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Subject: sci.math: Frequently Asked Questions [1/3]
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Version: 5.0
This is the list of Frequently Asked Questions for sci.math (version 5.0).
Any contributions/suggestions/corrections are most welcome. Please use
* email * on any comment concerning the FAQ list.
Section 1 of 3, questions 1Q to 5Q.
Table of Contents

1Q. Fermat's Last Theorem, status of ..
2Q. Values of Record Numbers
3Q. Formula for prime numbers...
4Q. Digits of Pi, computation and references
5Q. Odd Perfect Number
6Q. Computer Algebra Systems, application of ..
7Q. Computer Algebra Systems, references to ..
8Q. Fields Medal, general info ..
9Q. Four Colour Theorem, proof of ..
10Q. 0^0=1. A comprehensive approach
11Q. 0.999... = 1. Properties of the real numbers ..
12Q. There are three doors, The Monty Hall problem, Master Mind and
other games ..
13Q. Surface and Volume of the nball
14Q. f(x)^f(x)=x, name of the function ..
15Q. Projective plane of order 10 ..
16Q. How to compute day of week of a given date
17Q. Axiom of Choice and/or Continuum Hypothesis?
18Q. Cutting a sphere into pieces of larger volume
19Q. Pointers to Quaternions
20Q. Erdos Number
21Q. Why is there no Nobel in mathematics?
22Q. General References and textbooks...
23Q. Interest Rate...
24Q. Euler's formula e^(i Pi) =  1 ...
1Q: What is the current status of Fermat's last theorem?
and
Did Fermat prove this theorem?
Fermat's Last Theorem:
There are no positive integers x,y,z, and n > 2 such that x^n + y^n = z^n.
I heard that claimed to have proved it but later
on the proof was found to be wrong. ...
A: The status of FLT has remained remarkably constant. Every few
years, someone claims to have a proof ... but oh, wait, not quite.
UPDATE... UPDATE... UPDATE
Andrew Wiles, a researcher at Princeton, Cambridge claims to have
found a proof.
SECOND UPDATE...
A mistake has been found. Wiles is working on it. People remain
mildly optimistic about his chances of fixing the error.
The proposed proof goes like this:
The proof was presented in Cambridge, UK during a three day seminar
to an audience including some of the leading experts in the field.
The manuscript has been submitted to INVENTIONES MATHEMATICAE, and
is currently under review. Preprints are not available until the
proof checks out. Wiles is giving a full seminar on the proof this
spring.
The proof is long and cumbersome, but here are some of the first
few details:
*From Ken Ribet:
Here is a brief summary of what Wiles said in his three lectures.
The method of Wiles borrows results and techniques from lots and lots
of people. To mention a few: Mazur, Hida, Flach, Kolyvagin, yours
truly, Wiles himself (older papers by Wiles), Rubin... The way he does
it is roughly as follows. Start with a mod p representation of the
Galois group of Q which is known to be modular. You want to prove that
all its lifts with a certain property are modular. This means that the
canonical map from Mazur's universal deformation ring to its "maximal
Hecke algebra" quotient is an isomorphism. To prove a map like this is
an isomorphism, you can give some sufficient conditions based on
commutative algebra. Most notably, you have to bound the order of a
cohomology group which looks like a Selmer group for Sym^2 of the
representation attached to a modular form. The techniques for doing
this come from Flach; you also have to use Euler systems a la
Kolyvagin, except in some new geometric guise.
CLARIFICATION: This step in Wiles' manuscript, the Selmer group
bound, is currently considered to be incomplete by the reviewers.
Yet the reviewers (or at least those who have gone public) have
confidence that Wiles will fill it in. (Note that such gaps are
quite common in long proofs. In this particular case, just such
a bound was expected to be provable using Kolyvagin's techniques,
independently of anyone thinking of modularity. In the worst of
cases, and the gap is for real, what remains has to be recast, but
it is still extremely important number theory breakthrough work.)
If you take an elliptic curve over Q, you can look at the
representation of Gal on the 3division points of the curve. If you're
lucky, this will be known to be modular, because of results of Jerry
Tunnell (on base change). Thus, if you're lucky, the problem I
described above can be solved (there are most definitely some
hypotheses to check), and then the curve is modular. Basically, being
lucky means that the image of the representation of Galois on
3division points is GL(2,Z/3Z).
Suppose that you are unlucky, i.e., that your curve E has a rational
subgroup of order 3. Basically by inspection, you can prove that if it
has a rational subgroup of order 5 as well, then it can't be
semistable. (You look at the four noncuspidal rational points of
X_0(15).) So you can assume that E[5] is "nice." Then the idea is to
find an E' with the same 5division structure, for which E'[3] is
modular. (Then E' is modular, so E'[5] = E[5] is modular.) You
consider the modular curve X which parameterizes elliptic curves whose
5division points look like E[5]. This is a "twist" of X(5). It's
therefore of genus 0, and it has a rational point (namely, E), so it's
a projective line. Over that you look at the irreducible covering
which corresponds to some desired 3division structure. You use
Hilbert irreducibility and the Cebotarev density theorem (in some way
that hasn't yet sunk in) to produce a noncuspidal rational point of X
over which the covering remains irreducible. You take E' to be the
curve corresponding to this chosen rational point of X.
*From the previous version of the FAQ:
(b) conjectures arising from the study of elliptic curves and
modular forms.  The TaniyamaWeilShmimura conjecture.
There is a very important and well known conjecture known as the
TaniyamaWeilShimura conjecture that concerns elliptic curves.
This conjecture has been shown by the work of Frey, Serre, Ribet,
et. al. to imply FLT uniformly, not just asymptotically as with the
ABC conj.
The conjecture basically states that all elliptic curves can be
parameterized in terms of modular forms.
There is new work on the arithmetic of elliptic curves. Sha, the
TateShafarevich group on elliptic curves of rank 0 or 1. By the way
an interesting aspect of this work is that there is a close
connection between Sha, and some of the classical work on FLT. For
example, there is a classical proof that uses infinite descent to
prove FLT for n = 4. It can be shown that there is an elliptic curve
associated with FLT and that for n=4, Sha is trivial. It can also be
shown that in the cases where Sha is nontrivial, that
infinitedescent arguments do not work; that in some sense 'Sha
blocks the descent'. Somewhat more technically, Sha is an
obstruction to the localglobal principle [e.g. the HasseMinkowski
theorem].
*From Karl Rubin:
Theorem. If E is a semistable elliptic curve defined over Q,
then E is modular.
It has been known for some time, by work of Frey and Ribet, that
Fermat follows from this. If u^q + v^q + w^q = 0, then Frey had
the idea of looking at the (semistable) elliptic curve
y^2 = x(xa^q)(x+b^q). If this elliptic curve comes from a modular
form, then the work of Ribet on Serre's conjecture shows that there
would have to exist a modular form of weight 2 on Gamma_0(2). But
there are no such forms.
To prove the Theorem, start with an elliptic curve E, a prime p and let
rho_p : Gal(Q^bar/Q) > GL_2(Z/pZ)
be the representation giving the action of Galois on the ptorsion
E[p]. We wish to show that a _certain_ lift of this representation
to GL_2(Z_p) (namely, the padic representation on the Tate module
T_p(E)) is attached to a modular form. We will do this by using
Mazur's theory of deformations, to show that _every_ lifting which
'looks modular' in a certain precise sense is attached to a modular form.
Fix certain 'lifting data', such as the allowed ramification,
specified local behavior at p, etc. for the lift. This defines a
lifting problem, and Mazur proves that there is a universal
lift, i.e. a local ring R and a representation into GL_2(R) such
that every lift of the appropriate type factors through this one.
Now suppose that rho_p is modular, i.e. there is _some_ lift
of rho_p which is attached to a modular form. Then there is
also a hecke ring T, which is the maximal quotient of R with the
property that all _modular_ lifts factor through T. It is a
conjecture of Mazur that R = T, and it would follow from this
that _every_ lift of rho_p which 'looks modular' (in particular the
one we are interested in) is attached to a modular form.
Thus we need to know 2 things:
(a) rho_p is modular
(b) R = T.
It was proved by Tunnell that rho_3 is modular for every elliptic
curve. This is because PGL_2(Z/3Z) = S_4. So (a) will be satisfied
if we take p=3. This is crucial.
Wiles uses (a) to prove (b) under some restrictions on rho_p. Using
(a) and some commutative algebra (using the fact that T is Gorenstein,
'basically due to Mazur') Wiles reduces the statement T = R to
checking an inequality between the sizes of 2 groups. One of these
is related to the Selmer group of the symmetric square of the given
modular lifting of rho_p, and the other is related (by work of Hida)
to an Lvalue. The required inequality, which everyone presumes is
an instance of the BlochKato conjecture, is what Wiles needs to verify.
He does this using a Kolyvagintype Euler system argument. This is
the most technically difficult part of the proof, and is responsible
for most of the length of the manuscript. He uses modular
units to construct what he calls a 'geometric Euler system' of
cohomology classes. The inspiration for his construction comes
from work of Flach, who came up with what is essentially the
'bottom level' of this Euler system. But Wiles needed to go much
farther than Flach did. In the end, _under_certain_hypotheses_ on rho_p
he gets a workable Euler system and proves the desired inequality.
Among other things, it is necessary that rho_p is irreducible.
Suppose now that E is semistable.
Case 1. rho_3 is irreducible.
Take p=3. By Tunnell's theorem (a) above is true. Under these
hypotheses the argument above works for rho_3, so we conclude
that E is modular.
Case 2. rho_3 is reducible.
Take p=5. In this case rho_5 must be irreducible, or else E
would correspond to a rational point on X_0(15). But X_0(15)
has only 4 noncuspidal rational points, and these correspond to
nonsemistable curves. _If_ we knew that rho_5 were modular,
then the computation above would apply and E would be modular.
We will find a new semistable elliptic curve E' such that
rho_{E,5} = rho_{E',5} and rho_{E',3} is irreducible. Then
by Case I, E' is modular. Therefore rho_{E,5} = rho_{E',5}
does have a modular lifting and we will be done.
We need to construct such an E'. Let X denote the modular
curve whose points correspond to pairs (A, C) where A is an
elliptic curve and C is a subgroup of A isomorphic to the group
scheme E[5]. (All such curves will have mod5 representation
equal to rho_E.) This X is genus 0, and has one rational point
corresponding to E, so it has infinitely many. Now Wiles uses a
Hilbert Irreducibility argument to show that not all rational
points can be images of rational points on modular curves
covering X, corresponding to degenerate level 3 structure
(i.e. im(rho_3) not GL_2(Z/3)). In other words, an E' of the
type we need exists. (To make sure E' is semistable, choose
it 5adically close to E. Then it is semistable at 5, and at
other primes because rho_{E',5} = rho_{E,5}.)
Referencesm:
American Mathematical Monthly
January 1994.
Notices of the AMS, Februrary 1994.
2Q: What are the values of:
largest known Mersenne prime?
A: 2^8594331 is prime. It was discovered in 1994.
largest known prime?
A: The largest known prime is the Mersenne prime described above.
The largest known nonMersenne prime, is 391581*2^2161931.
See Brown, Noll, Parady, Smith, Smith, and Zarantonello, Letter to
the editor, American Mathematical Monthly, vol. 97, 1990, p. 214.
Throughout history, the largest known prime has almost always been
a Mersenne prime; the period between Brown et al's discovery in
Aug 1989 and Slowinski & Gage's in March 1992 is one of the few
exceptions.
largest known twin primes?
A: The largest known twin primes are 1691232 * 1001 * 10^4020 + 1,
which is a number with 4030 digits, found by H. Dubner.
The second largest known twin primes are 4650828 * 1001 * 10^3429 + 1.
They were found by H. Dubner
References:
B. K. Parady and J. F. Smith and S. E. Zarantonello,
Smith, Noll and Brown.
Largest known twin primes, Mathematics of Computation,
vol.55, 1990, pp. 381382.
largest Fermat number with known factorization?
A: F_11 = (2^(2^11)) + 1 which was factored by Brent & Morain in
1988. F9 = (2^(2^9)) + 1 = 2^512 + 1 was factored by
A.K. Lenstra, H.W. Lenstra Jr., M.S. Manasse & J.M. Pollard
in 1990. The factorization for F10 is NOT known.
Are there good algorithms to factor a given integer?
A: There are several that have subexponential estimated
running time, to mention just a few:
Continued fraction algorithm,
Class group method,
Quadratic sieve algorithm,
Elliptic curve algorithm,
Number field sieve,
Dixon's random squares algorithm,
Valle's twothirds algorithm,
Seysen's class group algorithm,
A.K. Lenstra, H.W. Lenstra Jr., "Algorithms in Number Theory",
in: J. van Leeuwen (ed.), Handbook of Theoretical Computer
Science, Volume A: Algorithms and Complexity, Elsevier, pp.
673715, 1990.
List of record numbers?
A: Chris Caldwell maintains a list called The Largest Known Primes.
Finger primes@math.utm.edu for a few record primes and ways to get the
email you the lists, but prefers you use the other methods.
different forms of this list. Currently the list is available
by anonymous ftp to math.utm.edu (directory /pub/math/primes) and by
gopher to unix1.utm.edu (directory 1/user/Public_FTP/pub/math/primes).
If nothing else works, you may send email to Chris for a copy
of the lists. (caldwell@utm.edu or caldwell@utmartn.bitnet).
What is the current status on Mersenne primes?
A: Mersenne primes are primes of the form 2^p1. For 2^p1 to be prime
we must have that p is prime. The following Mersenne primes are
known.
nr p year by

15 2,3,5,7,13 in or before the middle ages
67 17,19 1588 Cataldi
8 31 1750 Euler
9 61 1883 Pervouchine
10 89 1911 Powers
11 107 1914 Powers
12 127 1876 Lucas
1314 521,607 1952 Robinson
1517 1279,2203,2281 1952 Lehmer
18 3217 1957 Riesel
1920 4253,4423 1961 Hurwitz & Selfridge
2123 9689,9941,11213 1963 Gillies
24 19937 1971 Tuckerman
25 21701 1978 Noll & Nickel
26 23209 1979 Noll
27 44497 1979 Slowinski & Nelson
28 86243 1982 Slowinski
29 110503 1988 Colquitt & Welsh jr.
30 132049 1983 Slowinski
31 216091 1985 Slowinski
32? 756839 1992 Slowinski & Gage
33? 859433 1993 Slowinski.
The way to determine if 2^p1 is prime is to use the LucasLehmer
test:
Lucas_Lehmer_Test(p):
u := 4
for i from 3 to p do
u := u^22 mod 2^p1
od
if u == 0 then
2^p1 is prime
else
2^p1 is composite
fi
The following ranges have been checked completely:
2  355K,k 360K386K, and 430K  520K
More on Mersenne primes and the LucasLehmer test can be found in:
G.H. Hardy, E.M. Wright, An introduction to the theory of numbers,
fifth edition, 1979, pp. 16, 223225.
3Q. Formula for prime numbers...
Is there a polynomial which gives all the prime numbers?
No, there is not. This is a simple exercise to prove.
Is there a nonconstant polynomial that only takes on prime values?
It has been proved that no such polynomial exists.
The proof is simple enough that an high school student could probably
discover it. See, for example, Ribenboim's book _The Book of Prime
Number Records_.
Note, however, by the work of Jones, Sato, Wada, and Wiens, there *is* a
polynomial in 26 variables such that the set of primes coincides with
the set of *positive* values taken by this polynomial. See Ribenboim,
pp. 147150.
But most people would object to the term "formula" restricted to mean
polynomial. Can we not use summation signs, factorial, and the floor
function in our "formula"? If so, then indeed, there *are* formulas
for the prime numbers. Some of them are listed below.
If we can't, then exactly what operations do you allow and why?
Indeed, as I have previously argued, a reasonable interpretation of
the word "formula" is simply "Turing machine that halts on all inputs".
Under this interpretation, there certainly are halting Turing machines
which compute the n'th prime number. However, nobody knows how to
compute the n'th prime in time polynomial in log n. That's still
an open question.
Herb Wilf has addressed the question, "What is a formula?" in his
delightful article, "What is an answer?" which appeared in the
American Mathematical Monthly, 89 (1982), 289292. He draws the
distinction between "formula" and "good formula". Anyone who claims
"there is no formula for the prime numbers" should read this article.
Here are just a few articles that discuss "formulas" for primes. Almost
all of these do *not* require computation of the primes "ahead of time".
Most of them rely on standard mathematical functions such as summation,
factorial, greatest integer function, etc.
C. Isenkrahe, Math. Annalen 53 (1900), 4244.
W. H. Mills, Bull. Amer. Math. Soc. 53 (1947), 604.
L. Moser, Math. Mag. 23 (1950), 163164.
E. M. Wright, Amer. Math. Monthly 58 (1951), 616618. (Correction,
59 (1952), 99.)
E. M. Wright, J. Lond. Math. Soc. 29 (1954), 6371.
B. R. Srinivasan, J. Indian Math. Soc. 25 (1961), 3339.
C. P. Willans, Math. Gazette 48 (1964), 413415.
V. C. Harris, Nordisk Mat. Tidskr. 17 (1969), 82.
U. Dudley, Amer. Math. Monthly 76 (1969), 2328.
C. Vanden Eynden, Amer. Math. Monthly 79 (1972), 625.
S. W. Golomb, Amer. Math. Monthly 81 (1974), 752754.
For more references see
J.O. Shallit, E. Bach, _Algorithmic Number Theory_ (to be published,
MIT Press).
4Q: Where I can get pi up to a few hundred thousand digits of pi?
Does anyone have an algorithm to compute pi to those zillion
decimal places?
A: MAPLE or MATHEMATICA can give you 10,000 digits of Pi in a blink,
and they can compute another 20,0001,000,000 overnight (range depends
on hardware platform).
It is possible to retrieve 1.25+ million digits of pi via anonymous
ftp from the site wuarchive.wustl.edu, in the files pi.doc.Z and
pi.dat.Z which reside in subdirectory doc/misc/pi.
New York's Chudnovsky brothers have computed 2 billion digits of pi
on a homebrew computer.
How is pi calculated to many decimals ?
There are essentially 3 different methods.
1) One of the oldest is to use the power series expansion of atan(x)
atan(x)=xx^3/3+x^5/5... together with formulas like
pi=16*atan(1/5)4*atan(1/239). This gives about 1.4 decimals per term.
2) A second is to use formulas coming from ArithmeticGeometric mean
computations. A beautiful compendium of such formulas is given in the
book of Borwein and Borwein: Pi and the AGM, Canadian Math. Soc. Series,
John Wiley and Sons, New York, 1987. They have the advantage of converging
quadratically, i.e. you double the number of decimals per iteration.
For instance, to obtain 1 000 000 decimals, around 20 iterations are
sufficient. The disadvantage is that you need FFT type multiplication
to get a reasonable speed, and this is not so easy to program.
3) A third one comes from the theory of complex multiplication of elliptic
curves, and was discovered by S. Ramanujan. This gives a number of
beautiful formulas, but the most useful was missed by Ramanujan and
discovered by the Chudnovsky's. It is the following (slightly modified
for ease of programming):
Set k1=545140134;k2=13591409;k3=640320;k4=100100025;k5=327843840;k6=53360;
Then in AmsTeX notation
$\pi=\frac{k6\sqrt(k3)}{S}$, where
$$S=\sum_{n=0}^\infty (1)^n\frac{(6n)!(k2+nk1)}{n!^3(3n)!(8k4k5)^n}$$
The great advantages of this formula are that
1) It converges linearly, but very fast (more than 14 decimal digits per
term).
2) The way it is written, all operations to compute S can be programmed
very simply since it only involves multiplication/division by single
precision numbers. This is why the constant 8k4k5 appearing in the
denominator has been written this way instead of 262537412640768000.
This is how the Chudnovsky's have computed several billion decimals.
Question: how can I get a C program which computes pi?
Answer: if you are too lazy to use the hints above, you can use the
following 160 character C program (reportedly by Dik T. Winter) which
computes pi to 800 decimal digits.
int a=10000,b,c=2800,d,e,f[2801],g;main(){for(;bc;)f[b++]=a/5;
for(;d=0,g=c*2;c=14,printf("%.4d",e+d/a),e=d%a)for(b=c;d+=f[b]*a,
f[b]=d%g,d/=g,b;d*=b);}
References :
J. M. Borwein, P. B. Borwein, and D. H. Bailey, "Ramanujan,
Modular Equations, and Approximations to Pi", American Mathematical
Monthly, vol. 96, no. 3 (March 1989), p. 201  220.
P. Beckman
A history of pi
Golem Press, CO, 1971 (fourth edition 1977)
J.M. Borwein and P.B. Borwein
The arithmeticgeometric mean and fast computation of elementary
functions
SIAM Review, Vol. 26, 1984, pp. 351366
J.M. Borwein and P.B. Borwein
More quadratically converging algorithms for pi
Mathematics of Computation, Vol. 46, 1986, pp. 247253
J.M. Borwein and P.B. Borwein
Pi and the AGM  a study in analytic number theory and
computational complexity
Wiley, New York, 1987
Shlomo Breuer and Gideon Zwas
Mathematicaleducational aspects of the computation of pi
Int. J. Math. Educ. Sci. Technol., Vol. 15, No. 2, 1984,
pp. 231244
David Chudnovsky and Gregory Chudnovsky
The computation of classical constants, Columbia University,
Proc. Natl. Acad. Sci. USA, Vol. 86, 1989.
Y. Kanada and Y. Tamura
Calculation of pi to 10,013,395 decimal places based on the
GaussLegendre algorithm and Gauss arctangent relation
Computer Centre, University of Tokyo, 1983
Morris Newman and Daniel Shanks
On a sequence arising in series for pi
Mathematics of computation, Vol. 42, No. 165, Jan 1984,
pp. 199217
E. Salamin
Computation of pi using arithmeticgeometric mean
Mathematics of Computation, Vol. 30, 1976, pp. 565570
David Singmaster
The legal values of pi
The Mathematical Intelligencer, Vol. 7, No. 2, 1985
Stan Wagon
Is pi normal?
The Mathematical Intelligencer, Vol. 7, No. 3, 1985
5Q: Does there exist a number that is perfect and odd?
A given number is perfect if it is equal to the sum of all its proper
divisors. This question was first posed by Euclid in ancient Greece.
This question is still open. Euler proved that if N is an odd
perfect number, then in the prime power decomposition of N, exactly
one exponent is congruent to 1 mod 4 and all the other exponents are
even. Furthermore, the prime occurring to an odd power must itself be
congruent to 1 mod 4. A sketch of the proof appears in Exercise 87,
page 203 of Underwood Dudley's Elementary Number Theory, 2nd ed.
It has been shown that there are no odd perfect numbers < 10^300.
Copyright Notice
Copyright (c) 1993 A. LopezOrtiz
This FAQ is Copyright (C) 1994 by Alex LopezOrtiz. This text,
in whole or in part, may not be sold in any medium, including,
but not limited to electronic, CDROM, or published in print,
without the explicit, written permission of Alex LopezOrtiz.

Questions and Answers Edited and Compiled by:
Alex LopezOrtiz alopezo@maytag.UWaterloo.ca
Department of Computer Science University of Waterloo
Waterloo, Ontario Canada

Alex LopezOrtiz alopezo@neumann.UWaterloo.ca
Department of Computer Science University of Waterloo
Waterloo, Ontario Canada
http://daisy.uwaterloo.ca/~alopezo/home.html
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