Date: Tue, 22 Oct 96 17:13:44 CDT From: rusin (Dave Rusin) To: iain@stt.win-uk.net Subject: Re: Help with diophantine equ. Newsgroups: sci.math I see you're still considering the equation X^5+Y^5 = (X+Y)*Z^2. In article <271@stt.win-uk.net> you write: > >The equation is X^4 -X^3*Y + X^2*Y^2 -X*Y^3 + Y^4 = Z^2 > >X,Y,Z >0 and pairwise relatively prime. ... >There are solutions for X,Y,Z > >1,1,1 >8,11,101 >35,123,13361 Next is (20965,43993,161218641) = (5.7.599, 29.37.41, 11.2971.49331). If you drop the positivity, then interspresed among these are (0,1,1) (-1,3,11) (-627,808,1169341) etc. Of course you can interchange X and Y or change the sign of Z. As another poster suggested, you change variables: U=X/Y, Z=V/Y^2 You get rational numbers with 4 3 2 2 U - U + U - U + 1 = V Next change this to a cubic in Weierstrass normal form. Use 2 3 2 V - U + 2 - 4 V + 3 U - 2 U - 4 + U V + U {p = -----------, q = - ---------------------------------} 2 3 U U to get a new equation 3 2 2 p + p - 3 p - 2 = q which is an elliptic curve in canonical form. The rational points on this curve form a group isomorphic to Z x (Z/2Z). The element of order two is (p,q) = (-2,0), and the generator of infinite order is (-1,1). The first few elements of the infinite cyclic part are (1,-1) (2,2) (-17/9, -19/27), (89/16, -869/64), (-769/1225, -7421/42875), ...; their inverses are found by changing the sign of the second entry. Adding the element of order 2 to each of these gives more points: (3,-5) (-3/4, -5/8) (43,285), (-162/121, 1580/1331), ... (Again you change the sign of q to handle inverses). There is no "formula" for these pairs (p,q) but you can get them recursively: adding (-1,1) to any group element gives you the next one; if the first is (p,q), the next is 2 3 2 p + 4 p + 1 + 2 q - 2 q + 2 p q + p - 11 p - 5 - p [- ------------------, ----------------------------------] 2 3 (p + 1) (p + 1) Then you can substitute back to get U and V: If I've done it right, the effect of marching up one generator on the curve is to replace (U,V) with (U2,V2) where U2 is 2 - U + 4 U + 2 V - 1 -------------------- (U + 3) (U - 1) and V2 is 4 3 2 2 - 5 U + 16 U + 6 V U - 12 U + 8 U - 8 U V + 10 V + 1 - -------------------------------------------------------- 2 2 (U + 3) (U - 1) I guess you'd start with these: the first few powers of the generator give the following (U,V): (-1/3, 11/9) (11/8, 101/64) (123/35, -1331/1225) (-808/627, 1169341/393129) (20965/43993, -1612186411/1935384049) ... The first few powers of its inverse give (U,V)= (1,-1) * (-3,-11) (8/11,101/121) etc. (Twice the negative of the generator doesn't correspond to a pair (U,V) because of division by zero). Adding the element of order two to these elements gives some more: (-3,11) * (35/123, 13351/15129), ... for the first set, and (1,1) (0,-1) (-1/3, 11/9) ... for the second. Notice that whenever a pair (U,V) occurs, so must the pair (U,-V) and the pairs (1/U, V/U^2) and (1/U, -V/U^2); I haven't checked carefully but I'm pretty sure these give correspondences among the four columns. Thus you need only complete the first one and remember to throw the others in along the way. Then of course you recover your (X,Y,Z) by taking Y as the denominator of U, X its numerator, and Z=Y^2 V. You can see that's how I added to your list. ============================================================================== To: rusin@math.niu.edu Date: Mon, 28 Oct 1996 16:39:59 Subject: Re: Help with diophantine equ. From: Iain Davidson Thanks for useful reply. >I see you're still considering the equation X^5+Y^5 = (X+Y)*Z^2. Yes, I also tried linear factors of (X^5 + Y^5)/(X+Y) (X + rY)^2 x UNIT = P + rQ = square I am also interested in an impossibility proof for X^5 + Y^5 = Z^2. Someone has given me a reference on a proof using elliptic curves and winding numbers, but that would be too advanced for me to understand. Does this mean that there is no elementary proof on the lines of, say, X^4 + Y^4 = Z^2 ? >>There are solutions for X,Y,Z >> >>1,1,1 >>8,11,101 >>35,123,13361 > >Next is (20965,43993,161218641) = (5.7.599, 29.37.41, 11.2971.49331). > >If you drop the positivity, then interspresed among these are >(0,1,1) >(-1,3,11) >(-627,808,1169341) >etc. Of course you can interchange X and Y or change the sign of Z. Yes >As another poster suggested, you change variables: > U=X/Y, Z=V/Y^2 >You get rational numbers with > > 4 3 2 2 > U - U + U - U + 1 = V As at least three solutions (U0, U1, U2), say, have already been found by trial and error, you could find the quadratic V = aU^2 + bU + c, that goes through these three points then U^4 - U^3 + U^2 - U + 1 - (aU^2 + bU + c)^2 = 0 would have 4 solutions. The coefficient of the U^3 is then -(U0 + U1 + U2 + U3) and U3 can be found using rational operations and so must be another rational solution. This is perhaps more straightforward than using elliptic curves. Cheers Iain Iain Davidson Tel : +44 1228 49944 4 Carliol Close Fax : +44 1228 810183 Carlisle Email : iain@stt.win-uk.net England CA1 2QP ============================================================================== To: rusin@math.niu.edu Date: Wed, 30 Oct 1996 01:07:32 Subject: Re: Help with diophantine equ. From: Iain Davidson >I'm not sure what winding numbers have to do with this, but probably a >proof using elliptic curves will work. Yes I should have written "winding quotients". The title of the paper by Henri Darmon and Loic Merel with Appendix by Bjorn Poonen is "Winding Quotients and some variants of Fermat's Last Theorem". I think it's supposed to appear in Crelle's Journal. Poonen deals with impossibility of X^5 + Y^5 = Z^2 in the Appendix by means of elliptic curves and other special results. Other references that I have found but not looked at are: The equations X^n + Y^n = Z^2 and X^n + Y^n = Z^3 Darmon, H. Internat. Math. Res. Notices 10 (1993), 263 - 274 The equation X^4 - Y^4 = Z^p Darmon, H. C.R. Math. Rep. Acad. Sci. Canada 15 (1993) 286 - 290 On the equations X^p + Y^q = Z^r and Z^m = f(X,Y) Darmon, H., Granville, A., Bulletin of the London Math. Society, No. 129 27 part 6, Nov. 1995, pp. 513 - 544. Ueber die Diophantische Gleichung X^l + Y^l = cZ^l, Denes, P. Acta Math. 88 (1952) 241 - 251 >The 'elementary' proof for quartics really is an elliptic curve argument too! Fermat's method of infinite descent is proving that the elliptic curve x^4 + 1 = y^2 has no rational points; > >In just this way an elementary proof for quintics could have been found >by observing that x^4+x^3y+x^2y^2+xy^2+y^4 is never a square: >one would use a descent argument a la Fermat as an elementary mask for >the proof that the elliptic curve is the trivial group. Sadly, it >_isn't_ the trivial group, but rather has rank 1. So no, I would have >to say it's unlikely that a Fermat-like proof can be followed without >significant change. That is a very interesting and illuminating connection. Seems a bit paradoxical though that Dirichlet could deal with X^5 + Y^5 = Z^5 using descent and quadratic forms but X^5 + Y^5 = Z^2 is more intractible. >Cassel's book on elliptic curves does a nice job showing the connection >between Fermat's method on the structure of elliptic curves. Could you give me details of Cassel's book, it might be useful to have a look at it. What level is it pitched at ? Can you recommend any introductions to elliptic curves and diophantine equations ? Have you published anything yourself in this area ? Cheers Iain Iain Davidson Tel : +44 1228 49944 4 Carliol Close Fax : +44 1228 810183 Carlisle Email : iain@stt.win-uk.net England CA1 2QP ============================================================================== Date: Sat, 2 Nov 96 02:07:57 CST From: rusin (Dave Rusin) To: iain@stt.win-uk.net Subject: Re: Help with diophantine equ. >Other references that I have found but not looked at are: Thanks for the citations! I haven't read them, but Granville has invested quite a bit of time in this area; his work is likely to be authoritative. >Seems a bit paradoxical though that Dirichlet could deal with > X^5 + Y^5 = Z^5 using descent and quadratic forms but X^5 + Y^5 = >Z^2 is more intractible. I only know of Dirichlet's proof from secondary sources, but I was under the impression it was not descent along an elliptic curve but rather a proof by contradiction using factorization in the ring Z[w=exp(2pi i/5)]. If X and Y are relatively prime integers, then the factors X + w^n Y of Z^5 would all be relatively prime (unless they have in common a divisor of 5) which would mean X + w^n Y = U^5 for some U in this ring. At that point I don't think one continues to descend further but rather derives a contradiction just from the presumed existence of so many fifth powers among the linear combinations of X and Y. Dirichlet found many annoying features here though. For example, I lied about the U^5 since there could be any unit mutliplying U^5, too. And you have to prove that unique factorization continues to hold in this ring. By the way, I thought of proposing a proof like this for you when you first asked about Z^2=Z^5+Y^5. Factoring in the larger ring would make lots of expressions all be squares. But I didn't get a contradiction right away and so I quit. I should also say that the Dirichlet equation is _homogeneous_, which makes a lot of things nicer, even though the degree of Z^5 is higher than that of Z^2. >>Cassel's book on elliptic curves does a nice job showing the connection >>between Fermat's method on the structure of elliptic curves. >Could you give me details of Cassel's book, it might be useful to >have a look at it. What level is it pitched at ? Cassel's book is in the Undergrad Studies in Math series of (I think) Cambridge U press. It's only about 100 little pages, in softcover. Parts of it will be dicey if you haven't had sufficient abstract algebra courses, but I found it to be really very well done and includes a lot of material at an elementary level. >Have you published anything yourself in this area ? All the fun is gone when the problem is solved, so I publish little. Gets me in hot water. dave