From: rusin@math.niu.edu (Dave Rusin) Subject: Seeking counterexamples to FLT in other rings Date: 6 Sep 00 21:48:54 GMT Newsgroups: sci.math.numberthy Summary: x^n+y^n=z^n has nontrivial solutions in some rings. I often find myself trying to explain to students and amateurs why their putative proofs of Fermat's Last Theorem are incorrect. (On a good day I'll give the person a nice response, and watch to see whether they seem like a learner or a crank.) One technique which is useful is to produce counterexamples to FLT in other rings; this can point to places where they have made an algebra error or a leap in logic. Given my druthers I would love to have three elements x,y,z of a ring R for which x^p + y^p = z^p where the ring R is very similar to the ring of integers. Some useful features would be: R is associative, commutative, has no zero divisors R has unique factorization R has no units other than +- 1 (or maybe other roots of unity?) p (is not zero and) generates a prime ideal in R R/pR has exactly p elements xyz is relatively prime to p in R p > 3 (More suggestions also welcome.) Just what features ought to be present in an example depends on what gaps in a "proof" should be illustrated, so it could be useful to collect examples of several types. I have looked in particular at the case R is the ring of integers of a quadratic extension of the rationals. There are many examples of the form (a + b sqrt(d) )^3 + (a - b sqrt(d) )^3 = c^3 with a, b, and c rational integers. Such solutions exist precisely when the elliptic curve Y^2 = X^3 - 432 d^3 has rational points. Current technology makes it easy to decide whether or not there are such solutions for any particular d. However, these examples may not always suffice to locate quickly the errors in a particular proposed "proof"; these examples might not satisfy sufficiently many of the conditions I set above. Unfortunately, these are the only counterexamples to FLT I could generate for quadratic extension rings of Z. So I would like to ask the following: (1) Are there any known counterexamples to FLT in quadratic extensions of Z with exponent p >= 5 ? (2) Are there any solutions in these rings with p=3 besides the ones of the form above? (Of course I can hide the form in simple ways e.g. by multiplying through by a non-rational cube. Surely the answer to my query is yes; the question asks whether the group of Q(sqrt(d))-rational points of the curve Y^2 = X^3 - 432 d^3 has larger rank than the group of rational points. I don't have much experience computing ranks over other fields than Q.) Of particular interest to me would be the cases in which the ring R has unique factorization and only trivial units, that is, the rings of integers in Q(sqrt(d)) where d = -1, -2, -3, -7, -11, -19, -43, -67, -163. Can someone tell me the ranks of the Fermat curve a^3 + b^3 = 1 (that is, y^2 = x^3 - 432) over these nine fields? (Only when d=-2 and d=-11 are there points with a and b conjugate over Q as discussed above.) Learned discourse welcome as well as simple examples, of course :-) dave rusin@math.niu.edu ============================================================================== From: Franz Lemmermeyer Subject: Re: Seeking counterexamples to FLT in other rings Date: Thu, 7 Sep 2000 03:10:17 +0200 (METDST) To: Dave Rusin Hi Dave, > (2) Are there any solutions in these rings with p=3 besides the ones > of the form above? This question was studied by people like Fueter and Aigner in the 1930s and 1950s; I assume you have access to MathSciNet, so here goes Aigner: 14,452a; 14,621d; 17,464b; 17,945a; 19,120e I can't seem to find Fueter's papers, but I think Aigner may have given some references to preceding work. franz ============================================================================== From: Dave Rusin Subject: Re: Seeking counterexamples to FLT in other rings Date: Wed, 6 Sep 2000 21:11:47 -0500 (CDT) To: hb3@ix.urz.uni-heidelberg.de Great! These are just the pointers I need. (There are some good links to Fueter's work by searching for Anywhere="Fueter and Fermat" in MathSciNet, but since his work goes back much further it is more productive to check the online versions of Zentralblatt and the Jahrbuch uber die Fortschritte der Mathematik.) Guess it's time to head to the library... Thanks again for responding. dave ============================================================================== From: John Cremona Subject: Re: Seeking counterexamples to FLT in other rings Date: Thu, 07 Sep 2000 08:24:26 +0100 To: Dave Rusin Hi Dave, I passed your nmbrthry contribution on to Frazer Jarvis at Sheffield (which is where Neil Dummigan has a permanent position as of around now) as I know that he is working on this sort of question. I have some texed notes he sent me about it, but am not sure of their status. He is a student of R Taylor and is trying to use Wiles-type methods to extend FLT to some other fields -- but he has also been investigating some down-to-earth methods of finding nontrivial solutions over certain small degree number fields. [deletia --djr] John -- Prof. J. E. Cremona | University of Nottingham | Tel.: +44-115-9514920 School of Mathematical Sciences | Fax: +44-115-9514951 University Park | Email: John.Cremona@nottingham.ac.uk Nottingham NG7 2RD, UK | ============================================================================== From: John Cremona Subject: Re: Seeking counterexamples to FLT in other rings Date: Thu, 07 Sep 2000 08:27:05 +0100 To: Dave Rusin One more point on your posting: The Q(sqrt(d)) rank of Y^2=X^3-432 is equal to the rank of the curve over Q + the rank of its d-twist. Sine the curve has rank 0 over Q (by Fermat!) the rank over Q(sqrt(d)) is equal to the rank over Q of the twisted curve Y^2 = X^3 - 432 d^3. So all quadratic points on the p=3 Fermat curve arise this way. John -- Prof. J. E. Cremona | University of Nottingham | Tel.: +44-115-9514920 School of Mathematical Sciences | Fax: +44-115-9514951 University Park | Email: John.Cremona@nottingham.ac.uk Nottingham NG7 2RD, UK | ============================================================================== From: "A.F.Jarvis" Subject: Solutions to FLT in other rings Date: Thu, 7 Sep 2000 12:11:25 +0100 To: rusin@math.niu.edu Dear Dave, John Cremona kindly forwarded your message regarding solutions to FLT in other rings, as I am not a subscriber to the appropriate newsgroup! I was a student of Richard Taylor a few years ago, so am reasonably well acquainted with the proof of FLT, and as I've been working on Hilbert modular forms, it seemed natural to suggest to my PhD student that he try to prove FLT over certain totally real number fields, following the work of Ribet and Wiles, proving that the Frey curve is both not modular and modular. This will establish FLT over these fields for "sufficiently large exponent". We've been looking particularly at real quadratic fields Q(sqrt{p}) with p prime, although this is purely for convenience so far; other real quadratic fields can be treated in the same way. While my student, Paul Meekin, has been doing this, I have, in my spare moments, been thinking about solutions when the exponent is not "sufficiently large". I have a few results of relevance to your question, and some references from others. We intend to publish these in due course, probably combining his results on FLT for sufficiently large exponent (he has several examples of fields for which the Frey curve is not modular, and now we want to think about generalising Wiles's stuff, following Fujiwara's work) with my results on lower exponent. As you point out, the Fermat cubic has points in Q(sqrt{d}) precisely when Y^2 = X^3 - 432 d^3 has rational points. Perhaps surprisingly, there is an exact correspondence between rational points on this elliptic curve and points on the Fermat cubic with coordinates in Q(sqrt{d}). This is not completely obvious; it's a corollary of FLT for exponent 3 over Q! This means that one can compute the rank of the Fermat cubic over Q(sqrt{d}) by knowing the rank of these Mordell curves over Q. In particular, if d=2, one finds that the curve has rank 1, generated by (X,Y)=(28,136), leading to the solution: (378 + 357sqrt{2})^3 + 127^3 = (451 + 306sqrt{2})^3. If d=3, the Mordell curve has rank 0, so there are no points on the Fermat cubic over Q(sqrt{3}). For d=5, there is the point (9 + sqrt{5})^3 + (9 - sqrt{5})^3 = 12^3, which is the "easiest" solution I know of. Here's a way of constructing points on the Fermat cubic. Take a line with rational slope which passes through one of the three (in projective space) rational points on the cubic. Then this line necessarily intersects the Fermat cubic in two further points (by Bezout's theorem, if we work in P^2(C)); it is easy to see that these lie in some quadratic field and are conjugate. Some easy Galois theory leads to the conclusion that all quadratic points arise in this way. It is precisely the knowledge of FLT for exponent 3 over Q that means that we know all rational points on the Fermat cubic, and, by considering a general line of rational slope through these rational points, we can deduce the theorem I stated above. (The same method means that it is easy to compute the rank over any quadratic field of an elliptic curve with rank 0 over Q; X_0(15) would be interesting, as Wiles uses it in his work on FLT...) To construct points on the Fermat quartic, take a line with rational slope passing through a rational point; then this line will meet the quartic at three further points, all lying in some cubic field. If the line goes through two of the rational points (there are four rational points, so six such lines), we get points on the quartic lying in quadratic fields. If I remember, four of these are solutions like 0^4 + i^4 = 1, but the other eight lie in Q(sqrt{-7}). Faddeev (1960) proved that these were the only points on the Fermat quartic in quadratic fields, and further that all points in cubic fields arose in the manner above. For exponent 5, one can find two points in Q(sqrt{-3}) by considering the line going through all three rational points; here it is: (1 + sqrt{-3})^5 + (1 - sqrt{-3})^5 = 2^5; this is basically equivalent to observing that the sum of the two primitive 6th roots of unity is 1. So it is also a solution for all exponents congruent to plus or minus 1 mod 6. There is a theorem of Tzermias that these are the only points on the Fermat quintic lying in quadratic or cubic fields, if I remember correctly. But you should be able to construct points lying in quartic extensions using this intersection method. I believe that Tzermias has results also for exponent 7 and possibly 11 as well. You might also like to read a paper of Debarre and Klassen in which more applications of this intersection method are presented, and a general conjecture formulated. I've been writing up a few of these more naive thoughts on FLT for small exponent over small fields in some informal notes, not intended for publication; I can send you a copy if you are interested. Best wishes, Frazer Jarvis Dr A.F.Jarvis, Department of Pure Mathematics, Hicks Building, University of Sheffield, Sheffield S3 7RH (0114)2223845 http://www.shef.ac.uk/~pm1afj ============================================================================== From: Bill Daly Subject: Re: Seeking counterexamples to FLT in other rings Date: Wed, 13 Sep 2000 10:37:26 -0400 To: Dave Rusin This may not fit your requirements perfectly, but have you noticed that FLT has a (fairly trivial) nonzero solution in Z(w), for w a primitive 6th root of 1, for all odd primes p > 3, namely (1)^p + (w^2)^p = (w)^p If also p = 2 mod 3, then p is prime in Z(w). Regards, Bill ============================================================================== From: Dave Rusin Subject: Re: Seeking counterexamples to FLT in other rings Date: Wed, 13 Sep 2000 10:23:14 -0500 (CDT) To: bill.daly@tradition-ny.com Yes, thanks. This observation was made by others too: (1 + sqrt{-3})^5 + (1 - sqrt{-3})^5 = 2^5; and likewise for other exponents. I'm embarrassed I didn't spot this myself. Thanks again for responding. When I get a few minute's peace I'll summarize the other responses I received to the list. dave ============================================================================== Note [djr]: the solutions implied by Jarvis for n=4 are essentially (1+sqrt(-7))^4 + (1 - sqrt(-7))^4 = 2^4 allowing of course for sign changes absorbed by the fourth powers, multiplying through by constants in the ring, and so on. In cubic extensions we have solutions (x,y) = (m*t, 1+t) where t is a root of 4+6*t+4*t^2+(m^4+1)*t^3 . Here m could be any rational number (except +- 1, in which case the cubic has a rational root). Roughly speaking we understand that there are few solutions to the Fermat equations which are algebraic of low degree. More precise results are available for exponents 5 and 7, and are conjectured in general. See for example these articles and their reviews in Math Reviews: Tzermias, Pavlos Algebraic points of low degree on the Fermat curve of degree seven. Manuscripta Math. 97 (1998), no. 4, 483--488. 99j:11075 11G30 Sall, Oumar Points algébriques de petit degré sur les courbes de Fermat. C. R. Acad. Sci. Paris Sér. I Math. 330 (2000), no. 2, 67--70. 2001a:11104 11G30 (14G05) So the hope of the original poster -- to find nearly-rational solutions to a Fermat equation -- is pretty much limited to unimpressive cases! (Not nearly as impressive as Elkies' solution 3472073^7 + 4627011^7 = 4710868^7 which can be verified with any 20-digit calculator!...) ============================================================================== From: "Dik T. Winter" Subject: Re: FLT for Gaussian integers Date: Wed, 31 Jan 2001 01:24:49 GMT Newsgroups: sci.math In article <94th88$p0s$1@agate.berkeley.edu> chernoff@math.berkeley.edu (Paul R. Chernoff) writes: > There has been a lot of discussion on this newsgroup of "elementary" > proofs of FLT for the usual integers. I want to change the subject a > bit. Is FLT true (or false) for the *Gaussian integers* Z[i], i.e. > the set of all complex numbers of the form a = m + ni, where m and n > are ordinary integers. I do not think much work has been done specifically for the Gaussian integers. And I think it is still much more difficult and intractable than FLT for the integers themselves. Research has been done for particular exponents in arbitrary quadratic number fields: 2, 3, 4, 6 and 9. I am not aware of other exponents being researched. I understand results are complete for 2, 4, 6 and 9, and possibly complete for 3. And as far as I know in most cases only those numbers in Q(sqrt(m)) (with m square free, != 0 and 1) are considered of the form a + b.sqrt(m) with a and b integer, so not always all algebraic integers. Considering the results I can only say that they are looking, eh, complex. See: n = 2 provides additional non-trivial solutions only when m is not divisible by a prime factor of the form +-3 mod 8. n = 4 provides solutions when m = -7 (but only one essential primitive solution). n = 6 and n = 9 have never any solution. n = 3 is bizarre. Aigner has done much work on that one, but class numbers of the field come in here again. On the other hand Gauss' proof of FLT3 for the Eisenstein integers is closely related to finding rational points on elliptic curves. So there is hope. Another generalisation is x^k + y^k = n.z^k; which also has not yet been solved in the integers. The example I gave uses k = 5 and n = 68101. It is known that 68101 is the smallest n with k = 5 such that there are solutions (and only one primitive solution). More such n are known, but it is not known whether there are infinitely many or not. But I think it has been shown (Faltings?) that if there are solutions there are only finitely many for any particular n. So in this direction Wiles' proof might offer opportunities. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ============================================================================== From: Jan Kristian Haugland Subject: Re: FLT for Gaussian integers Date: Mon, 29 Jan 2001 00:10:01 +0100 Newsgroups: sci.math Jamas Enright wrote: [deletia -- djr] > Another related question might be: what conditions do you need for FLT to > be false? > > Numerous finite rings fail FLT, pick a number n and a power p > 2, then in > Z mod n^p, you get n^p + n^p = n^p (=0). :) > > Also p-adics fall over, such as Plutonium's much loved 10-adic solutions. > > In many ways it looks like that if the field of interest is too large or > too small then FLT fails, so it's needs to be the right size. (For an as > yet to be defined value of 'right size' of course :). Size isn't everything: With w = (-1 + i sqrt(3))/2, x^n + y^n = z^n have non-zero solutions in Z[w] for all integers n not divisible by 2 or 3, for instance w^p + (-1-w)^p = (-1)^p. -- Jan Kristian Haugland http://home.hia.no/~jkhaug00 ============================================================================== From: Bill Daly Subject: FLT in algebraic rings Date: Tue, 23 Oct 2001 08:52:12 -0700 (PDT) To: Dave Rusin I was reminded of your earlier question concerning solutions of FLT in algebraic rings by a reference from mathpuzzle.com. I have a few simple comments. If there is a solution of x^n + y^n = z^n with x,y,z in an algebraic ring R, then w = norm_R(z)/z is also in R, thus if x' = xw, y' = yw, z' = zw = norm_R(z), we obtain a solution x'^n + y'^n = z'^n with x,y in R and z in Z. Thus, it is sufficient to look for solutions with z in Z. This means that all solutions in quadratic rings may be obtained from solutions of (u + sqrt(v))^n + (u - sqrt(v))^n = w^n. For n = 3, this expands to u^3 + 3uv = w^3 / 2. This is linear in v, thus for arbitrary u and w, there is a solution with v = (w^3 / 2 - u^3) / (3u). For n = 4, we get u^4 + 6u^2 v + v^2 = w^4 / 2. As a quadratic in v, the discriminant is 32u^4 + 2w^4, which must be a square if there is to be a solution with v in Z. This is equivalent to solving a^4 + b^4 = 2c^2, corresponding to [u,w] = [a,2b]. An obvious solution is [a,b] = [1,1], giving [u,w] = [1,2], from which we obtain the solutions v = 1, leading to 2^4 + 0^4 = 2^4, and v = -7, leading to (1+sqrt(-7))^4 + (1-sqrt(-7))^4 = 2^4. I assume that there are no other solutions, but I haven't checked it. We should also consider solutions of x^4 + y^4 + z^4 = 0, which has no nontrivial solutions in real quadratic rings, but which may be solvable in imaginary quadratic rings. This leads to the solvability of a^4 - b^4 = 2c^2, with again the obvious solution [a,b] = [1,1], giving [u,w] = [1,2]. This leads to a single solution v = -3, corresponding to (1+sqrt(-3))^4 + (1-sqrt(-3))^4 + 2^4 = 0. I'm fairly sure that this is the only solution. For n = 5, we get u^5 + 10u^3 v + 5uv^2 = w^5 / 2, again a quadratic in v whose discriminant 80u^6 + 10uw^5 must be a square. The obvious [u,w] = [1,2] gives solutions with v = 1 and v = -3, corresponding to the trivial 2^5 + 0^5 = 2^5 and (1+sqrt(-3))^5 + (1-sqrt(-3))^5 = 2^5. Presumably, there are no other solutions. Regards, Bill [sig deleted --djr] ============================================================================== From: rusin@math.niu.edu Subject: Re: FLT in algebraic rings Date: Sat, 27 Oct 2001 22:50:00 -0500 (CDT) To: billdaly99@yahoo.com You wrote me recently about solving FLT, especially in quadratic extension fields of Q. >If there is a solution of x^n + y^n = z^n ... >Thus, it is sufficient to look for solutions with z in Z. OK >This means that all solutions in quadratic rings may >be obtained from solutions of (u + sqrt(v))^n + (u - sqrt(v))^n = w^n. Not OK. It's probably "almost true" somehow but at the very least you have to allow for the possibility that y/(x_bar) is some other root of unity besides 1. (I'm mostly thinking of n=3 here, but even in the general case I'm not 100% sure I see what to do.) >For n = 3, this expands to u^3 + 3uv = w^3 / 2 OK: this constructs solutions to FLT3 in quadratic fields. Modulo your claim in the previous paragraph, these are (close to) the only solutions. Usually people ask me about specific quadratic fields, and it's less clear from your construction whether it will give elements in any pre-assigned field, but it has been proved that the case n=3 for quadratic number fields reduces to finding rational points on a particular elliptic curve, so (for d not too terribly large) this is a solvable problem. Your comments about n=4 and n=5 are also OK, although it looks like you missed the solutions for n=4 which happen to have x y z = 0. More generally, it seems people have made progress classifying solutions to FLT with exponent n in number fields of degree less than n, for n up through about 11. You can see some of what I learned about this at http://www.math-atlas.org/newstuff/flt_rings Unless you have an objection I will probably tack on your message to that file. OK? dave ==============================================================================