Proofs of divisibiliy of Fermat numbers of order m
                                                       Cino Hilliard  07/07/07

A Fermat number F(n) is a number that is 2 raised to a power of 2 + 1 = 2^(2^n) + 1.

A Fermat number of order m is a  number that is 2 raised to a power of 2 + m = 2^(2^n) + m.

Fermat numbers do not have special divisibility properties by a given number such as 3,7,11 etc
whileFermat numbers of order m do. For example m=3 we have 2^(2^n)+3 is always divisible
by 7 for odd values of n. Actually, Fermat numbers are Fermat numbers of order 1.

We prove this and other cases by induction making use of the following expansions where
appropriate.

Remark: The n used here is local to these expressions and is not to be confused with the n in
the theorems.

For any integers a and b, 
(1.1)   a^n - b^n = (a-b)(a^(n-1) + a^(n-2)b + . . . + b^(n-1))  and
(1.2)   a^n - b^n = (a+b)(a^(n-1) + a^(n-2)b + . . . + b^(n-1))  for even n.
(1.3)   a^n + b^n = (a+b)(a^(n-1) + a^(n-2)b + . . . + b^(n-1))  for odd n.

Remark: All variables used in the theorems are local to the theorems.

Theorem  1.
If n is odd, r=3,  2^(2^n)+3 = 7h for some integer h.
Proof:
Since we are restricted to n odd, we re-write 2^(2^n)+3 = 2^(2^(2m+1))+3 = 4^(2m) + 3 =
(2)  4^(4^m) + 3.
For m = 1, this expression  is 4^4+3 = 7*37. So the statement is true for m = 1. Assume the
statement is true for some integer k and show the statement is true for k+1. So 4^(4^k) +3 = 7h
for some h. Let 4^(4^(k+1))+ 3 = h1. Now consider the difference h1 - 7h. If this difference is
a multiple of 7 then so is h1. So we have 4^(4^(k+1))+ 3 - (4^(4^k) + 3)) = 256^(4^k)-4^(4^k).
This is of the form (1.1) where n = 4^k, a = 256 and b = 4. So the difference as in (1.1), a^n-b^n
is  divisible by (a-b) = (256-4) = 252 = 7*36. This implies 4^(4^(k+1)+3 is divisible by 7. So we
hav assumed the statement was true for k and have shown it to be true for k+1. Therefore, by the 
induction hypothesis, the statement is true for all m.

Theorem  2.
If n is even, r=5,  2^(2^n)+5 = 7h for some integer h.
Proof:
Since we are restricted to n even, we re-write 2^(2^n)+5 = 2^(2^(2m))+5 = 2^(4^m) + 5.
For m = 1, this expression  is 2^4+5 = 7*3. So the statement is true for m = 1. Assume the
statement is true for some integer k and show the statement is true for k+1. So 2^(4^k) +5 = 7h
for some h.Let 2^(4^(k+1))+ 5 = h1. Now consider the difference h1 - 7h. If this difference is a
multiple of 7 then so is h1. So we have 2^(4^(k+1))+ 5 - (2^(4^k) + 5)) = 16^(4^k)-2^(4^k).
This is of the form (1.1) where n = 4^k, a = 16 and b = 2. So the difference as in (1.1), a^n-b^n
is divisible by (a-b) = (16-2) =14=7*2. This implies 2^(4^(k+1)+5 is divisible by 7. So we have
ssumed the statement was true for k and have shown it to be true for k+1. Therefore, by the
induction hypothesis, the statement is true for all m.