I have one rather simple explanation for why I

think that the number of Fermat primes is finite.

Conjecture:

If a Fermat number of the form F(k,m)= 2^(2^((2^k)*m)) +1

is prime, then [2^(2^(2^k)*m) +1 == 2^q (mod 2^(2^k*m) +1)]

where 'k' is /-1 or a whole number/ and 'k' <= 'm' which is

a /whole number/ for some /odd number/ 'q'.

This argument further reduces to...[2^(2^k)*m) == (2n +/-1) mod

(2^k*m -k)] and is only satisfied when a combination of 'k' and 'm'

can be found to confirm that n= 0. The modulo equation is nasty,

but the results are wonderful!

Examine the following modulo equations:

expanded...

k=-1, m=0; [(2^((2^(-1))*0)) == (2n+1) mod ((2^(-1))*0) -(-1)]

=>[2^0 == (2n+1) mod (0+1)] => [1 == (2n+1) mod 1]

=> [0 == (2n) mod 1]=> [2n= 0] => [n= 0]; F0= 3 is prime,

continued...

k= 0, m=1; [2 == (2n-1) mod 1] => [0 == (2n) mod 1]

=> [2n= 0] => [n= 0]; F1= 5 is prime,

k= 1, m=1; [4 == (2n+1) mod 1] => [0 == (2n) mod 1]

=> [2n= 0] => [n= 0]; F2= 17 is prime,

k= 0, m=3; [8 == (2n-1) mod 3] => [0 == (2n) mod 3]

=> [2n= 0] => [n= 0]; F3=257 is prime,

k= 1, m=2; [16 == (2n+1) mod 3] => [0 == (2n) mod 3]

=> [2n= 0] => [n= 0]; F4= 65537 is prime.

but...

k= 0, m=5; [32 == (2n-1) mod 5] => [3 == (2n) mod 5]

=> [2n= 3] => [n= fraction]; and F5 is composite!

k= 0, m=6; [64 == (2n+1) mod 6] => [3 == (2n) mod 6] =>

[2n= 3] => [n= fraction]; and k= 1,m=3; [64 == (2n+1) mod 5]

=> [3 == (2n) mod 5] => [2n= 3] => [n= fraction]; & F6 is composite!

k= 0, m=7; [128 == (2n-1) mod 7]=> [3 == (2n) mod 7]

=> [2n= 3] => [n= fraction]; so F7 is composite!

and...

k= 0, m=10;[256 == (2n+1) mod 10]=> [5 == (2n) mod 10

=> [2n= 5]=> [n= fraction]; and k= 2,m=5; [256 == (2n+1) mod 8]

=> [7 == (2n) mod 8]=> [2n= 7] => [n= fraction]; & F10 is composite!

etc.

Hence, the number of Fermat primes is finite because the pattern

ends abruptly on the 5th number in the sequence, and thereafter.

Thanks, Bill