Problems & Puzzles:
Conjecture 4. Fermat primes are finite.
A Fermat number Fm is
defined this way : Fm=2^(2^m)+1. The
first 5 Fermat numbers are : F0=3, F1=5,
F2=17, F3=257 & F4=65537,
all primes !
Its known that Pierre Fermat
believed and tried to prove that all the Fermat numbers
are primes, but he didnt succeed.
Soon Euler not only factorised F5=641x6700417,
but proved that all the factors of the composite Fermat
numbers are of the form k.2^n+1.
After that we know that some Fermat
numbers are primes (five) and others are composites.
In 1877 Pepin devised a rigorous
proof for testing the primality of the Fermat numbers,
but it hasnt been used with a positive result
for any other Fermat number greater than F4,
that is to say, the Pepin test has never proved
that another Fermat number Fm>F4
is prime. The larger Fm that has been tested using the Pepin's
proof is F24, as recent as 1999 by Mayer, Papadopoulus & Crandall (see
the whole story -according to Leonid Durman - here).
On the other hand nobody has proved
that all the Fn>F4 are
and different prime hunters have found
one or several factors of many Fermat numbers
Then we have an open
question : "are F0
to F4 the only Fermat prime
numbers ?" or "are composite all the
Fermat numbers Fm>F4 ?"
The following argument to the finiteness of the Fermat
prime numbers has been sent by Leonid Durman (18/8/01):
"Certainly strict proofs are necessary, But...
As it is not revealed of anomalies Mersenne prime and Fermat from
others prime numbers yet. Therefore it is possible to tell, that they
submit to the law of allocation prime numbers. function pi(x), the
number of primes below x. Is proved, that at x ==> infinity, pi(x)==>x/ln
x -known function.
Then for enough large x to find anyone prime makes ==> 1/ln x
Then to receive an amount expected prime numbers we should summarize for
all numbers C*sum (1/ln x, x=x1.. x2). Where C coefficient, I shall take
1.57, but I read the large theories on this coefficient. Sense not in
The sums which we obtain: For Mersenne x= 2^n:
sum(C/ln((2.^x)),x=1..1e7) = 37.81540138
sum(C/ln((2.^x)),x=1..infinity) = infinity
As you can see up to x < 10^7 is obtained equally 38 known
Mersenne prime. But for x= infinity amount Mersenne prime - is infinite!
Further. For Fermat number x=2^2^m
sum(C/ln(2^(2.^x)),x=33..infinity) = 0.0000000005273686755
From F0-F4 all Fermat prime them 5. The formula is not mistaken.
For all remaining numbers with unknown character F33... We obtain very
small number. Infinite series converges also it gives the basis to
speak, that probability to detect another Fermat prime practically is
Needles to say that this is not a proof but an
acceptable argument. Does anybody has another approach-argument, while the
real proof comes?
Leonid Durman has sent a follow up to this
Conjecture 4 stating a stronger conjecture about the Fermat numbers:
"All Fermat numbers m
> 4 have divisor and anyone first smaller divisor k*2^n+1 <
Here is how he arrives to this statement:
"I would like to continue reviewing Conjecture 4. I am
independent has offered the formula for quantifying divisors in some
range. The similar formula has output Paul Jobling integrating
probability of a divisor offered Yves Gallot: probability 1/k.
Here obtained formula by the various people: C • ln (nMax /
(nMin-1)) • ln (kMax / kMin) Where C some coefficient. I would tell,
that he is equal C~1 for very large n, which we can consider on
But at smaller ranges, n's and k's he changes 0.7<C<1.6,
that is characterizes some discontinuities of allocation. Therefore we
admit having taken the obviously underestimated value C=0.1, we shall
try to estimate limits, what maximum value k's to us is necessary for
taking to find even one divisor:
at: kMin=3, C=0.1, nMax=nMin+1
n=16, F_14, k<10^72 !!!
So far should be checked k to be as much as possible sure
99.99999% :-)) but is not absolute, that we shall detect out a divider
n=22, F_20, k<10^98
n=24, F_22, k<10^107
n=26, F_24, k<10^115
n=35, F_33, k<10^154
n=402, F_400, k<10^1750
n=1000002, F_1000000, k<10^4340000
Or very roughly, but it is possible to tell. k<10^(4.5*n) or
divisor Fermat k*2^n+1=> ~10^(4.5*n)*2^n =>~2^(15*n)*2^n=>
This value most not favorable, which is improbable, is more
probable, that we shall discover a divider much earlier. Thus, I give
new Conjecture for Conjecture 4:
"All Fermat numbers m > 4 have divisor and anyone first
smaller divisor k*2^n+1 < 2^(16*n)"
But Leonid is well aware that Emil
Artin has a conjecture in the opposite direction, that is to
say, conjecturing that there must be more Fermat primes, and ask us to
find and publish that conjecture:
"I would be very pleased, if you could find in
libraries or on Web, "proof" Artine, (is written for me
on russian), where there are serious arguments about others prime Fermat.
It would be the good counterbalance to my guesses."
Does anybody has that
conjecture in his books?
It was the same Leonid that found a very interesting
argument elaborated by John B. Cosgrave ( Mathematics
Department, St. Patrick’s College, Drumcondra, Dublin 9, IRELAND)
in a document titled "Could
there exist a sixth Fermat prime? I believe it is not impossible",
available in his own site.
In short (and with all the risks of the shortening): Cosgrave
proceeds the following way:
- He reformulate the compositeness conjecture for the Fermat
numbers larger than F4 in the following elegant statement: "ifFm is composite thenFm+1 is also composite"
- He defines a new numbers class (unhappily also named Generalised
Fermat Numbers, name that has been used for a very different class
of other numbers) that contains the classical Fermat numbers as one of its
- He shows that the compositeness conjectural statement is
not valid for these new class of numbers in certain domain ( that he calls
- He hopes then that then probably that statement is also
not valid in the rank where these new class of numbers coincides with the
In his own (Cosgrave) words:
"What I have done is to place -
in, I believe, a very natural way -
the Fermat numbers in a larger setting, and point
out that in that larger setting-almost
certainly at the 17th rank-the
corresponding behavior is different. If that can happen at the 17th
rank, then surely it is fair to note that it could happen at any
rank, and therefore that it is not impossible (until proven otherwise) for a
sixth Fermat prime to exist."
Pepin tests story, according
to Leonid Durman
Morehead & Western (independently) using Pepin's test with a=3 verified
that F7 is composite. J. C. Morehead, Note on Fermat's numbers, Bull. Amer.
Math. Soc. vol. 11 (1905) pp. 543-545.
Morehead & Western (by a very long computation) verified that F8 is
composite. (Pepin test a=3) J. C. Morehead and A. E. Western, Note on Fermat's
numbers, Bull. Amer. Math. Soc. vol. 16 (1909) pp. 1-6
F_13 Paxson 6 hour on IBM 7090 G. A. PAXSON, "The compositeness of the
thirteenth Fermat number," Math. Comp., v. 15, 1961, p. 420
F_14 Selfridge & Hurwitz on IBM 7090, the outcomes Paxson for F13 were
checked, the attempt is made to check up F17, but have fulfilled only 20
operations modulo from necessary 131071. The complete operation would occupy
then 128 complete weeks ~ 2 years. see: A. HURWITZ & J. L. SELFRIDGE,
"Fermat numbers and perfect numbers," Notices Amer. Math. Soc., v.
8, 1961, p. 601, abstract 587-104. and J.L.Selfridge and A.Hurwitz, "Fermat
numbers and Mersenne numbers," Math. Comp. 18 (1964), 146-148
F_20 Buell & Young see: J.Young and D.Buell, "The Twentieth Fermat
Number is Composite" Math. Comp 50 (1988), 261-263
F_22 Crandall, Doenias, Norrie & Young see: V.Trevisan and J.B.
Carvalho, "The composite character of the twenty- second Fermat
number" J.Supercomputing 9 (1995), 179-182
F_24 Mayer, Papadopoulos & Crandal see http://www.perfsci.com/F24
and letters :-)