2^(2^m) + 1 mod prime

3 cannot divide any number in this series because (2 -1)^2 + 1 = 2

5 cannot divide any number in this series except 5 because (2 - 1)^2 + 1 = 2

7 cannot divide any number in this series because (5 - 1)^2 + 1 = 3 mod

7 and (3-1)^2 + 1 = 5

mod 11 repeats the sequence 5,6,4,10

mod 13 repeats the sequence 4,10

17 cannot divide any number in this series because (2 - 1)^2 + 1 = 2

In general no number in this series divides any other number in the series

In order for 2^(2^m) + 1 to be divisible by p,

the equation (x - 1)^2 + 1 = 0 must have a solution mod p.

(x - 1)^2 + 1 = x^2 - 2 x + 1 + 1 = x^2 - 2x + 2

x^2 - 2 x + 2 = 0

x = 1 + i or x = 1 - i

In order for 2^(2^m) + 1 to be divisible by p, p must be = 1 mod 4.

In order for 2^(2^m) + 1 to be divisible by p,

one of the equations

x^2 - 2x + 2 = 1 + i

or

x^2 - 2 x + 2 = 1 - i

must have a solution mod p

One of the equations

x^2 - 2x + 1 - i = 0

or

x^2 - 2 x + 1 + i = 0

must have a solution mod p.

x1 = 1 + sqrt(i) or x1 = 1 - sqrt(i)

x2 = 1 + sqrt( -i) or x2 = 1 - sqrt(-i)

Is it true that if any of these exist, mod p, then all of them exist?

In mod 641, we may take i = 155, and we may take sqrt(i) = 257.

In general, which primes, p, divide a number of the form 2(2^m) + 1

will be related to the existence of

sqrt(-1), sqrt(sqrt(-1)), sqrt(sqrt(sqrt(-1))) , etc mod p.

Kermit <

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