

I dedicate this page to my friend Wilfrid Keller
"
Dr. John Cosgrave's talk on number theory was for many people the best thing they've been to all year. He gave a wonderfully lucid
account of his recent discoveries and ideas about Fermat numbers  accessable to all & extremely interesting for those of us (i.e. everyone)
who hasn't actually seen a number theorist at work. The talk [which was about the Fermat 6 work on this page] actually managed to
stretch on for more than two (very short & extremely enjoyable) hours, and it was well after twelve [midnight] before we let him leave
the maths department. "
(Quoted from the Trinity
College Dublin (Student) Mathematical Society web site.
Thank you to whoever wrote the above; no bribes were offered!)
This page contains material relating to an email which I sent to the
Number Theory mailing list
on 5^{th} October 1999, following a
question posted by Mingzhi Zhang
on 4^{th} October 1999 re primes of the form
(2 ^{ k } + 2^{ i } + 1).
Following this I submitted a paper (now, in Sept 2006, available in pdf format:
Fermat_6),
'Could there exist a sixth Fermat prime? I believe it is not impossible',
to the American Mathematical Monthly on 19^{th} March 1999, and it was rejected.
I have been asked why did I not submit it elsewhere, and the simple answer is that I lacked the emotional energy to do so; my paper made a certain
obvious and elementary point, which some open minded people listened to, while others didn't.
The email of 20^{th} March 1999 in which I made an announcement to a few friends and colleagues about my (new, I then thought;
see later) unification of Fermat and Mersenne numbers :
From: John B. Cosgrave
To: ...
Subject: a unification of Mersenne and Fermat numbers
Date: 20 March 1999 22:36
Dear Colleagues (and, in some cases, friends),
I am writing to tell you of some work I have done in recent weeks, which, in a way, unifies into a whole the Fermat and Mersenne numbers. This work may (actually, should) throw doubt on the frequent assertion that the only Fermat primes are the first five (3, 5, 17, 257 and 65537). That was not, of course, what I set out to do; rather I was just making some routine preparations for my third year Number Theory and Cryptography
course (in connection with improving the statement of Proth’s theorem, and its proof, so as to make it more accessible to my students), when a
number of things took a turn ... (the actual details are in a paper that I have just completed, and submitted).
Here, then, is my discovered linking of the Mersenne and Fermat numbers (if it is already buried out there in the literature, then obviously I apologise
for wasting your time in reading this, and I end up with egg on face. Anyway, at the age of 53 I am not worried anymore about making a fool of myself):
The Mersenne numbers are the M[p] with M[p] = 2^p  1, p prime; the Fermat numbers are the F[n] = 2^(2^n) + 1, n = 0, 1, 2, 3, 4, 5, ... .
Currently there are 37 Mersenne primes (the last three found by George Woltman’s wonderful GIMPS project (<
http://www.mersenne.org>),
and just those 5 Fermat primes.
I propose (as a pictorial image) that instead of thinking of those two sequences
horizontally (as it were), one should *see* the Fermat numbers as being the vertical side,
and the Mersenne numbers the horizontal side of the doubly infinite matrix of numbers:
. . . . .
(‘level’ 3) F[3, 1], F[3, 2],
F[3, 3], F[3, 4], ...
(‘level’ 2) F[2, 1], F[2, 2],
F[2, 3], F[2, 4], ...
(‘level’ 1) F[1, 1], F[1, 2],
F[1, 3], F[1, 4], ...
(‘level’ 0) F[0, 1], F[0, 2],
F[0, 3], F[0, 4], ...
(‘rank’ 1) (‘rank’ 2) (‘rank’ 3)
(‘rank’ 4) ...
where the left vertical side is comprised of the Fermat numbers (F[n, 1] = F[n]), the
bottom horizontal side is comprised of the Mersenne numbers (F[0, r] = M[p], p the rth.
prime), and the general entry F[n, r] is given by the value of the cyclotomic polynomial
(x^{p1} + x^{p2} + ... + x + 1), evaluated at x = 2^(p^{n}).
Thus the ‘ranks’ are comprised of the numbers:
rank 1: 2^(2^n) + 1 (n = 0, 1, 2, 3, 4, ... ), the Fermat numbers,
rank 2: 2^(2*3^n) + 2^(3^n) + 1 (n = 0, 1, 2, ... ),
rank 3: 2^(4*5^n) + 2^(3*5^n) + 2^(2*5^n)
+ 2^(5^n) + 1 (n = 0, 1, 2, ... ),
...
Level 0 is comprised of the numbers {F[0, r]}, and with F[0, r] = 2^{p  1} + 2^{p
 2} + ... + 2 + 1 = 2^{p}  1, then those are the Mersenne numbers.
I would like to call those at level 1 the ‘first cousins’ of the Mersenne
numbers; they are given by
F[1, r] = 2^((p  1)*p) + 2^((p  2)*p) + ... + 2^p + 1 ( = (2^(p^{2})  1)/(2^{p}  1))
These {F[n, r]} are (proposed) ‘generalised Fermat numbers’ (though it might be
better to call them the MersenneFermat numbers?), and are quite different from the
already named ‘generalised Fermat numbers’ (as in, e.g., Hans Riesel’s
Prime Numbers and Computer Methods of Factorization), namely the F[n](a, b) = a^(2^n) +
b^(2^n), with gcd(a, b) = 1, and a, b of opposite parity.
The {F[n, r]} have the following (easily proved) properties:
1. They satisfy a functional equation (similar to F[0]*F[1]*F[2]*...*F[n1] + 2 = F[n]),
2. They are pairwise relatively prime (within and across the ranks, meaning: gcd(F[n, i],
F[n, j]) = 1, for i <> j, and gcd(F[m, i], F[n, j]) = 1, for m <> n),
3. (the most interesting one, in view of the fact that the Fermat numbers  as known  all
pass theLucasFermat test to the base 2, and indeed the same was known of the
Mersenne numbers)
Every F[n, r] passes the LucasFermat test to the base 2.
And here, now, is how some doubt must be cast over the belief (?) (assertion?) that the
Fermat numbers are all composite after the 5^{th} of them. That belief may be
equivalently restated as follows:
a Fermat prime may not follow a Fermat composite
(meaning: if F[n] is composite then F[n+1] is also composite?)
But the Fermat numbers are simply the left vertical side of the above
square, and what about the next rank along from it?, and the one alongside it?, and ... .
Those numbers (in each rank) are like the Fermat numbers (the only
difference is that they grow  of course  so much more quickly in size:
F[5] = F[5, 1] = 2^(2^5) + 1 = 2^32 + 1, has 10 digits,
F[5,2] has 147 digits, F[5,3] has 3763 digits, F[5,4] has 30357 digits,
F[5,5] has 484812 digits, ... ).
Can one not also assert for each of those ranks that a prime may not follow a composite
(meaning: if F[n, r] is composite, then F[n, r+1] is also composite)?
But I have found (there is a caveat, below) that at the 17^{th} rank,
where F[0, 17] is M[59] (and, as is known  M[59] = 2^59  1, is composite) is composite,
is followed by the 1031 digit prime F[1, 17].
I have not been able to prove that F[1, 17] is prime by a classical method (say
Pocklington, or Selfridge’s ‘change of base theorem’), and those of you who
are familiar with such methods will immediately see why (I have explained this point in my
paper). Maple (which I use) declares F[1, 17] to be prime, but as you will know that
cannot be relied upon for a number of 1031 digits. F[1, 17] passes the LucasFermat test
to all prime bases up to 137 (I had to stop somewhere), and passes Miller’s test for
the first 25 primes (again I had to stop somewhere), and so is almost certainly prime (if
not it will be the first counterexample to the Maple test – but no consolation!!) I
have pointed all of this out to the Editor of the journal to which I have submitted my
paper, and will (if he is agreeable) submit a request about proving (or disproving)
primality to the Number Theory List. In the meantime I would ask you to regard all of the
above as being a private communication.
I hope it has been of some interest to you.
With best wishes,
John
4. Part of Yves Gallot's almost immediate return email to
me:
From: Yves Gallot
To: John Cosgrave
Subject: Re: a unification of Mersenne and Fermat numbers
Date: 20 March 1999 23:13
... ... ...
Wilfrid Keller wrote an unpublished paper in 1991: "Factors of Generalized Fermat
and Mersenne Numbers".
In 1962, D. Shanks proved a general divisibility theorem for a wide class of numbers
denoted:
L_p, m(a, b) = (a^(p^(m+1))b^(p^(m+1))) / (a^(p^m)b^(p^m))
Wilfrid Keller studied the L_p, m = L_p, m(2, 1). L_p, m = (2^(p^(m+1))1) /
(2^(p^m)1) With his notation, F_m = L_2, m and M_p = L_p, 0.
All the divisors of some L_p, m are of the form 2.h.p^n+1. A lot of factors are related
in the paper and today the search of them is being extented with my program.The result " The 1031digit number L_59, 1 proved to be a probable prime "
was also related in the paper. I don’t know if the primality of this number is proved
today.
I am sure that Wilfrid Keller will be pleased to give to you more details.
Best Regards,
Yves
I wrote to Wilfrid Keller, and I received a very friendly reply from him, together with
his fine, unpublished paper.
5. An email from Francois Morain:
From: Francois Morain
To: John Cosgrave
Subject: Re: a unification of Mersenne and Fermat numbers
Date: 25 March 1999 17:04F[1,17] is prime. It took less than two days on a DecAlpha 500 MHz (V5). The
certificate is available on request.
F. Morain
6. A later email from me:
From: John B. Cosgrave
Subject: a conjectured primality test for Mersenne primes
Date: 13 April 1999 08:38
Dear Colleagues/Friends,
I am writing to all of you since each of you commented in some way on my recent
circular note re ‘a unification of Mersenne and Fermat numbers’ (and it was of
great value to me to hear from one of you  Yves Gallot  that my proposal was already out
there in the literature. YG referred me to an unpublished paper of Wilfrid Keller’s,
and in turn Wilfrid’s paper drew my attention to work of Shanks, Ligh and Jones in
which my rediscovery was to be found).
I am now circulating the following, just discovered (rediscovered?  please inform me
if you think so, and I will just give up), conjectured primality test for Mersenne primes:
let ‘p’ be an odd prime, then M[p] ( = 2^p  1) is prime if and only if 3^((M[p] 
1)/p) = 2^a (mod M[p]), for some nonnegative integer ‘a,’ and 2^a < M[p].
I have proved the easy part of that, namely:
if M[p] is prime then 3^((M[p]  1)/p) = 2^a (mod M[p]), 2^a < M[p].
And here is a proof: First note that the
congruence x^p = 1 (mod M[p]) ... (i), has at most p
mutually incongruent solutions. Then note that x = 1, 2, 2^{2}, 2^{3}, ...
, 2^{(p  1)} (mod M[p]) are p mutually
incongruent solutions of (i) (this is obvious: just note that 2 ^{p} = 1
mod (M[p]), and form all the other powers: i.e., square both sides, cube
both sides, etc), and so are the p solutions of (i). (Thus the numbers 1, 2, 2^{2},
2^{3}, ... , 2^{(p  1)} are the pth roots of unity mod
M[p].) Now, by Fermat's 'little' theorem we have 3^(M[p1]
 1) = 1 (mod M[p]), and since (M[p]  1)/p
is an integer (again by Fermat's 'little' theorem) then 3^((M[p]  1)/p)
is an integer, which is a pth root of unity mod M[p], and so
we have 3^((M[p]  1)/p) = 2^a (mod M[p]),
2^a < M[p].
Comment. Of course this is trivial; just replace '3' by any integer 'b,'
not divisible by p, and the corresponding result is true. Indeed in the case
where b = 2, one has that M[p] = 2 ^{p}  1
passes the LucasFermat test to the base b = 2, even when M[p]
is not prime. In short, the conjecture (question, really) is placing a lot of faith in the
case where b = 3.
Incidentally, here are the values of ‘a’ for the known values of
‘p’ (up to 44497); I give them as (p, a) pairs:(3, 1), (5, 4), (7, 2), (13, 12), (17, 15), (19, 8), (31, 10), (61, 50), (89, 36),
(107, 40), (127, 90), (521, 122), (607, 182), (1279, 599), (2203, 1634), (2281, 672),
(3217, 1724), (4253, 813), (4423, 2385), (9689, 5644), (9941, 6272), (11213, 318), (19937,
18742), (21701, 21424), (23209, 19413) and (44497, 44095).This happened in the search for a primality test for the ‘MersenneFermat’
numbers (the F[n, r] of my notation). In my Monthly note I had asked, at the end, this
single question: is F[n, r] prime ([n, r] not [0, 1]) if and only if F[n, r] passes
Fermat’s test to the base 3 (JB gently chided me for calling this ‘Fermat’s
test,’ maintaining that it ignores Lucas’ contribution, and we have agreed to
start calling it the LucasFermat test)? I would dearly love to know if that is true, and
you will see how the above conjectured Mersenne test is related to my Monthly question.Actually this  in general  is what I now believe to be true:
Let F[n, r] be the ‘level n’, ‘rank r’ MersenneFermat number, and
let‘p’ be the rth prime; then F[n, r] is prime ([n, r] not [0, 1]) if and only
if 3^((F[n, r]  1)/p) = 2^a (mod F[n, r]), for some nonnegative integer ‘a,’
and 2^a < F[n, r].(Pepin’s test for the Fermat numbers  F[n] is prime (n not 0) if and only if
3^((F[n]  1)/2) = 1 (mod F[n]))  is clearly a special case of this; you have to see,
e.g., 3^((F[4]  1)/2) = 1 (mod F[4]) as 3^((F[4]  1)/2) = 2^16 (mod F[4]) My request to prove the primality of the 1031digit number F[1, 17], was answered by
Francois Morain ("F[1, 17] is prime. It took less than two days on a DecAlpha 500 MHz
(V5). The certificate is available on request"). If the above were true, then the
primality of F[1, 17] may be established in a minute or so on an ordinary Pentium (my
166MHz), with this computation: 3^((F[1, 17] 1)/59) = 2^1888 (mod F[1, 17]).
(‘1888’ is 59 times 32.)
I hope there is something here to interest you, and I will appreciate any comments
(even ones that say, e.g., John, some 12year old published that in a school Math. mag. in
1872 ... in Kurdish (that is not a slur on Kurdish; rather I hope you know the Erdos
limerick ... )). I would especially appreciate comment  about implementation time
estimates – from those of you who are experts in the use of the FFT. Of course I
would especially appreciate hearing of a proof of the hard part ... .
With best wishes, John
Finally, I received some friendly communications in connection with this,
and eventually a rejection of my paper by the Monthly.
 
