- #1

- 36

- 0

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:

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.