Re: A possible argument for no more Fermat primes.

You seem to be looking for primes of the form 2^n +
1

No

or 2^(4n) + 1.

yes

But your results are a consequence of the fact that
if >n=rs where s is
odd

Nothing is odd except after (+1)

then 2^n+1 can be divided by 2^r+1.

So for example, 2^96 + 1 is divisible by 2^32 + 1

Yes---

The search for bigger Fermat primes (or a proof that
no .more exist)
can be restricted to those of the form
2^(2^n) + 1.

Henry,

I am just showing the pattern of all 2^n+1 where (n)
will always be divisible by 4.

Its in this pattern that if ever another Fermat prime

is found it will throw off this pattern.

2^36+1 = has as one of its factors (17)
2^40+1 = " " " " " " (257)
2^44+1 = " (17)
2^48+1 = " (65537)
2^52+1 = " (17)
2^56+1 = " (257)
2^60+1 = " (17)
2^64+1 = Has 2 prime factors and none are Fermat
primes.

I will skip too 2^100 for the sake of simplicity
but this pattern is consistent.

(")= has as one of its prime factors

2^100+1 = (") (17)
2^104+1 = (") (257)
2^108+1 = (") (17)
2^112+1 = (") (65537)
2^116+1 = (") (17)
2^120+1 = (") (257)
2^124+1 = (") (17)
2^128+1 = (") a composite with no Fermat primes as
factors.

My point is, say at some very large (n) where
2^(2^n) +1 = a prime, making it the 6th Fermat
prime.

The cyclic pattern =

2^(2^(n-28)+1 = (") 17
2^(2^(n-24)+1 = (") 257
2^(2^(n-20)+1 = (") 17
2^(2^(n-16)+1 = (") 65537
2^(2^(n-12)+1 = (") 17
2^(2^(n-8)+1 = (") 257
2^(2^(n-4)+1 = (") 17
2^(2^n)+1 = the 6th Fermat prime.

Will any later cycles include the 6th Fermat prime
as one of the (8) composite factors shown above?
Right now the cycle picks up the last 3 Fermat
primes 17,257,65537, as one of the factors in 7
of these composites.
Will there now be 4 Fermat primes after the sixth
Fermat prime is found and it would only fit in the
last position of the cycle?


Dan

look at the none-fermat prime factors of 2^n + 1.
.

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