Problems & Puzzles: Conjectures

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 !…

It’s known that Pierre Fermat believed and tried to prove that all the Fermat numbers are primes, but he didn’t 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 hasn’t 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 composites… and different prime hunters have found one or several factors of many Fermat numbers

(See: )

Then we have an open question : "are F0 to F4 the only Fermat prime numbers ?" or "are composite all the Fermat numbers Fm>F?"


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 him. C:=1.57:

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=0..4) =4.4

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 not present."


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 < 2^(16*n)"

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 computers.

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 for F14.

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=> ~2^(16*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:

  1. 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"
  2. 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 members
  3. He shows that the compositeness conjectural statement is not valid for these new class of numbers in certain domain ( that he calls "rank")
  4. He hopes then that then probably that statement is also not valid in the rank where these new class of numbers coincides with the classical Fermat numbers.

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 

and letters :-)


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