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Show that for each a < b a, b ∈ N we have the following

  1. Jan 27, 2015 #1
    • Missing template due to originally being posted in different forum.
    1) 3^(2^a) + 1 divides 3^(2^b) -1

    2) If d > 2, d ∈ N, then d does not divide both 3^(2^a) + 1 and 3^(2^b) -1

    Attempt:

    Set b = s+a for s ∈ N

    m = 3^(2^a). Then 3^(2^b) - 1 = 3^[(2^a)(2^s)]-1 = m^(2^s) -1

    Thus, m+1 and m-1 divides m^(2^s) -1 by induction.

    If s = 1, then m^(2^s) -1 = m^2 - 1 = (m+1)(m-1)

    For s>= 1, m^(2^s) = (m^(2^(s-1))+1)(m^(2^(s-1))-1). The induction hypothesis approves.

    I'm confused on how to prove with the second condition.
     
  2. jcsd
  3. Jan 27, 2015 #2

    mfb

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    Are you sure the second problem statement is correct?
    3^(2^1)+1 = 10, 3^(2^2)-1 = 80.
    d=5 divides both 10 and 80.
     
  4. Jan 27, 2015 #3
    Yes. This is why it was confusing to me.

    I tried,

    d = m+1>2 which shows that it can divide both m+1 and m^(2^(s-1)) -1.
     
  5. Jan 28, 2015 #4

    mfb

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    There is something missing in the problem statement then.
     
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