Nabble - Number Theory

Coppersmith--Howgrave-Graham proof at 221071 digits

2008-10-07T13:42:25Z

When at least 1/3 of the digits of N-1 or N+1 are completely
factorized into proven primes, proof of the primality of a
probable prime N becomes a routine matter. This is currently
the case for all but 7 of the 5000 largest primes recorded
at http://primes.utm.edu/primes/home.php

With a factorization fraction between 3/10 and 1/3, the
methods of Konyagin and Pomerance furnish a proof, as is the
case for 5 of the primes in the top-5000, recently proven by
Chris Caldwell and myself. These are cyclotomic numbers with
sizes ranging from 110573 to 110578 decimal digits.

There remain two primes in the top-5000 for which only the
methods of Coppersmith and Howgrave-Graham (CHG) are at
present adequate to provide a proof. One of these is
2^389335+2^97837+1, with 117202 digits, for which I used
algebraic methods to obtain polynomials for a CHG proof, as
developed in http://physics.open.ac.uk/~dbroadhu/chg_alg.pdf

The largest prime with a proof not accessible by the methods
of Brillhart, Lehmer and Selfridge (BLS) alone was recently
recorded in http://primes.utm.edu/primes/page.php?id=85584

It has 221071 digits and is a cyclotomic number of the form
N=(X^5-1)/(X-1), with X=A*B^7 containing the proven primes
A=3668*P-1 and B=378266*P/5+1, where P is the product of the
1863 primes not exceeding 16001. N-1 is divisible by
F=28*26893*P*X, which contains 1866 distinct primes. The
quotient (N-1)/F is the product of two composite numbers,
5*(X+1)/(14*P) and (X^2+1)/268930, with 48364 and 110531
digits. Neither has a prime divisor less than 10^9.

While testing the Open University's IMPACT cluster, I
identified N as a probable prime, by running concurrent
searches on 48 CPUs, using the PrimeForm programme PFGW,
which also performed the 1866 BLS tests to prove that every
divisor of N is congruent to 1 modulo F. The witness a=16067
gives a^(N-1)=1 mod N and gcd(a^((N-1)/q)-1,N)=1 for each
prime q that divides F, as recorded in the output file
http://physics.open.ac.uk/~dbroadhu/cert/phi5big_pfgw.txt

Since log(F)/log(N) is very close to 9/32=0.28125, this
candidate is well suited to the CHG method. Polynomials
proving that the only divisors of N congruent to 1 modulo F
are 1 and N were obtained with the LLL procedure of Pari-GP.
The resulting CHG certificate has a size of 8.8 MB and was
validated in 385 seconds, as shown in the output file
http://physics.open.ac.uk/~dbroadhu/cert/phi5big_cert.out

I thank Geoff Bradshaw, for encouragement to test the IMPACT
cluster, and John Renze, for checking my CHG code. Advice
from Don Coppersmith and Carl Pomerance informed this work.

David Broadhurst
The Open University
Milton Keynes MK7 6AA, UK

Primality Proof for Wagstaff numbers (1/3)(2^p+1)

2008-10-07T09:54:25Z

Ladies & Gentlemen,

I am sure that I have found the proof for a primality test
for Wagstaff numbers of the form (1/3)(2^p+1), p>3.

Please be so kind and use the following link
your feedback would be highly appreciated.

http://www.mersenneforum.org/showpost.php?p=144516&postcount=1

below an abstact

{ It is now possible to prove Wagstaff numbers, of the
form $W_p = \frac{1}{3} \left(1+2^p\right)$, prime! Proof for a test
is based on properties of groups and of the iteration $s\rightarrow
s^2-2$. Primality is given if the result of the second iteration
equals the result of the $p^{th}$ iteration. The 'order of the
group' is discussed briefly to show that the order of an element
$\omega=a+b\sqrt{c}$ can be determined even though no solution
$\omega^k=1$ exist; the order can be evaluated if
$\omega^n=\bar\omega^j$.}

Re: Sieve of Zakiya

2008-10-07T09:54:24Z

Date: Thu, 18 Sep 2008 16:27:51 -0400
From: Jabari Zakiya <jzakiya@...>
Subject: Sieve of Zakiya

Hi

Greetings.:)

I am new to this list, which was recommended to me
as a place that would be interested in my work on
prime sieves I have developed.

I've been on the list quite a while myself but most of it is way over
my head since I'm also not a mathematician per se, but primarily a
programmer, though I have been interested in number theory for at
least 30 years now and have participated in and programmed for GIMPS
(http://www.mersenne.org) since Jan. 1996.

On June 7, 2008 I released a paper, and code, for
a prime sieve method that I designated the
Sieve of Zakiya (SoZ), to differentiate if from the
Sieve of Eratosthenes (SoE) and Sieve of Atkin (SoA).

Here is the site you can download the paper and various
implementations code sets in 3 languages (Forth, Ruby, Python).

http://www.4shared.com/dir/7467736/97bd7b71/sharing.html

Basically, I have found a way to find a class of prime number
generators of the form:

Pj = m*k + (1+ri)

where m is a modulus value, and ri are what I call residues mod m.

I list 9 generators in my paper, but here are the first 4:

P3 = 6*k+(1,5)
P5 = 30*k+ (1, 7, 11, 13, 17, 19, 23, 29)
P60 = 60*k+(1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59)
P110 = 110*k+(1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 37, 39, 41,
43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 79, 81, 83, 87, 89,
91, 93, 97, 99, 101, 103, 107, 109)

These look like a superset or variant of work I did as an
undergraduate and are also related to code I found in the
implementation of /usr/games/primes for BSD (UCBerkeley) UNIX
available at the time (Nov. 1987). I called them "Tables of
Eratosthenes" since they are based on his sieve and may have been
known to him.

The basic idea of the tables is to continue the thought of "Why test
even numbers?" to "Why test multiples of 3?" (matching your P3
exactly) to "Why test multiples of 5?" (matching your P5 exactly).
Your P60 and P110 are new to me, however; using your notation, my next
table would be:

P7 = 210*(1 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83
89 97 101 103 107 109 113 121 127 131 137 139 143 149 151 157 163 167
169 173 179 181 187 191 193 197 199 209)

The multiplier is the primorial (as factorials but only prime numbers
are used). Each of the tables is built from the prior table by
repeating the prior table p times where p is the next prime and then
striking out the values that are a multiple of p. E.g., P3 = 6*(1, 5)
is equivalent to P3 = 30*(1, 5, 7, 11, 13, 17, 19, 23, 25, 29), so P5
= 30*(1, 7, 11, 13, 17, 19, 23, 29). Since P3 has 2 members (1, 5),
P5 has 2*(5 - 1) = 8 members, P7 has 8*(7 - 1) = 48 members, etc. P17
has 92,160 entries from 1 thru 510,509 and is 616 KB on disk in ASCII;
the next larger one was too big for me to retain on disk at the time.

I put quite a bit of thought into storing them efficiently (as
differences, as half of the differences, using the largest table for
which half-differences fit in one byte, etc.) and implemented one of
them in C as part of my early (pre-GIMPS) efforts to find Mersenne
primes.

The BSD UNIX implementation of /usr/games/primes in 1987 used P11, as
I recall, so this technique was obviously already known but perhaps
never published per se.

-- Will Edgington, wedgingt@...
-- http://www.garlic.com/~wedgingt/mersenne.html (Mersenne numbers
-- and related programs, links, etc.)

J Number Theory call for papers

2008-10-03T10:20:13Z

Dear Colleagues,

I'm appending a call for papers for a special issue of the Journal of
Number Theory. Please also bring it to the attention of anyone else you
think might be interested.

Thank you.

Regards,
Neal Koblitz

Call for Papers for a Special Issue of the Journal of Number Theory

The Journal of Number Theory invites researchers to submit articles for
a special issue devoted to elliptic curve cryptography. Authors should
consult the journal's website~ http://ees.elsevier.com/jnt~
for instructions on electronic submission. A precise deadline has not
yet been set; however, it is expected to be in late spring or summer of 2009.

Please note the following policy of the journal:
``Original papers only will be considered. Manuscripts are accepted for
review with the understanding that the same work has not been and will not
be nor is currently submitted elsewhere, and that the submission for
publication has been approved by all of the authors and by the institution
where the work was carried out.''

In addition, please send a copy of your submission to either of the guest
editors Victor Miller or Neal Koblitz, indicating that it is intended for
the special issue. Their email addresses are:

victor@...

koblitz@...

update on video abstracts

2008-10-03T10:20:12Z

Dear All: In April JNT announced the beginnings of video abstracts for
accepted papers. This endeavor is gaining momentum and we now have
some video abstracts from around the world. The latest is by Alain Connes
on his paper with Caterina Consani and Matilde Marcolli entitled
"Fun with F_1$. You can see them all at
http://www.youtube.com/user/JournalNumberTheory
Connes' video is a terrific example of what is possible with this
technology...
David Goss

2 year Lecturer position in Pure Mathematics at UEA

2008-09-27T16:03:41Z

University of East Anglia
School of Mathematics

Fixed-term Lectureship in Pure Mathematics Ref:ATS311
UK 28,290 to UK 33,780 per annum (pay award pending October 2008).

The University of East Anglia invites applications for a
24 month fixed-term lectureship in Pure Mathematics. The
successful candidate will be expected to have the skills
to enable him/her to contribute to the teaching of Pure
Mathematics modules at all levels of the undergraduate
programme. Preference may be given to a candidate whose
area of interest fits with one of the School's existing
groupings in Logic, Algebra, Number Theory and Dynamical
Systems.

The starting date for the appointment is 1 January 2009.

Information about the School can be found at www.mth.uea.ac.uk

Closing date: 12 GMT on 17 October 2008.

Further particulars and an application form can be obtained from the
University's web page at: www.uea.ac.uk/hr/jobs/ or by e-mail at:
hr@... or by calling the answerphone on 01603 593493 or by mail
to the Human Resources Division, UEA, Norwich NR4 7TJ.

Re: a galois group question through number theory

2008-09-27T16:03:40Z

Kent asked
> Can it be shown that there does not exist relatively prime positive
> integers a, b, and c such that
> R2:=x (2(a^2+b^2) + c^2 – x)^2 – 4(a^2-b^2)^2 c^2
>
> is a square for a positive integer root x of the
> cubic
>
> R1:=x^3 - (2d^2 + c^2) x^2 + (2c^2 + d^2) d^2 x - (a^2 - b^2)^2 c^2= 0?

I am not able to solve this problem, but I may reformulate it in a way
which shows that there is certainly no elementary answer.

The Descartes resolvent R1 is a resolvent for the group D4 of 8
elements. The polynomial y^2-R2 is a relative resolvent for the groups
Z4 and D4. It follows that R3:= resultant(y^2-R2,R1,x) is a resolvent
for Z4.

Thus, without considering the various cases of Kent's mail, to prove
that the Galois group is never Z4 it "suffices" to prove that the
Diophantine equation R3=0 has no integer solution in the variables
y,a,b,c. Kent's question amounts to say that it has no solution for
which x is also integer; so the two problems are not so much different.

The polynomial R3 is quasi-homogeneous of degree 18 when y,a,b,c have
the respective weights 6,1,1,1. I do not imagine an elementary solution
for a Diophantine equation of such a high degree.

Sincerely

Daniel Lazard
------------------------------------------------------
LIP6 (Laboratoire d'Informatique de Paris VI)
et Projet SALSA (INRIA Rocquencourt)
104 Avenue du Président Kennedy, F-75016 Paris
Tel.: +33 1 44 27 62 40 GSM: +33 6 73 18 79 29
------------------------------------------------------

Kent Holing a écrit :

> For the quartic equation
> (*) x^4 - 2c x^3 + (c^2-a^2-b^2) x^2 + 2a^2c x - a^2c^2 = 0
>
> we assume that a, b and c are integers. We also assume that a, b, c
> are positive and
>
> gcd(a,b,c) = 1.
>
> It is known that the Galois group G of this quartic can never be Z4.
>
> In the case a = b this can be proved elementary and Kurt Foster gave a
> proof in the a /= b case, see
> http://www.mathforum.com/kb/message.jspa?messageID=1696921&tstart=720.
>
> I have tried to prove the a /= b case more elementary than Kurt, using
> the following result:
>
> A reduced quartic equation
> (1) Q(x) = x^4 +B x^2 + C x + D = 0
>
> with integer coefficients and the auxiliary equation (Descartes
> resolvent)
>
> (2) R(t) = t^3 + 2B t^2 + (B^2-4D) t - C^2 = 0 are given.
>
> If (1) has no integer roots, C ≠ 0 and (2) has one and only one
> integer root t0, the Galois Group G of (1) can be found as follows:
>
> If t0 is a square then G = Z2 x Z2; otherwise G = Z4 if f =
> t0(t0+2B)^2 – 4C^2 is a square, and G = D4 if not. Note that G = Z4 is
> possible only when t0 > 0 and all of the roots of (2) are real (i.e. f
> is needed to distinguish between Z4 and D4).
>
> (We have only to discuss the cases where G may be Z4. We also left
> out the case C = 0 because it easy.)
>
> Now, removing the cubic term of (*), we find that the Descartes
> resolvent of the reduced quartic is R(t) = t^3 - (2d^2 + c^2) t^2 +
> (2c^2 + d^2) d^2 t - (a^2 - b^2)^2 c^2 = 0 where d > 0 is given by d^2
> = a^2 + b^2. (Note that C /= 0 for a /= b.)
>
> Using the theorem above we ask:
>
> Can it be shown that there does not exist relatively prime positive
> integers a, b, and c such that
> X (2(a^2+b^2) + c^2 – X)^2 – 4(a^2-b^2)^2 c^2
>
> is a square for a positive integer root X of the
> cubic
>
> x^3 - (2d^2 + c^2) x^2 + (2c^2 + d^2) d^2 x - (a^2 - b^2)^2 c^2= 0?
>
> We know that the answer is no, but can we prove the above, we have
> another proof that G is never Z4 for the case a /=b of the quartic
> (*). And, may be it is easier than Kurt’s original proof?
>
> Best regards,
>
> Kent Holing
>
> NORWAY
>
>
>

Re: a galois group question through number theory

2008-09-27T16:03:39Z

For ease of reference, I give a condensed version of my argument.
Things are elementary unless otherwise indicated.

The ladder-box quartic is

f(x) = x^4 - 2*c*x^3 + (c^2 + (-a^2 - b^2))*x^2 + 2*a^2*c*x - a^2*c^2

where a, b, and c are positive integers. Since f(x) is homogeneous of
degree 4 in a, b, c, and x, we may assume that gcd(a,b,c) = 1. The
discriminant is

D = poldisc(f) = (4*a*b*c)^2*((c^2-a^2-b^2)^3 - 27*(a*b*c)^2).

If N is a nonzero integer, let v_p(N) denote the exponent of the exact
power of p which divides N.

Then v_p(D) can be odd only if

(oddv) 3*v_p(c^2 - a^2 - b^2) <= v_p(27*(a*b*c)^2).

Let P(x) be a monic irreducible quartic in Z[x] with discriminant D.
Then the Galois group if P(x) is cyclic of order 4 if and only if P(x)
splits into conjugate quadratic factors q1(x) and q2(x) in Q(sqrt(D))
[x].

Using the standard formula for the discriminant of product, we have

disc(P(x)) = disc(q1(x))*disc(q2(x))*(Res(q1(x),q2(x)))^2.

It is easily shown that the resultant is a rational integer. Thus,
disc(q1(x))*disc(q2(x)) differs from D by a square factor. If d is
the squarefree core of D, we then see that d is a norm from Q(sqrt(d))
to Q, from which it easily follows that d is a sum of two squares, so
that -1 is a quadratic residue of each prime factor of d.

Now for something non-elementary: if P(x) is monic and irreducible of
degree n in Z[x], P(r) = 0, K = Q(r), R is the integral closure of Z
in K, and p is totally ramified in K/Q, pR = P^n, then P(x) (mod p) =
(x - k)^n. [This can be shown by localizing and using Eisenstein
polynomials.]

If p does not divide 2*c, then the ladder-box quartic does not factor
in this way (mod p).

If p is odd and p|c, then the ladder quartic is x^4 (mod p) only if p
divides a^2 + b^2.

Now for more non-elementary things: If we assume that f(x) is
irreducible with Galois group C4, and K is the splitting field, then p
is totally ramified in K/Q if and only if v_p(D) is odd. Thus the
inequality (oddv) must hold. If also p is odd, p divides c, and pR =
P^4, then [using v_P() analogously to v_p()], for a nonzero rational
integer N, v_P(N) = 4*v_p(N). The zeroes of f(x) are conjugate, and P
is stable under conjugation, so each zero r of f(x) has v_P(r) =
2*v_p(c). Since the coefficient of x^2 is the sum of the products of
the zeroes taken two at a time, we obtain

v_P(c^2 - a^2 - b^2) >= 4*v_P(c), i.e.

v_p(c^2 - a^2 - b^2) >= v_p(c).

Since gcd(a,b,c) = 1, p does not divide a or b, and since p divides
a^2 + b^2, p is not 3. Thus

v_p(27*(a*b*c)^2) = 2*v_p(c),

and (oddv) is violated. Thus if we assume f(x) has Galois group C4,
no odd prime can be totally ramified in the splitting field.

The case p = 2 can be split into the subcases 2|c and 2 does not
divide c. As with odd p, v_2(D) must be odd so (oddv) must hold. But
in each subcase, the 2-value of the coefficient of x^2 again gives a
violation of (oddv).

Announce: Conference "Random matrices, L-functions and primes", Oct. 27-31, 2008

2008-09-23T13:19:55Z

Ashkan Nikeghali (Universität Zürich) and myself would like to announce a
conference

"Random matrices, L-functions and primes"

that will be held at the Forschungsintitut für Mathematik of ETH Zürich from
October 27 to 31, 2008.

For more information (poster, list of speakers, and soon detailed program),
please see the conference web page

http://www.math.ethz.ch/~kowalski/fim-08/fim-2008.html

(For other FIM activities, see also http://www.fim.math.ethz.ch )

Emmanuel Kowalski and Ashkan Nikeghbali

a galois group question through number theory

2008-09-23T13:19:55Z

For the quartic equation
(*) x^4 - 2c x^3 + (c^2-a^2-b^2) x^2 + 2a^2c x - a^2c^2 = 0

we assume that a, b and c are integers. We also assume that a, b, c
are positive and

gcd(a,b,c) = 1.

It is known that the Galois group G of this quartic can never be Z4.

In the case a = b this can be proved elementary and Kurt Foster gave a
proof in the a /= b case, see
http://www.mathforum.com/kb/message.jspa?messageID=1696921&tstart=720.

I have tried to prove the a /= b case more elementary than Kurt, using
the following result:

A reduced quartic equation
(1) Q(x) = x^4 +B x^2 + C x + D = 0

with integer coefficients and the auxiliary equation (Descartes
resolvent)

(2) R(t) = t^3 + 2B t^2 + (B^2-4D) t - C^2 = 0 are given.

If (1) has no integer roots, C ≠ 0 and (2) has one and only one
integer root t0, the Galois Group G of (1) can be found as follows:

If t0 is a square then G = Z2 x Z2; otherwise G = Z4 if f =
t0(t0+2B)^2 – 4C^2 is a square, and G = D4 if not. Note that G = Z4 is
possible only when t0 > 0 and all of the roots of (2) are real (i.e. f
is needed to distinguish between Z4 and D4).

(We have only to discuss the cases where G may be Z4. We also left
out the case C = 0 because it easy.)

Now, removing the cubic term of (*), we find that the Descartes
resolvent of the reduced quartic is R(t) = t^3 - (2d^2 + c^2) t^2 +
(2c^2 + d^2) d^2 t - (a^2 - b^2)^2 c^2 = 0 where d > 0 is given by d^2
= a^2 + b^2. (Note that C /= 0 for a /= b.)

Using the theorem above we ask:

Can it be shown that there does not exist relatively prime positive
integers a, b, and c such that
X (2(a^2+b^2) + c^2 – X)^2 – 4(a^2-b^2)^2 c^2

is a square for a positive integer root X of the
cubic

x^3 - (2d^2 + c^2) x^2 + (2c^2 + d^2) d^2 x - (a^2 - b^2)^2 c^2= 0?

We know that the answer is no, but can we prove the above, we have
another proof that G is never Z4 for the case a /=b of the quartic
(*). And, may be it is easier than Kurt’s original proof?

Best regards,

Kent Holing

NORWAY

2008 Western Number Theory Conference

2008-09-23T13:19:54Z

Dear Colleagues,

The 42nd Western Number Theory Conference (*) will be at Colorado
State University, starting the evening of Monday, December 15 and
ending Thursday evening.

Talks are welcome in any area of number theory, and graduate students
and recent PhDs are especially encouraged to come and speak. With
generous funding from the NSA and the Number Theory Foundation, we're
pleased to be able to support the travel of junior participants.

Information is available at

http://wntc.org

and I'd be happy to answer any questions you have about the
conference.

Have a great semester, and I hope to see you in December,

Jeff Achter
j.achter@...
http://www.math.colostate.edu/~achter

(*) nee West Coast Number Theory Conference

Re: Sieve of Zakiya

2008-09-21T09:31:45Z

Dear Jabari,

If you're seeking a candidate for a minimal prime generating sieve, take a
look at:

http://alixcomsi.com/A_Minimal_Prime.pdf

I programmed the Compact Algorithm in the Sinclair ZX+ Spectrum assembly
language way back in 1986, and the machine code ran amazingly fast.

I'd be interested in passing on that effort to anyone who wants to try
programming it directly in 386+ or equivalent assembly language.

Sincerely,

Bhup

======================
-----Original Message-----
From: Number Theory List [mailto:NMBRTHRY@...] On Behalf Of
Jabari Zakiya
Sent: Friday, September 19, 2008 1:58 AM
To: NMBRTHRY@...
Subject: Sieve of Zakiya

Hi

I am new to this list, which was recommended to me
as a place that would be interested in my work on
prime sieves I have developed.

On June 7, 2008 I released a paper, and code, for
a prime sieve method that I designated the
Sieve of Zakiya (SoZ), to differentiate if from the
Sieve of Eratosthenes (SoE) and Sieve of Atkin (SoA).

Here is the site you can download the paper and various
implementations code sets in 3 languages (Forth, Ruby, Python).

http://www.4shared.com/dir/7467736/97bd7b71/sharing.html

Basically, I have found a way to find a class of prime number
generators of the form:

Pj = m*k + (1+ri)

where m is a modulus value, and ri are what I call residues mod m.

I list 9 generators in my paper, but here are the first 4:

P3 = 6*k+(1,5)
P5 = 30*k+ (1, 7, 11, 13, 17, 19, 23, 29)
P60 = 60*k+(1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59)
P110 = 110*k+(1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 37, 39, 41,
43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 79, 81, 83, 87, 89,
91, 93, 97, 99, 101, 103, 107, 109)

My paper is NOT A FORMAL MATH DOCUMENT but an explanation of the process
I used to find how to create these generators and how to code them in
software
and their relative performance. One person has coded some of my
generators in
C on multi-core Intel systems, and has shown some are faster than the
SoA.

My interest in posting here is to get people interested in figuring out
the
implications of the underlying mathematics, in a formal way.

Presently, I know (and explain) how to computationally find all the
residues for
a given generator, and conjecture on the properties and form of larger
ones.

What would be really revealing is to figure out an analytical way to
generate
the modulus/residues values for any generator. I know how to find the
residues
computationally but not analytically.

As an example, the SoA is based on quadratic forms, from this paper:

"Prime Sieves Using Binary Quadratic Forms"
by A.O.L. Atkin and Daniel J. Bernstein. Download the pdf of the paper
here:

http://thedjbway.org/scientific/primegen.html

They use three equations of eliptic curves. I showd how to combine the
residues of
these 3 curves to form my generator P60, so this suggests there are
similar
representations for my other generators.

Of particualr interest is the properties of what I call the Modular
Compliments (MC)
which always come in pairs which add to equal the modulus value m.
Especailly of
interest is where do the MCs which are compliments of prime composites
come from.
I suspect they come from the complex plane, and maybe they have something
to do
with the Riemann Hypotheses?

Anyway, I AM NOT A FORMAL MATHEMATICIAN, I'm an engineer (BS/MSEE
Cornell/Ga Tech) who worked 15 years at NASA. This stuff is just a
hobby/fun interest of mine.

I figure serious mathematicians can answer some of these questions.

Thanks for your indulgence.

Jabari Zakiya

Three conjectures (Mersenne-Fermat-Wagstaff) looking for a proof ! 100Euro reward

2008-09-21T09:31:44Z

Hi,

I've put on the http://mersenneforum.org/showthread.php?t=10670
GIMPS/Mersenne forum the description of 3 conjectures that are waiting for
a proof.

I've already done half the proof for one of them (the easy part...).
I've provided PARI/gp code that exercises the 3 conjectures.

I'll give 100Euro for the first guy/lady who will provide one proof out of
the three.

These conjectures deal with properties of a Digraph under x^2-2 modulo a
Mersenne, a Wagstaff or a Fermat prime, and they use cycles of length q-1
(Mersenne and Wagstaff) or 2^n-1 (Fermat). Look at the 3 papers I give link
to in the thread, and to the
http://www.cs.uwaterloo.ca/~shallit/Papers/vas.ps Shallit&Vasiga paper on
Digraph .

The final idea/dream is to use cycles of length (q-1)/n for proving that
such numbers are NOT prime, in order to speed up the search and proof. The
main goal is to improve the speed of the GIMPS project, which is searching
for gigantic Mersenne primes.

Come and have a look, have fun, and compete !

(And come look how big are the two new Prime Mersenne numbers we found at
http://www.mersenne.org/ GIMPS , and that I've verified.)

I've summarized the information in this
http://tony.reix.free.fr/Mersenne/SummaryOfThe3Conjectures.pdf paper .
And here is a link to a
http://www.cs.uwaterloo.ca/~tmjvasig/papers/newvasiga.pdf .pdf version of
Shallit&Vasiga paper .

Tony

Re: Three conjectures (Mersenne-Fermat-Wagstaff) looking for a proof ! 100Euro reward

2008-09-20T13:35:21Z

I've summarized the information in this paper.
And here is a link to a .pdf version of Shallit&Vasiga paper.
Tony

Three conjectures (Mersenne-Fermat-Wagstaff) looking for a proof ! 100Euro reward

2008-09-20T02:38:48Z

Hi,

I've put on the GIMPS/Mersenne forum the description of 3 conjectures that are waiting for a proof.

I've already done half the proof for one of them (the easy part...).
I've provided PARI/gp code that exercises the 3 conjectures.

I'll give 100Euro for the first guy/lady who will provide one proof out of the three.

These conjectures deal with properties of a Digraph under x^2-2 modulo a Mersenne, a Wagstaff or a Fermat prime, and they use cycles of length q-1 (Mersenne and Wagstaff) or 2^n-1 (Fermat). Look at the 3 papers I give link to in the thread, and to the Shallit&Vasiga paper on Digraph.

The final idea/dream is to use cycles of length (q-1)/n for proving that such numbers are NOT prime, in order to speed up the search and proof. The main goal is to improve the speed of the GIMPS project, which is searching for gigantic Mersenne primes.

Come and have a look, have fun, and compete !

(And come look how big are the two new Prime Mersenne numbers we found at GIMPS, and that I've verified.)

Tony

Sieve of Zakiya

2008-09-18T13:27:51Z

Hi

I am new to this list, which was recommended to me
as a place that would be interested in my work on
prime sieves I have developed.

On June 7, 2008 I released a paper, and code, for
a prime sieve method that I designated the
Sieve of Zakiya (SoZ), to differentiate if from the
Sieve of Eratosthenes (SoE) and Sieve of Atkin (SoA).

Here is the site you can download the paper and various
implementations code sets in 3 languages (Forth, Ruby, Python).

http://www.4shared.com/dir/7467736/97bd7b71/sharing.html

Basically, I have found a way to find a class of prime number
generators of the form:

Pj = m*k + (1+ri)

where m is a modulus value, and ri are what I call residues mod m.

I list 9 generators in my paper, but here are the first 4:

P3 = 6*k+(1,5)
P5 = 30*k+ (1, 7, 11, 13, 17, 19, 23, 29)
P60 = 60*k+(1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59)
P110 = 110*k+(1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 37, 39, 41,
43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 79, 81, 83, 87, 89,
91, 93, 97, 99, 101, 103, 107, 109)

My paper is NOT A FORMAL MATH DOCUMENT but an explanation of the process
I used to find how to create these generators and how to code them in
software
and their relative performance. One person has coded some of my
generators in
C on multi-core Intel systems, and has shown some are faster than the
SoA.

My interest in posting here is to get people interested in figuring out
the
implications of the underlying mathematics, in a formal way.

Presently, I know (and explain) how to computationally find all the
residues for
a given generator, and conjecture on the properties and form of larger
ones.

What would be really revealing is to figure out an analytical way to
generate
the modulus/residues values for any generator. I know how to find the
residues
computationally but not analytically.

As an example, the SoA is based on quadratic forms, from this paper:

"Prime Sieves Using Binary Quadratic Forms"
by A.O.L. Atkin and Daniel J. Bernstein. Download the pdf of the paper
here:

http://thedjbway.org/scientific/primegen.html

They use three equations of eliptic curves. I showd how to combine the
residues of
these 3 curves to form my generator P60, so this suggests there are
similar
representations for my other generators.

Of particualr interest is the properties of what I call the Modular
Compliments (MC)
which always come in pairs which add to equal the modulus value m.
Especailly of
interest is where do the MCs which are compliments of prime composites
come from.
I suspect they come from the complex plane, and maybe they have something
to do
with the Riemann Hypotheses?

Anyway, I AM NOT A FORMAL MATHEMATICIAN, I'm an engineer (BS/MSEE
Cornell/Ga Tech) who worked 15 years at NASA. This stuff is just a
hobby/fun interest of mine.

I figure serious mathematicians can answer some of these questions.

Thanks for your indulgence.

Jabari Zakiya

Postdoc positions at Univ. of Copenhagen

2008-09-18T13:27:50Z

A number of postdoc positions at the University of Copenhagen, Denmark,
become available on 1 September, 2009. The positions are across all fields
including number theory.

Application deadline is 1 December, 2008.

The following link provides the official announcement as well as more
information.

http://www.math.ku.dk/english/research/postdoc_apply/

Ian Kiming

Factoring workshop in Nancy, Oct 7-9

2008-09-18T13:27:49Z

[ please accept apologies for multiple postings ]

CADO workshop on integer factorization
INRIA Nancy-Grand Est (INRIA Lorraine), Villers-lès-Nancy, France
October 7 – 9, 2008

http://cado.gforge.inria.fr/workshop/

** second announcement, registration opening **

The CADO workshop on integer factorization will take place at INRIA
Lorraine (LORIA), Villers-lès-Nancy, France, from October 7th to 9th,
2008.

The topic of the workshop is integer factorization and in particular the
Number Field Sieve algorithm and its implementations. The CADO workshop
takes place just before Sage Days 10, scheduled in the same place from
Oct. 10th to 15th.

Confirmed speakers:

* Kazumaro Aoki, NTT ;
(title to be announced)
* Daniel J. Bernstein, University of Illinois at Chicago ;
Predicting NFS time (tentative title)
* Thorsten Kleinjung, EPFL-LACAL ;
(title to be announced)
* Reynald Lercier, CELAR & Université de Rennes 1 ;
(title to be announced)
* Peter L. Montgomery, Microsoft Research ;
Preliminary Design of Post-Sieving Processing for RSA-768
* Jason S. Papadopoulos, ViaSat Inc ;
A Self-Tuning Filtering Implementation for the Number Field Sieve

Besides invited talks, some contributed talks on the topics of the
workshop will be included. As there are still a few slots left,
participants willing to present work on integer factorizations are kindly
requested to contact the organizers as soon as possible. Note that no
proceedings are planned for the workshop.

Registration

The registration is open yet. The registration fee is 50 euros, and
includes lunches. Registration is accessible from the workshop web page.
The registration deadline is October 1st, 2008.

About CADO

The CADO project is an initiative from the CACAO project-team at INRIA
Lorraine, the TANC project-team at INRIA Saclay, and the number theory
group at IECN. CADO and the CADO workshop are funded by ANR, the French
research agency.

Organisation and contacts

* Scientific organization: Pierrick.Gaudry at loria.fr,
Francois.Morain at lix.polytechnique.fr, Gerald.Tenenbaum at
iecn.u-nancy.fr Emmanuel.Thome at normalesup.org, Paul.Zimmermann
at loria.fr

* Local arrangements: Anne-Lise.Charbonnier at loria.fr

Acknowledgements

The organizers acknowledge the support of the following funding and
hosting institutions:

ANR -- Agence Nationale de la Recherche
INRIA -- Institut National de Recherche en Informatique et Automatique
CNRS -- Centre National de la Recherche Scientifique
Nancy Université
École polytechnique
LORIA -- Laboratoire Lorrain de Recherche en Informatique et Applications
IECN -- Institut Élie Cartan, Nancy

Almost universal forms of the type ax^2+by^2+cT_z

2008-09-08T15:04:21Z

Dear number theorists,

In 1997 Ken Ono and K. Soundararajan [Invent. Math. 130(1997)] proved
that under the generalized Riemann hypothesis any positive odd integer
greater than 2719 can be represented by the famous Ramanujan form
x^2+y^2+10z^2, equivalently the form 2x^2+5y^2+4T_z represents all
integers greater than 1359, where T_z denotes the triangular number
z(z+1)/2.

In the following preprint posted to arXiv on August 20, 2008,

Ben Kane and Zhi-Wei Sun,
On almost universal mixed sums of squares and triangular numbers,
http://arxiv.org/abs/0808.2761

we used modular forms and the theory of quadratic forms to determine
completely when the form ax^2+by^2+cT_z (with a,b,c positive integers)
is almost universal (i.e., it represents sufficiently large integers)
and also establish similar results for the forms ax^2+bT_y+cT_z and
aT_x+bT_y+cT_z.

Given a positive integer a, we have the following consequences of our
main theorems:

(i) All sufficiently large odd numbers have the form 2ax^2+y^2+z^2
if and only if all prime divisors of a are congruent to 1 modulo 4.

(ii) The form ax^2+y^2+T_z is almost universal if and only if each odd
prime divisor of a is
congruent to 1 or 3 modulo 8.

(iii) ax^2+T_y+T_z is almost universal if and only if all odd prime
divisors of a are
congruent to 1 modulo 4.

(iv) When v_2(a) is not three, the form
aT_x+T_y+T_z is almost universal if and only if all odd prime
divisors of a are congruent to 1 modulo 4 and
v_2(a) does not belong to {5,7,...}, where v_2(a) is the 2-adic order
of a.

Perhaps the above information might interest some of you. Any comments
are welcome!

Zhi-Wei Sun (Nanjing University, China)
http://math.nju.edu.cn/~zwsun

latticechallenge.org

2008-09-04T05:01:06Z

We would like to announce latticechallenge.org.

Building upon a popular paper by Ajtai, we have constructed lattice
bases for which the solution of SVP implies a solution of SVP in all
lattices of a certain smaller dimension. This does not mean that one
can solve all instances simultaneously, but rather that one can solve
even the worst case instances. We think these lattice bases are hard
instances and most fitting to test and compare modern lattice
reduction algorithms.

The contestants are given lattice bases of lattices L_m, together with a
norm bound b. Initially, we set b=n(m).

The goal is to find a vector v in L_m, with |v| < b.
Each solution v to the challenge decreases b to |v|.

The challenge is hosted at

http://www.latticechallenge.org

Re: A question about perfect powers

2008-09-04T05:01:06Z

> Does then NO "n" exist, such that
> ( n! + prime(n) )
> yields integral m^k where k>1 ?
>
> prime(n) denotes the n-th prime.

Here's an easy proof that at least

n! + p_n = m^2

has no solutions if n is "large enough." First, 1! + 2 = 3, 2! + 3 =
7, and 3! + 5 = 11 are not squares. For n > 3, n! is divisible by 8, so

n! + p_n is congruent to p_n modulo 8.

This can only be a square if p_n is congruent to 1 modulo 8, so 2 is a
quadratic residue mod p_n; (2/p_n) = +1. Assume n > 3, p_n is
congruent to 1 modulo 8, and

n! + p_n = m^2.

Then if q <= n is an odd prime, p_n is congruent to m^2 modulo q, so
(p_n/q) = +1. Since p_n is congruent to 1 modulo 8,

(q/p_n) = +1 by quadratic reciprocity.

Since also (2/p_n) = +1, we have (k/p_n) = +1 for all k <= n, so that

SUM, k = 1 to n, (k/p_n) = n.

But this sum cannot exceed sqrt(p_n) * ln(p_n) by the Polya-
Vinogradov inequality. Since p_n is only about n*ln(n), for "large
enough" n we have a contradiction.

I'm sure that how large is "large enough" can be found, and that it
isn't terribly large, and the thing can be checked up to that point,
but I'm too lazy to work out the details
:-D

A question about perfect powers

2008-09-02T21:25:43Z

Does then NO "n" exist, such that
( n! + prime(n) )
yields integral m^k where k>1 ?

prime(n) denotes the n-th prime.

Alexander R. Povolotsky

Lucas-Lehmer Test analogue for repunits

2008-08-29T17:09:54Z

Dear number theorists;
Does there exist a known analogue to Lucas-Lehmer Test for repunits (or
generalized repunits)? I would be really grateful for a link or a short
description like:
Lucas_Lehmer_Test(p):
s := 4;
for i from 3 to p do s := s^2-2 mod 2^p-1;
if s == 0 then
2^p-1 is prime
else
2^p-1 is composite;
(copied from http://primes.utm.edu/mersenne/)
Thanks in advance,
Max

CFP: AMS Special Session on Computational Algebraic and Analytic Geometry

2008-08-29T17:09:53Z

Dear Colleagues,

We would like to solicit your participation in the AMS Special Session
titled "Computational Algebraic and Analytic Geometry for
Low-dimensional Varieties" to be held on Tuesday, January 6, 2009,
as part of the AMS/MAA Joint Mathematics Meetings in Washington, DC.
The session webpage is here:

http://www.ams.org/amsmtgs/2110_program_ss48.html .

The purpose of this session is to bring together researchers
interested in algorithms for algebraic curves, Riemann surfaces, and
algebraic surfaces. We strive to have talks covering algebraic,
arithmetic, and analytic aspects of this area. This session is the
sixth in a biennial series that began in 1999. The web page for the
most recent session in New Orleans (2007) is here:

http://algcurves.albmath.org/files/New_Orleans.html .

We welcome your proposals for twenty-minute talks. Contact us directly
with any questions or suggestions. Please submit your abstract
by the deadline of September 16 using the AMS web form here:

http://www.ams.org/cgi-bin/abstracts/abstract.pl .

Because we receive more proposals than the time allotted to us can
accommodate, we regret that we can not accept all proposals. As in
previous years, we expect to publish abstracts in the "ACM
Communications in Computer Algebra" (CCA). On the occasion of the
tenth anniversary of our session, we are considering the possibility
of preparing a collection of papers based on this session for a volume
of "Contemporary Mathematics".

We look forward to seeing you in Washington, DC.

Best regards,

Mika Seppala <seppala@...>
Tony Shaska <shaska@...>
Emil Volcheck <volcheck@...>

Close, but no cigar...

2008-08-27T23:22:39Z

[I'm resending this message because my redistribution program managed
to mangle it -- your editor]

I was hoping to "fill in" an exceptional sequence of three units to
get an exceptional sequence of four units. Unfortunately, the attempt
failed. However, it produced a couple of curious examples, of which
the following is one. With Pari-GP bludgeoning the algebra into
submission...

? f = x^8 - 3*x^6 + 2*x^4 + 1
%1 = x^8 - 3*x^6 + 2*x^4 + 1
? f1 = x^8 - 8*x^7 + 25*x^6 - 38*x^5 + 27*x^4 - 4*x^3 - 5*x^2 + 2*x + 1
%2 = x^8 - 8*x^7 + 25*x^6 - 38*x^5 + 27*x^4 - 4*x^3 - 5*x^2 + 2*x + 1
? lift(Mod(subst(f1,x,x+1),f))
%3 = 0
? tx = x^5-2*x^3
%4 = x^5 - 2*x^3
? lift(Mod(subst(polcyclo(12),x,tx),f))
%5 = 0
? lift(Mod(subst(f,x,x+tx),f))
%6 = 0
? g = x^8 - 4*x^7 + 7*x^6 - 4*x^5 - 2*x^3 + 16*x^2 - 2*x + 1
%7 = x^8 - 4*x^7 + 7*x^6 - 4*x^5 - 2*x^3 + 16*x^2 - 2*x + 1
? lift(Mod(subst(g,x,x+1-tx^2),f))
%8 = 0

This says that if f(r) = 0, then t = t(r) = r^5 - 2*r^3 is a primitive
twelfth root of unity. Also,

r + 1 is a zero of f1, so is also a unit,
r + t is another zero of f, so is also a unit, and
r + 1 - t^2 is a zero of g, so is a unit as well.

Now 1 - t^2 = -t^4 = t^10 is a primitive sixth root of unity.

Thus r, r + 1, r + t and

r, r + 1, r + 1 - t^2

are exceptional sequences of three units.

However, 1 - t - t^2 is not a unit (its norm from Q(t) to Q is 4), so

r, r + 1, r + t, r + 1 - t^2 is NOT an exceptional sequence of four
units.

The Galois group is _{8}T_{9} and Q(r) has class number 1:

? polgalois(f)
%9 = [16, 1, 9, "E(8):2=D(4)[x]2"]
? k=bnfinit(f);
? k.clgp.no
%11 = 1

In case anyone is curious, the factorization of field discriminant is

? factor(k.disc)
%12 =
[2 8]

[3 4]

[13 2]

----------
At the request of the Moderator, for the benefit of the list I am
appending an explanation of the term "exceptional sequence of N
units." Nagell [1] originated the term "exceptional units" to mean
two units (in the ring of integers of a number field) whose sum is 1.

Lenstra [2] did the following: Given any sequence in R

r_1, r_2, ... r_N

for which N > 1 and r_i - r_j a unit of R for i \ne j, then taking w_i
= (r_i - r_1)/(r_2 - r_1) gives a sequence of w's with w_1 = 0, w_2 =
1, and w_i - w_j a unit for i \ne j.

If N > 2, then any w_i for i > 2 is an exceptional unit. Lenstra
showed that if the supremum M of the length of such sequences is
"large enough" (M is always finite), then R is Euclidean.

Leutbecher and Martinet [3] called sequences, the difference of any
two terms of which is a unit, "exceptional sequences."

An exceptional sequence of N units in R is a sequence of N units of R,

u1, u_2, ..., u_N

which is an exceptional sequence in R. That is, a sequence of N units
in R, the difference of any two of which is a unit.

I had inquired whether there was a "standard" terminology for such
sequences in this group on September 16, 2007 and received several
replies. Thanks again to all.

Note that it is possible to have exceptional sequences of M terms in
R, without R having any exceptional sequences of M units. For example
in Z[i] the sequence 0, 1 is an exceptional sequence, but no two of
the units 1, -1, i, and -i differ by a unit, so Z[i] has no
exceptional units. Likewise, if w is a primitive cube root of unity,
the sequence 0, 1, -w is an exceptional sequence in R = Z[w]. But
there are only two nonzero residue classes mod (1 - w)R, so there
cannot be an exceptional sequence of three units in R.

[1] T. Nagell, Sur un type particulier d’unit\{�´}es alg\
{�´}ebriques, Arkiv f. Matem 8 #18 (1969), 163–164

[2] H. W. Lenstra, jr., Euclidean number ﬁelds of large degree,
Invent. Math. 38 (1977) 237–254

[3] A. Leutbecher and J. Martinet, Lenstra’s constant and Euclidean
number ﬁelds, Ast\{�´}erisque 94 (1982) 87–131

Close, but no cigar...

2008-08-27T15:32:20Z

I was hoping to "fill in" an exceptional sequence of three units to=20=20
get an exceptional sequence of four units. Unfortunately, the attempt=20=20
failed. However, it produced a couple of curious examples, of which=20=20
the following is one. With Pari-GP bludgeoning the algebra into=20=20
submission...

? f =3D x^8 - 3*x^6 + 2*x^4 + 1
%1 =3D x^8 - 3*x^6 + 2*x^4 + 1
? f1 =3D x^8 - 8*x^7 + 25*x^6 - 38*x^5 + 27*x^4 - 4*x^3 - 5*x^2 + 2*x + 1
%2 =3D x^8 - 8*x^7 + 25*x^6 - 38*x^5 + 27*x^4 - 4*x^3 - 5*x^2 + 2*x + 1
? lift(Mod(subst(f1,x,x+1),f))
%3 =3D 0
? tx =3D x^5-2*x^3
%4 =3D x^5 - 2*x^3
? lift(Mod(subst(polcyclo(12),x,tx),f))
%5 =3D 0
? lift(Mod(subst(f,x,x+tx),f))
%6 =3D 0
? g =3D x^8 - 4*x^7 + 7*x^6 - 4*x^5 - 2*x^3 + 16*x^2 - 2*x + 1
%7 =3D x^8 - 4*x^7 + 7*x^6 - 4*x^5 - 2*x^3 + 16*x^2 - 2*x + 1
? lift(Mod(subst(g,x,x+1-tx^2),f))
%8 =3D 0

This says that if f(r) =3D 0, then t =3D t(r) =3D r^5 - 2*r^3 is a primitiv=
e=20=20
twelfth root of unity. Also,

r + 1 is a zero of f1, so is also a unit,
r + t is another zero of f, so is also a unit, and
r + 1 - t^2 is a zero of g, so is a unit as well.

Now 1 - t^2 =3D -t^4 =3D t^10 is a primitive sixth root of unity.

Thus r, r + 1, r + t and

r, r + 1, r + 1 - t^2

are exceptional sequences of three units.

However, 1 - t - t^2 is not a unit (its norm from Q(t) to Q is 4), so

r, r + 1, r + t, r + 1 - t^2 is NOT an exceptional sequence of four=20=20
units.

The Galois group is _{8}T_{9} and Q(r) has class number 1:

? polgalois(f)
%9 =3D [16, 1, 9, "E(8):2=3DD(4)[x]2"]
? k=3Dbnfinit(f);
? k.clgp.no
%11 =3D 1

In case anyone is curious, the factorization of field discriminant is

? factor(k.disc)
%12 =3D
[2 8]

[3 4]

[13 2]

----------
At the request of the Moderator, for the benefit of the list I am=20=20
appending an explanation of the term "exceptional sequence of N=20=20
units." Nagell [1] originated the term "exceptional units" to mean=20=20
two units (in the ring of integers of a number field) whose sum is 1.

Lenstra [2] did the following: Given any sequence in R

r_1, r_2, ... r_N

for which N > 1 and r_i - r_j a unit of R for i \ne j, then taking w_i=20=20
=3D (r_i - r_1)/(r_2 - r_1) gives a sequence of w's with w_1 =3D 0, w_2 =3D=
=20=20
1, and w_i - w_j a unit for i \ne j.

If N > 2, then any w_i for i > 2 is an exceptional unit. Lenstra=20=20
showed that if the supremum M of the length of such sequences is=20=20
"large enough" (M is always finite), then R is Euclidean.

Leutbecher and Martinet [3] called sequences, the difference of any=20=20
two terms of which is a unit, "exceptional sequences."

An exceptional sequence of N units in R is a sequence of N units of R,

u1, u_2, ..., u_N

which is an exceptional sequence in R. That is, a sequence of N units=20=20
in R, the difference of any two of which is a unit.

I had inquired whether there was a "standard" terminology for such=20=20
sequences in this group on September 16, 2007 and received several=20=20
replies. Thanks again to all.

Note that it is possible to have exceptional sequences of M terms in=20=20
R, without R having any exceptional sequences of M units. For example=20=20
in Z[i] the sequence 0, 1 is an exceptional sequence, but no two of=20=20
the units 1, -1, i, and -i differ by a unit, so Z[i] has no=20=20
exceptional units. Likewise, if w is a primitive cube root of unity,=20=20
the sequence 0, 1, -w is an exceptional sequence in R =3D Z[w]. But=20=20
there are only two nonzero residue classes mod (1 - w)R, so there=20=20
cannot be an exceptional sequence of three units in R.

[1] T. Nagell, Sur un type particulier d=E2=80=99unit\{=C2=C2=B4}es alg\=20
{=C2=C2=B4}ebriques, Arkiv f. Matem 8 #18 (1969), 163=E2=80=93164

[2] H. W. Lenstra, jr., Euclidean number =EF=AC=81elds of large degree,=20=
=20
Invent. Math. 38 (1977) 237=E2=80=93254

[3] A. Leutbecher and J. Martinet, Lenstra=E2=80=99s constant and Euclidean=
=20=20
number =EF=AC=81elds, Ast\{=C2=C2=B4}erisque 94 (1982) 87=E2=80=93131

Re: class number of IQ(2^{1/n})

2008-08-24T09:22:07Z

> Are there statements about the class number of IQ(2^{1/n}) (beside
> general
> theorems like the Minkowski bound)?
>
> Using PARI, I found that it equals 1 for n <= 42.
>
> Does someone know a n for which the class number is > 1?
>
> It can't be known that it is always 1 since this would solve an open
> question.

The following doesn't answer the question, but it may be of interest
as a nice example:

If p is an odd prime, and

T(x) = x^p - 2,

then (easy calculation) disc(T(x)) = (-1)^(p-1)/2 * p^p * 2^(p-1).

If T(r) = 0, then using Dedekind's Criterion, we find that

the maximal order R in F = Q(r) is Z[r], UNLESS p^2 divides 2^p - 2,

i.e. R = Z[r] unless p is a Wieferich prime WRT the base 2.

If p is NOT a Wieferich prime, p is totally and wildly ramified in R/Z.

If p IS a Wieferich prime, Z[r] has index p in R, the field
discriminant is 1/p^2 times the polynomial discriminant, and

pR = P_{1} * P_{2}^(p-1)

where P_{1} and P_{2} are both of relative degree 1. So if p IS a
Wieferich prime, p is only tamely ramified in R/Z.

If p is a Wieferich prime, L = F(exp(2*i*pi/p)) is the splitting field
of T(x), and S the maximal order of Lthen P_{1} is totally ramified in
L/F, and P_{2} splits completely in S/R.

If p is NOT a Wieferich prime, p is totally ramified in S/Z.

Kurt Foster

class number of IQ(2^{1/n})

2008-08-23T21:43:16Z

Are there statements about the class number of IQ(2^{1/n}) (beside general
theorems like the Minkowski bound)?

Using PARI, I found that it equals 1 for n <= 42.

Does someone know a n for which the class number is > 1?

It can't be known that it is always 1 since this would solve an open question.

Thue equation F(X,Y)=d

2008-08-13T22:54:22Z

Dear number theoristic:

I would like to know if there is some result about the following Thue=20
equation problem.

PROBLEM: Let F(X,Y) an irreducible polynomial with integer coefficients.=20
And let be {F(X,Y)=3Dd} a family of Thue equation depending on the=20
variable d (an integer). Is there a solution of all this Thue equation=20
depending on d?

All the best and thanks in adavanced,

Enrique

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D

Enrique Gonzalez-Jimenez
Departamento de Matem=C3=A1ticas
Facultad de Ciencias
Universidad Aut=C3=B3noma de Madrid
Campus de Cantoblanco
28049 Madrid, Spain

Office: C-XV-610
Phone: +34 914975236
Fax: +34 914974889
http://www.uam.es/enrique.gonzalez.jimenez

=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D

Re: References for Diophantine Equation of the form MX^p + NY^p = QZ^p

2008-08-09T06:27:58Z

Yes, of course, all of you are right. I believe I should have specified N > 1, M > 1, Q > 1, and nontrivial solutions in X, Y, Z, meaning integer solutions in X, Y, Z other than 0 or 1. Sorry.
Anyway thank you very much (and humbly!) for all the responses on this and for the Faltings reference that is related to this.
Respectfully,

Robert Betts
University of Massachusetts Lowell
Department of Computer and Electrical Engineering
One University Avenue
Lowell, MA 01854
....................................................................................................................................
[Previous]
That proof would be wrong. One can pick p,M,N,X,Y arbitrary, take Z=1 and
solve for Q.

The best result one can hope for without specifying N,M,Q follows from
Faltings' proof of the Mordell-conjecture:

For any p>3 (not necessarily prime) and any N,M,Q >= 1, there are only
finitely many integral solutions X,Y,Z with GCD(X,Y,Z)=1.

............................................................
On Tue, 5 Aug 2008, Robert Betts wrote:

> Let X, Y, Z, M, N, Q all be pairwise relatively prime positive
> integers where M >= 1, N >= 1, Q >= 1 and p an odd prime. Does anyone
> know if it has been proved or disproved that
>
> MX^p + NY^p = QZ^p,
>
> has no solutions X, Y, Z for all odd primes p? I would appreciate any
> references on this Diophantine equation, if anyone can think of any.

That proof would be wrong. One can pick p,M,N,X,Y arbitrary, take Z=1 and
solve for Q.

The best result one can hope for without specifying N,M,Q follows from
Faltings' proof of the Mordell-conjecture:

For any p>3 (not necessarily prime) and any N,M,Q >= 1, there are only
finitely many integral solutions X,Y,Z with GCD(X,Y,Z)=1.

Respectfully,

Robert Betts

Re: Special Session on Continued Fractions

2008-08-09T06:27:58Z

Some additional details:
- talks will be 20 minutes in length with 5 minutes for questions;
- the deadline for submitting an abstract is September 16th;
- abstracts must be submitted online at
http://www.ams.org/cgi-bin/abstracts/abstract.pl
- some more information is available at
http://math.wcupa.edu/~mclaughlin/ss_letter_for_submissions-3.doc

Jimmy Mc Laughlin

On Mon, 28 Jul 2008 23:42:13 -0400, James Mc Laughlin <clashton@...>
wrote:

>I would like to draw the attention of list members to
> a special session on continued fractions,
>http://www.ams.org/amsmtgs/2110_program_ss50.html#title
>which will be held at the Joint Meetings in Washington, January 2009.
>
>This will be similar in nature to a similar special session held in
>San Antonio, Texas, in January 2006, where the aim was to bring people
>who work with algebraic/number theoretic aspects of continued fractions
>in contact with people who work on the analytic side.
>See this webpage for the list of talks given in San Antonio.
>http://math.wcupa.edu/~mclaughlin/cfspecialsession.htm
>
>If you are interested in giving a talk on a topic with connections to
>continued fractions,
>please send the title and abstract to
>Nancy Wyshinski, nancy.wyshinski@...
>
>Jimmy Mc Laughlin
>=========================================================================

Re: References for Diophantine Equation of the form MX^p + NY^p = QZ^p

2008-08-09T06:27:57Z

One can easily find solutions by inspection.

Let (x,y,z) = (1,1,A) for any A > 1.

Then mx^p + ny^p = z^p simply for m+n = z^p,
e.g. 125 * 1^7 + 3* 1^7 = 2^7 etc.

(x,y,z) = (1,2,4) is a solution to e.g. 24x^3 + 5y^3 = z^3

BTW, I tried to contact you privately.
UMASS Lowell does not list Robert Betts on its website
and the switchboard could not find you either......

Seemingly simple question...

2008-08-09T06:27:56Z

If n > 2, then k <--> n-k defines a one-to-one correspondence between
the numbers k in (0,n/2) which are relatively prime to n, and the
numbers in (n/2,n) which are relatively prime to n. Thus, there are
phi(n)/2 numbers relatively prime to n in each interval.

Now suppose n > 1 is odd. I want to know when the number of k
relatively prime to n in the interval (n/4, n/2) is even, and when the
number of such k is odd.

When n = p^e, p > 2 prime, e >= 1, then by exact count the number in
question is even if the quadratic character (2/p) = +1, and odd if (2/
p) = -1.

If n is odd and has two or more prime factors, phi(n) is divisible by
4, so I know that the number of numbers relatively prime to n in the
intervals (0,n/4) and (n/4,n/2) are either both even or both odd. I
suspect they're always both even. But how to prove it? I'm sure
there's a simple argument, known for centuries, which I'm overlooking...

Re: References for Diophantine Equation of the form MX^p + NY^p = QZ^p

2008-08-09T06:27:56Z

On Tue, 5 Aug 2008, Robert Betts wrote:

> Let X, Y, Z, M, N, Q all be pairwise relatively prime positive
> integers where M >= 1, N >= 1, Q >= 1 and p an odd prime. Does anyone
> know if it has been proved or disproved that
>
> MX^p + NY^p = QZ^p,
>
> has no solutions X, Y, Z for all odd primes p? I would appreciate any
> references on this Diophantine equation, if anyone can think of any.

That proof would be wrong. One can pick p,M,N,X,Y arbitrary, take Z=1 and
solve for Q.

The best result one can hope for without specifying N,M,Q follows from
Faltings' proof of the Mordell-conjecture:

For any p>3 (not necessarily prime) and any N,M,Q >= 1, there are only
finitely many integral solutions X,Y,Z with GCD(X,Y,Z)=1.

--------------------------------------------------------
Nils Bruin Department of Mathematics
telephone: (778) 782 3794 Simon Fraser University
fax: (778) 782 4947 Burnaby, BC
e-mail: nbruin@... CANADA, V5A 1S6
WWW: http://www.cecm.sfu.ca/~nbruin

Re: References for Diophantine Equation of the form MX^p + NY^p = QZ^p

2008-08-09T06:27:55Z

Thanks for the additional responses. But after the responses I did
follow up with the second posting that I meant to state M > 1, N > 1,
Q > 1, and nontrivial solutions in X, Y, Z, meaning solutions in X, Y,
Z other than 0 or 1. There is some 1995 literature on MX^p + NY^p =
QZ^p having finitely many solutions where X, Y, Z are pairwise
relatively prime. I was actually inquiring if anyone had proved or
disproved this had no solutions.

The UML switchboard operators had no listing for me? More than likely
one reason is because I am not due on the UML campus until
Sept. 2008. I am affiliated with two universities and two campuses,
one of which is UMASS/UML here in the states.

Again, thanks for the helpful responses.
Respectfully,

Robert Betts
University of Massachusetts Lowell
Department of Computer and Electrical Engineering
One University Avenue
Lowell, MA 01854

[Previous]

One can easily find solutions by inspection.Let (x,y,z) =3(1,1,A) for
any A > 1.Then mx^p + ny^p = z^p simply for m+n = z^p,e.g. 125 * 1^7
+ 3* 1^7 = 2^7 etc.(x,y,z) = (1,2,4) is a solution to e.g. 24x^3 +
5y^3 = z^3BTW, I tried to contact you privately.UMASS Lowell does not
list Robert Betts on its websiteand the switchboard could not find you
either......