The Fermat Numbers are of the form: 22n+1 (3, 5, 17, 257, 65537...) The conjecture is whether a prime of this form exists past the number F4 or 224 + 1 = 65537.
First I have to digress to base ten and the Least Positive Exponent that creates 1 modulo m, m relatively prime to ten. In other words 10a - 1 = 0 mod(m). m must be odd since it is relatively prime to 2*5. "a" is also the number of repeating digits in the decimal expansion of m. (This tidbit thanx to Joe Cutrone at Boston College)
If a is even or equal to 2n then m divides 10n + 1, or 10n + 1 = 0 mod(m)
Proof:
By definition a or 2n is the Least Positive Exponent, compare to Euler's Totient Function, that makes 10a - 1 = 0 mod(m) or
102n = 1 mod(m)
Factoring 102n - 1 = (10n - 1)(10n + 1)
Therefore m must divide either (10n - 1) or (10n + 1)
since 102n = (10n)(10n)
102n = 1 mod(m) and (10n)(10n) = 1 mod(m)
10n = -1 mod(m) since (-1)(-1) = 1 the congruence is satisfied and
10n + 1 = 0 mod(m)
10n = 1 mod(m) cannot be true since that would contradict the original premise that 102n is the Least Positive Exponent.
A few examples:
106 - 1 = 0 mod(7), 103 + 1 = 0 mod(7), (7)(143)
106 - 1 = 0 mod(13), 103 + 1 = 0 mod(13), (13)(77)
Now for base 2 or 2a - 1
F5 = 225+1 = 232 + 1 = 4294967297
if 232 + 1 is composite, then there exists a number who's least positive exponent is 64 or has a repeating decimal in base two of 64 digits. 264 - 1 = 0 mod(x) then 232 + 1 = 0 mod(x)
64110 = 10100000012 it has a repeating decimal of:
.00000000011001100011110110000000111111111001100111000010011111112 or 64 digits
with the above proof, 232 + 1 = 0 mod(641)
(more to come, like what happens with Mersenne Primes, etc.)