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

  • #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.
 

Answers and Replies

  • #2
34,056
9,922
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.
 
  • #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.
 
  • #4
34,056
9,922
There is something missing in the problem statement then.
 

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