfortunately not. All I see is that 4^k can be written as 2^2k.
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    (Original post by thers)
    Unfortunately not. All I see is that 4^k can be written as 2^2k.
    Which can further be written as 2\times 2^{2k-1}.
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    (Original post by Farhan.Hanif93)
    Which can further be written as 2\times 2^{2k-1}.
    Ah right, got it now. Thank you.
 
 
 
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