Newsgroups: sci.math.research From: dbl@godel.msri.org (Daniel Lieman) Subject: Re: Fermat's Last Theorem has been proved Summary: Ribet's summary of the proof Date: Wed, 23 Jun 1993 18:24:10 GMT In article mckay@alcor.concordia.ca (John McKay) writes: >I have heard that an announcement to this effect has been >made in Cambridge by Andrew Wiles today, June 23rd. 1993. > Below is Ken Ribet's summary of Wiles' lectures on his proof. Daniel ----cut here I imagine that many of you have heard rumours about Wiles's announcement a few hours ago that he can prove Taniyama's conjecture for semistable elliptic curves over Q. This case of the Taniyama conjecture implies Fermat's Last Theorem, in view of the result that I proved a few years ago. (I proved that the "Frey elliptic curve" constructed from a possible solution to Fermat's equation cannot be modular, i.e., satisfy Taniyama's Conjecture. On the other hand, it is easy to see that it is semistable.) 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. If you take an elliptic curve over Q, you can look at the representation of Gal on the 3-division 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 3-division 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 non-cuspidal 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 5-division 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 parametrizes elliptic curves whose 5-division 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 3-division structure. You use Hilbert irreducibility and the Cebotarev density theorem (in some way that hasn't yet sunk in) to produce a non-cuspidal 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. -ken ribet &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Newsgroups: sci.math.research From: wjcastre@magnus.acs.ohio-state.edu (W. Jose Castrellon G.) Subject: Fermat's Last Theorem Date: Wed, 23 Jun 1993 17:25:11 GMT Well known number theorists e-mailed us the following, about the events in Cambridge: Andrew Wiles [ of Princeton University ] just announced, at the end of his 3rd lecture here, [ at the Newton Institute in Cambridge ] that he has proved Fermat's Last Theorem. He did this by proving [ a large part of the Taniyama-Weil conjecture for elliptic curves ] that every semistable elliptic curve over Q (i.e. square-free conductor) is modular. The curves that Frey writes down, arising from counterexamples to Fermat, are semistable and by work of Ribet they cannot be modular, so this does it. [ It has been known for several years, by work of Frey and Ribet, that this implies Fermat's Last Theorem ] It's an amazing piece of work. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Newsgroups: sci.math.research From: wjcastre@magnus.acs.ohio-state.edu (W. Jose Castrellon G.) Subject: Fermat's Last Theorem. Sketch Date: Thu, 24 Jun 1993 20:54:06 GMT Local number theorist Prof. Karl Rubin was present at the Wiles lectures in Cambridge. He has posted the following outline of the proof to our math newsgroup: ============================================================================ From K.C.Rubin@newton.cam.ac.uk Thu Jun 24 14:50:52 EDT 1993 Article: 535 of math.announce Path: math.ohio-state.edu!gateway From: K.C.Rubin@newton.cam.ac.uk Newsgroups: math.announce Subject: sketch of Fermat Date: 24 Jun 1993 09:19:10 -0400 Organization: The Ohio State University, Department of Mathematics Lines: 103 Sender: daemon@math.ohio-state.edu Message-ID: NNTP-Posting-Host: mathserv.mps.ohio-state.edu Several people have asked for more details about Andrew's proof. Here is a lengthy sketch. Enjoy. Karl ____________________________________________________________ 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(x-a^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 p-torsion E[p]. We wish to show that a _certain_ lift of this representation to GL_2(Z_p) (namely, the p-adic 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 sqaure of the given modular lifting of rho_p, and the other is related (by work of Hida) to an L-value. The required inequality, which everyone presumes is an instance of the Bloch-Kato conjecture, is what Wiles needs to verify. He does this using a Kolyvagin-type 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 non-semistable 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 mod-5 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 5-adically close to E. Then it is semistable at 5, and at other primes because rho_{E',5} = rho_{E,5}.) ====================================================================== &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Newsgroups: sci.math From: cet1@cus.cam.ac.uk (Chris Thompson) Subject: Re: FLT: Not quite finished? Date: Sat, 26 Jun 1993 15:21:20 GMT In article , kellyfj@unix1.tcd.ie (Q) writes: |> |> I'm not a maths-head or anything, but didn't fermat state that he had a |> 'nice, short' proof of the theorem, no corrolary or whatever, but a |> head on proof? |> |> It seems as though the race for the 'simple, direct' proof may still be |> on, but it's good to know that someone has cracked a major chunk off of |> it. Perhaps it would be as well if we had the wording of Fermat's original marginal note to bring speculation like this down to earth a bit. Can anyone supply the original Latin (or was it in French)? Struik gives the following English translation, but I suspect it of having been modernised (wouldn't Fermat have said "biquadrate" rather than "fourth power"?): To divide a cube into two other cubes, a fourth power, or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too small to contain it. Chris Thompson Internet: cet1@phx.cam.ac.uk JANET: cet1@uk.ac.cam.phx &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Newsgroups: sci.math From: de11@cunixb.cc.columbia.edu (David Epstein) Subject: Re: FLT: Not quite finished? Date: Sat, 26 Jun 1993 17:22:47 GMT In article muttiah@thistle.ecn.purdue.edu (Ranjan S Muttiah) writes: >>In article , kellyfj@unix1.tcd.ie (Q) writes: >>|> >>|> I'm not a maths-head or anything, but didn't fermat state that he had a >>|> 'nice, short' proof of the theorem, no corrolary or whatever, but a >>|> head on proof? > >That may have been the case if he had some nifty new method (such as his >infinite descent; did Fermat discover this thing the first ?). Given Fermat's >genius, this is a possibility too. The story of Fermat's notation is one of the most wonderful stories in math folklore. However, it seems that even Fermat might have realized that his proof was incorrect. The fact is that Fermat wrote that annotation quite *early* in his life. The book itself and the annotation were only brought to light after Fermat's death, when some relative was going through his papers. Actually, Fermat asked some mathematician friends to solve the N=3 and N=4 cases as puzzles well after he originally wrote the annotation. If he had really found a complete proof, then it stands to reason he would have offered the full problem as a challenge. He never again made reference to his "marvelous proof" again. It seems most probable that he thought the method of infinite descent (which I think he did invent, but I'm not sure) could handle all cases through generalization. Then he realized a flaw in his proof and never went back to correct the margin annotation. For all we know, he might have forgotten about it entirely! His one published proof, as has been pointed out, is on the n=4 case, and is a quite ingenious example of proof by infinite descent. David Epstein This is not a .sig! &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Newsgroups: sci.math From: bs@gauss.mitre.org (Robert D. Silverman) Subject: Re: FLT: Not quite finished? Date: Sat, 26 Jun 1993 21:03:51 GMT In article muttiah@thistle.ecn.purdue.edu (Ranjan S Muttiah) writes: :>In article , kellyfj@unix1.tcd.ie (Q) writes: :>|> :>|> I'm not a maths-head or anything, but didn't fermat state that he had a :>|> 'nice, short' proof of the theorem, no corrolary or whatever, but a :>|> head on proof? : :That may have been the case if he had some nifty new method (such as his :infinite descent; did Fermat discover this thing the first ??). Given Fermat's :genius, this is a possibility too. Infinite descent will not work for n>4. This follows from the fact that the non-triviality of Sha for the associated elliptic curve blocks both the Hasse-Minkowski theorem and descent arguments. We (the collective group of modern mathematicians) know so much more than Fermat about the underlying structure that it is extremely unlikely that Fermat had a simple proof. Indeed, the fact that Fermat did not even make a single comment about his general approach makes it likely that he had no proof at all. -- Bob Silverman These are my opinions and not MITRE's. Mitre Corporation, Bedford, MA 01730 "You can lead a horse's ass to knowledge, but you can't make him think" &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& From: Charles Yeomans[x2317] Newsgroups: sci.math Subject: Re: Has Fermat's Last Theorem Really Been Proved? Date: 26 Jun 1993 12:06:01 -0400 In <1993Jun26.141456.17682@austin.onu.edu> Charles Yeomans[x2317] wrote: >In article mayer@sono.uucp (Ron Mayer) writes: >> >>len@schur.math.nwu.edu (Len Evens) <20ca1l$joi@news.acns.nwu.edu> writes: >>> >>> It had been shown previously by others that the Fermat Theorem >>> follows from a conjecture by Taniyama about elliptic curves. >>> Wiles proved the Taniyama conjecture. >> >>Hmm, I wonder if this is the same method Fermat used... >> >>Ok, to rephrase this in a more serious way, does conventional wisdom >>believe that Fermat actually have a proof for his theorem. If so, >>will the search continue for the simple solution or will people just >>decide the problem no longer of any interest (assuming, of course, >>that this proof is real). >> >For informed speculation, a good place to look is Andre Weil's >history of number theory, as he probably knows more about it >than anyone else. > >Charles Yeomans > No better source than that. In case it happens to be checked out [not very unlikely these days...], people might want to try Andre Weil's _Two lectures on Number Theory: Past and Present_. If I remember well, among other things, he says why Fermat's Infinite Descent Method [ which works for some exponents ] doesnt work in the general case. I think it appeared in the Monthly ~1971. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& From: andrew@rentec.com (Andrew Mullhaupt) Newsgroups: sci.math Subject: Re: Has Fermat's Last Theorem Really Been Proved? Date: 26 Jun 93 14:50:33 GMT In article <1993Jun25.090403.4997@sun0.urz.uni-heidelberg.de> gsmith@lauren.iwr.uni-heidelberg.de (Gene W. Smith) writes: >In article gsc@cairo.anu.edu.au (Sean Case) writes: > >>I see a problem here: no matter who has proved it, it's still going to >>be Fermat's Last Theorem. Nope. Hasn't been for a while. The correct term has been 'the Fermat Conjecture' for quite a while. >We can't call it "Wiles' theorem" anyway, for that surely will refer >to Taniyama-Weil for semistable elliptic curves, or semistable curves >plus the others which his proof covers. Presumably the official name >will be something like "the theorem of Frey-Serre-Ribet-Wiles". If you were to take some tack other than Wiles' theorem, you can distinguish his recent result from the resolution of the Fermat Conjecture, and use language like "Wiles' proof of the Fermat Conjecture". Some people will add or subtract authors from the list of people on the theorem, but it appears that _nobody_ can get past who put the final nail in the coffin. Either way would be consistent with mathematical tradition. >I asked Ribet about this and he said "Ask the spin doctors". Is that >us? If you have to ask... Later, Andrew Mullhaupt &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& From: wjcastre@math.ohio-state.edu (W. Jose Castrellon G.) Newsgroups: sci.math Subject: Re: Has Fermat's Last Theorem Really Been Proved? Date: 27 Jun 1993 00:27:26 -0400 In <20hs59$l1h@math.mps.ohio-state.edu> I wrote: >_Two lectures on Number Theory: Past and Present_. [...] >I think it appeared in the Monthly ~1971. It has been kindly pointed out to me that it actually appeared in L'Enseignement de Mathematiques 1974. [ also, for trivial reasons, infinite descent works for infinitely many exponents, since if it works for 3, it also does for 3n for all n; but of course primes are the only cases of interest ]. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Newsgroups: sci.math From: lip@mundoe.maths.mu.OZ.AU (Lawrence Ip) Subject: Original Latin for FLT (with translation) Keywords: Latin, Fermat, FLT Date: Sun, 27 Jun 1993 05:23:47 GMT Someone asked for the original margin note in Latin. Well here it is : Cubem autem in duos cubos, aut quadrato-quadratum in duos quadrato- quadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duas ejusdem nominis fas est dividere : cujas rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caparet. translated ... On the other hand, it is impossible to separate a cube into two cubes, or a biquadrate (fourth power) into two biquadrates, or in general, any power except a square into two powers with the same exponent. I have discovered a truely marvellous proof of this, which however the margin is not large enough to contain. _________________________________________________________________________ Lawrence Ip lip@mundoe.maths.mu.oz.au c/o Dept of Maths, University of Melbourne, Parkville VIC 3052, Australia &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Newsgroups: sci.math Subject: Re: Has Fermat's Last Theorem Really Been Proved? From: gouvea@zariski.harvard.edu (Fernando Gouvea) Date: 27 Jun 93 16:21:40 EDT In article <20hs59$l1h@math.mps.ohio-state.edu> wjcastre@math.ohio-state.edu (W. Jose Castrellon G.) writes: > >No better source than that. In case it happens to be checked out [not >very unlikely these days...], people might want to try Andre Weil's >_Two lectures on Number Theory: Past and Present_. If I remember well, >among other things, he says why Fermat's Infinite Descent Method >[ which works for some exponents ] doesnt work in the general case. >I think it appeared in the Monthly ~1971. The lectures appeared in Enseignement Mathematique, and are reprinted in his collected works. But there's an irony here. Weil says that infinite descent will only work for equations defining elliptic curves, and that as soon as n is 5 or more, the curve in question is not elliptic. (It isn't for n=4, either, but there's a neat little work-around.) That's certainly true... But it's also true that the final proof as we have it now depends on proving something (much deeper and subtler, of course) about elliptic curves. So Weil's comments seem curiously off the mark. I still believe that Fermat did not have a proof, but this shows how hard it is to predict what will turn out to be relevant in a case like this. -- ============================================================================ Fernando Q. Gouvea fqgouvea@colby.edu Dept. of Math/CS (207)-872-3278 Colby College Mudd 407 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& From: gerry@macadam.mpce.mq.edu.au (Gerry Myerson) Newsgroups: sci.math Subject: Two questions on elliptic curves Date: 28 Jun 1993 00:56:42 GMT What's a semistable elliptic curve? I know what an elliptic curve is, but what does semistable mean in this context? What's the history of Shimura-Taniyama-Weil? When I was a kid, the conjecture "elliptic implies modular" was just attributed to Weil. I take it that in the interim someone pointed out that all three mathematicians had (independently? simultaneously?) come to the same hypothesis. How old is the conjecture? Gerry Myerson &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Newsgroups: sci.math Subject: Re: Two questions on elliptic curves From: gouvea@zariski.harvard.edu (Fernando Gouvea) Date: 28 Jun 93 00:16:17 EDT In article <20lfka$9oj@sunb.ocs.mq.edu.au> gerry@macadam.mpce.mq.edu.au (Gerry Myerson) writes: >What's a semistable elliptic curve? I know what an elliptic curve is, >but what does semistable mean in this context? Take the equation of the curve, with integer coefficients, aand reduce modulo some prime p. You'll get either a smooth curve or one of two types of singular curves. The cases are called good reduction, multiplicative reduction, and additive reduction. The curve is semistable if its reduction modulo any prime is either good or multiplicative. > >What's the history of Shimura-Taniyama-Weil? When I was a kid, the >conjecture "elliptic implies modular" was just attributed to Weil. >I take it that in the interim someone pointed out that all three >mathematicians had (independently? simultaneously?) come to the same >hypothesis. How old is the conjecture? The conjecture seems to have been formulated, in vaguer terms, by Taniyama in the fifties, then made more precise by Shimura and Weil. When I first met it, it tended to be referred to as "Weil-Taniyama". There have been rather acrimonious debates about the details, and the most politic way out is to attach all three names to the conjecture. -- ============================================================================ Fernando Q. Gouvea fqgouvea@colby.edu Dept. of Math/CS (207)-872-3278 Colby College Mudd 407 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& From: allenk@ugcs.caltech.edu (Allen Knutson) Newsgroups: sci.math Subject: Privacy flame/new FLT news WAS Re: Where is Wiles? Date: 28 Jun 1993 23:40:19 GMT >In article slim@cybernet.cse.fau.edu (Nick Glenos) writes: >>Ok, where is Dr. Wiles now? I'm SURE he is on Internet, and probally >>even reading these messages. Why doesn't he post his 200 page report? :) mingo@panix.com (Charles Mingo) writes: [Wiles' email address deleted] Christ, that was singularly thoughtful of you, Charlie. Why don't you post his home phone number too, so that amateur mathematicians from all over the world have their well-deserved chance to complain to him personally that his proof is longer than the one Fermat had? I think the rule of thumb is pretty simple here: if Wiles had wanted his address spread across the net, he would have posted to sci.math.research. Imagine all the FLT provers in the world (or even just those on the net), whose efforts used to at least be distributed among all the universities and journals, now concentrated on one poor person! No wonder he sat on a big piece of the result for two years! ObNews: After Wiles' second lecture in Cambridge, the audience figured out how to avoid usage of the full p-adic Hodge-Tate that Faltings hasn't yet written up to everyone's satisfaction. (As I posted before, this sounded like it might be the most dangerous step.) With an audience that helpful, it becomes more understandable why there won't be any preprint circulated until the end of the summer. Actually, I hear the main reason to wait on preprints is the same as that of software developers - it's annoying to have people repeatedly tell you about errors that have already been corrected in later versions. Allen K. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Newsgroups: sci.math From: mckay@alcor.concordia.ca (John McKay) Subject: FLT? Date: Tue, 29 Jun 1993 20:47:21 GMT New York Times Headline Tuesday, June 29, 1993 U.S. SAID IT WAITED FOR CERTAIN PROOF BEFORE IRAQ RAID -- Deep ideas are simple. Odd groups are even. Even simples are not. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&