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Divisibility Proofs

  1. Jul 8, 2012 #1
    Can anyone help me confirm if I have solved this correctly?

    Many thanks.

    1. The problem statement, all variables and given/known data

    Q. [itex]9^n-5^n[/itex] is divisible by 4, for [itex]n\in\mathbb{N}_0[/itex]

    3. The attempt at a solution

    Step 1: For [itex]n=1[/itex]...
    [itex]9^1-5^1=4[/itex], which can be divided by [itex]4[/itex].
    Therefore, [itex]n=1[/itex] is true...

    Step 2: For [itex]n=k[/itex]...
    Assume [itex]9^k-5^k=4\mathbb{Z}[/itex], where [itex]\mathbb{Z}[/itex] is an integer...1
    Show that [itex]n=k+1[/itex] is true...
    i.e. [itex]9^{k+1}-5^{k+1}[/itex] can be divided by [itex]4[/itex]
    [itex]9^{k+1}-5^{k+1}[/itex] => [itex]9^{k+1}-9\cdot5^k+9\cdot5^k-5^{k+1}[/itex] => [itex]9(9^k-5^k)+5^k(9-5)[/itex] => [itex]9(4\mathbb{Z})+5^k(4)[/itex]...from 1 above => [itex]36\mathbb{Z}+5^k\cdot4[/itex] => [itex]4(9\mathbb{Z}+5^k)[/itex]

    Thus, assuming [itex]n=k[/itex], we can say [itex]n=k+1[/itex] is true & true for [itex]n=2,3,[/itex]... & all [itex]n\in\mathbb{N}_0[/itex]
  2. jcsd
  3. Jul 8, 2012 #2
    That's good. Some remarks on notation:

    Using [itex]\mathbb{Z}[/itex] is not good here. That is the symbol for the set of all integers. You should write [itex]9^k-5^k=4z[/itex] where z is an integer.

    The => should be =
  4. Jul 8, 2012 #3
    Ok. Thank you.
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