From: Dave Rusin Date: Fri, 4 Sep 1998 17:03:52 -0500 (CDT) Newsgroups: sci.math.numberthy Subject: Re: Some Mistaken Identities and a Diophantine Question M.Cooker wrote: > Students mistakenly equate the reciprocal of a sum with the sum of >reciprocals. Nevertheless some of these mistaken identities are true >sometimes. For example: [...] >the non-zero complex numbers, of the next problem: > > 1 1 1 1 1 > ------------- = --- + --- + --- + --- (3) > A + B + C + D A B C D [...] > Are there any non-zero INTEGER solutions of equation (3) ? Clear denominators; scale so that D=1. Let s1, s2, s3 be the symmetric functions of {A,B,C}. Then (3) is equivalent to s3=-(s1+1)/s2, so the question is, are there (rational) s1, s2 so that the polynomial 3 2 (s1 + 1) s2 x - s1 x + s2 x + ----------- (4) s1 has three rational roots? This requires in particular that the discriminant 4 4 3 2 2 3 2 4 s1 (s1 + 1) s2 + (s1 - 18 s1 - 45 s1 - 54 s1 - 27) s2 - 4 s2 s1 be square. (Conversely, if the discriminant _is_ a square, we have a solution to (3) iff (4) has at least one rational root.) For each s1, this becomes a question about an elliptic curve. Fire up APECS; s1=1,2,3,6: rank is zero. s1=4: rank is 1, generator has s2=45/64. This gives infinitely many chances to make (4) have a rational root. But indeed s1=4, s2=45/64 makes (4) factor over Q! Scaling back from the roots gives a solution to (3): A=-3 B=5 C=8 D=30 Want more? Twice the generator on that same elliptic curve has s2=-4348215481/111513600; again (4) factors over Q giving a solution {A,B,C,D}={-17296, -35075, 10560, 94611} I must be failing to see an algebraic identity, because (4) factors again when taking s2 from the point 3*P on the same elliptic curve, and when using the generator of the curves with s1=5 and s1=7 (also of rank 1). That is, it appears to be true that (4) factors over the function field Q(Y,s1,s2)/(Y^2-discriminant). I suspect there is a parameterized solution to (4). dave ============================================================================== From: Dave Rusin Date: Mon, 7 Sep 1998 09:03:26 -0500 (CDT) To: rusin@math.niu.edu Subject: phony reciprocals Problem: find integers {a_i} with Sum(a_i) = Sum(1/a_i) = 0. Motivation: equivalent to student errors 1/(a_1 + a_2 + ... ) = 1/a_1 + 1/a_2 + ... Equivalently: find integers a_i roots of a polynomial X^n + s_2X^{n-2} + ... + s_{n-2}X^2 + s_n = 0. n=2: x^2 - s_2, i.e. the two are negatives of each other. Leads to 1/a = 1/a only n=3: X^3 - s_3 = 0, i.e. the three differ by powers of omega 1/a + 1/a omega + 1/a omega^2 = 0 n=4: X^4 - s_2 X^2 + s_4 = 0, i.e. the four come in pos/neg pairs. Leads to 1/a + 1/b = 1/a + 1/b only n=5: X^5 - s_2 X^3 + s_3 X^2 + s_5 = 0. Better strategy: Eliminate E, get one equation in a b c d . Quadratic in a: solveable iff discrim is a square. That discrim is a quartic in b whose constant term is a square. So we do EC calcs. Final substitutions lead to using b = U/V and either a = - U cd( V (c+d) + U^2(c+d)^2 + U ) / V( V cd + U (c+d) )(c+d) or a = - (U + V (c+d) )( U cd(c+d)^2 - (c^2+cd+d^2) ) / V( V cd + U (c+d) )(c+d)^3 [U/V, -U*c*d*(V*(c+d)+U^2*(c+d)^2+U)/V/(V*c*d+U*(c+d))/(c+d), -(U+V*(c+d))*(U* c*d*(c+d)^2-c^2-c*d-d^2)/V/(V*c*d+U*(c+d))/(c+d)^3] Here we further simplify with U = U1 - 1/(c+d)^2 V = V1 - (c^2+cd+d^2)(1 + U(c+d)^2) / 2cd(c+d)^3 and then U1 = U2/ (2cd(c+d))^2 V1 = V2/ (2cd(c+d))^3 so [b, a, a'] is [2*c*d*(4*d^2*c^2-U2)*(c+d)/(U2*c^2+U2*c*d-V2+U2*d^2), -(c+d)*(4*d^2*c^2-U2)*( 2*U2*c^3*d+6*U2*c^2*d^2-2*c*V2*d+2*U2*c*d^3-U2^2)/(U2*c^2+U2*c*d-V2+U2*d^2)/(8 *d^2*c^4+16*d^3*c^3-U2*c^2+8*d^4*c^2-3*U2*c*d-V2-U2*d^2), -2*(c+d)*d*c*(8*d^3* c^3+U2*c^2-U2*c*d-V2+U2*d^2)*(-U2+4*c^3*d+8*d^2*c^2+4*c*d^3)/(U2*c^2+U2*c*d-V2 +U2*d^2)/(8*d^2*c^4+16*d^3*c^3-U2*c^2+8*d^4*c^2-3*U2*c*d-V2-U2*d^2)] The relationship among abcd then becomes this one among U2 V2 c d: V2^2 = U2 ( U2^2 + [s^4 - 6ps^2 - 3p^2] U2 + 16s^2p^3 ) where we have set s = c+d, p=cd. Since by scaling WMA d=1, we get s= (c+1), p = s and have an elliptic surface over Q(c). For each c there is a torsion subgroup of order 6 with element of order 2 at U2=0, elements of order 3 at U2 = 4c^2 and elements of order 6 at U2 = 4c(c+a)^2 (that is, 4(cd)^2 and 4cd(c+d)^2 respectively.) Note that by rescaling to make c=1 instead, we see E_c and E_{1/c} are the same curve, so we may assume |c|>1 in what follows. (No negative c were tested). A number of rational c were tested. We got rank 2 just once (c=7/5), and rank=0 in general (c=3/2, 5/2, 5/3, 6/5, 7/3, and integral c <= 30 not shown below). The cases with rank=1 found were those with c=4/3, 5/4, 7/2, 7/4, 7/6, and the following integral c (with U2 of a generator given): 3 12 6 336 9 -300 11 256 14 -291312/529 16 16 17 -18252/49 19 8656875 23 8054344516/3025 24 3825936/529 27 yuk 28 yuk 29 21498909/2500 No pattern is clear to me (even when trying the other generators +-g + Tors). Still it is of some interest to know there are infinitely many equations e.g. of the form 1/(a+b+3+1) = 1/1 + 1/3 + 1/a + 1/b. ============================================================================== From: jbuddenh@texas.net (James Buddenhagen) Newsgroups: sci.math.numberthy Subject: Re: Some Mistaken Identities and a Diophantine Question Date: 10 Sep 98 22:05:09 GMT Mark Cooker asks for quadruples (A,B,C,D) of integers such that (1) 1/A + 1/B + 1/C + 1/D = 1/(A+B+C+D). Here is an explicit 1-parameter family of solutions: A = -3*(5*m-21)*(m-5)*(m^2-21) B = 12*(3*m-14)*(m-5)*(19*m^2-174*m+399) C = -4*(5*m-21)*(2*m-9)*(19*m^2-174*m+399) D = (m^2-21)*(19*m^2-174*m+399). Both sides of (1) equal 1/(-12*(2*m-9)*(3*m-14)*(m^2-21)). The parameter m can be any integer unequal to 5. The A above may be replaced by 12*(2*m-9)*(3*m-14)*(m^2-21) making both sides of (1) equal to 1/(3*(5*m-21)*(m-5)*(m^2-21)). Combined with the elliptic curve approach described by Joe Wetherell these solutions may be thought of as giving a family of positive rank elliptic curves (over Q), each of which gives via its rational points infinitely many solutions of (1). Not all solutions of (1) arise in this way. --Jim Buddenhagen (jbuddenh@texas.net) ============================================================================== From: jb1556@daditz.sbc.com (Jim Buddenhagen) Newsgroups: sci.math.numberthy Subject: 1/a + 1/b + 1/c + 1/d = 1/(a+b+c+d) another family Date: 10 Sep 98 22:06:21 GMT Recently I sent the list a 1-parameter family of integer solutions to: (1) 1/a + 1/b + 1/c + 1/d = 1/(a+b+c+d). After looking a little more I found another, slightly simpler, 1-parameter family of solutions which appears to be inequivalent to the previous. It is: a = -2*n*(2*n+3)*(n^2+3) (or a=-6*(n-2)*(n^2+3)). b = 3*n*(n-2)*(n^2+3*n-3) c = 3*(2*n+3)*(n^2+3*n-3) d = (n^2+3)*(n^2+3*n-3). With the first a both sides of (1) equal 1/(6*(n-2)*(n^2+3)). With the second a both sides of (1) equal 1/(2*n*(2*n+3)*(n^2+3)). n can be any integer other than 0 or 2. --Jim Buddenhagen (jbuddenh@texas.net) ============================================================================== Subject: Re: 1/a + 1/b + 1/c + 1/d = 1/(a+b+c+d) another family To: rusin@math.niu.edu (Dave Rusin) Date: Wed, 16 Sep 98 15:34:50 CDT From: Jim Buddenhagen Dave, >Good job finding those rational curves embedded in the elliptic surface! >How did you go about finding them? Starting with (1) 1/a + 1/b + 1/c + 1/d - 1/(a+b+c+d) = 0 clear of fractions an set numerator to 0. Result is homogeneous so instead of searching for integer solutions to (1) can set any variable equal to 1 and search for rational solutions. The resulting eqn is quadratic in each variable so, for rational solutions the discriminant wrt any variable, say wrt a, must be a perfect square: 2 2 2 2 2 Y = (c + b + b c) (b + c + 1) (b + b c - b c + b + b c + c + c). This is a family of ellptic curves, i.e. an elliptic surface Look on b+c=k which makes middle factor a constant. (k is in Q). 2 2 2 2 2 2 (2) Y = (- k + b - b k) (k + 1) (- 3 b + b k + 3 b k - b k - k - k) First and third factors are quadratic in b. Each can be made square (times a const.) by finding k so discriminant is 0 discrim of first factor wrt b = k (4 + k) 2 discrim of third factor wrt b = (- 3 + k) k (k + 4 + k ) k=0 no good but k=3 or k=-4 work. So substitute k=3 into (2) and get: 2 2 Y = - 48 b + 144 b + 144 After dividing by 16: 2 2 y = - 3 b + 9 b + 9 Find all rational points by standard methods e.g. look at lines of rational slope m through (b,y)=(0,3). Work backward to get the a,b,c,d for (1): 2 a = - 2 (2 m - 3) m (m + 3) 2 b = - 3 (2 m - 3) (m - 3 m - 3) 2 c = 3 (2 + m) m (m - 3 m - 3) 2 2 d = (m - 3 m - 3) (m + 3) Similarly, substitute k=-4 into (2) and get: 2 2 2 Y = 3 (b + 2) (7 b + 28 b + 12) 2 2 2 y = 21 b + 84 b + 36 To get rational points look at lines of rational slope m through (b,y)=(0,6). Working backward and tranforming to the original variables get second family of solutions to (1): 2 a = 12 m (m - 21) 2 b = - 12 (m - 7) (21 + m - 6 m) 2 c = - 4 (- 3 + m) m (21 + m - 6 m) 2 2 d = (21 + m - 6 m) (m - 21). > (I've been mulling over the general situation of arithmetic on algebraic > surfaces for a while, and I'm anxious to collect all possible tricks.) I hope you'll share all these tricks, perhaps at your web site. > dave Jim -- Jim Buddenhagen jbuddenh@texas.net ============================================================================== From: Dave Rusin Date: Thu, 17 Sep 1998 11:44:27 -0500 (CDT) To: jb1556@daditz.sbc.com Subject: Re: 1/a + 1/b + 1/c + 1/d = 1/(a+b+c+d) another family Jim, Thanks for responding. Allow me to comment on your method; it lets us try to generalize, but we'll see you've already found the best (only?) rational curve in the surface. Still, this exercise has driven home for me a lesson in finding embedded rational curves. >>Good job finding those rational curves embedded in the elliptic surface! >>How did you go about finding them? >Starting with [...] [Natural change to quartic-in-b-and-c = square.] >This is a family of ellptic curves, i.e. an elliptic surface >Look on b+c=k which makes middle factor a constant. (k is in Q). This is a key idea. You did indeed have an affine algebraic surface S. But it's only an elliptic surface if you specify a map f:S -> C to some curve, with the fibres being elliptic curves embedded in S. You didn't specify your f, nor did you need to; but then you don't have a family of elliptic curves yet, you just have their union, the surface S -- you haven't broken it down into curves. But now your introduction of k is important: the map (a,b,c) |--> b+c is an affine map f: S -> Q. The fibres are the subvarieties (curves) in S given by the additional equation b+c=k (constant). That's roughly how I found points on S too: intersect S with some linear hyperplane in [a:b:c:d]-space to get an elliptic curve; with the right hyperplane (I used k=6, I think) we get a curve with lots of points. (Actually we both lie a little: the fibres are varieties specified by an equation of the form y^2=quartic in x. That's not really an elliptic curve, but rather a homogeneous space for an elliptic curve. You don't have an elliptic curve until you have a distinguished point to serve as the identity element. We don't even know there are _any_ points on y^=quartic, in general, if we want them rational.) >k=0 no good but k=3 or k=-4 work. "work" means that one factor or another of the quartic (in b) is a square. This is also significant. No single fibre is in general going to give a parameterized family of solutions to the original problem: there is no nonconstant rational map from Q into a curve of positive genus. But, while it's clear the _generic_ fibre is of genus 1, we -- OK, I'll admit it, "I" -- have not checked for the possibility that some of the fibres are not of genus 1. Since any nonsingular curve y^2=quartic has genus 1, the condition for success is precisely that the curve be singular, i.e., that the quartic have multiple roots. So you took the natural step and looked for rational solutions to discrim(quartic)=0; you found k=3, -4 are the only ones. So you found two fibres of genus zero, which are then parameterizable. Works just great! Let's see if we can generalize to anything similar. Write the original variety as the projective surface S in P^5 given by equations a + b + c + d + e = 0 1/a + 1/b + 1/c + 1/d + 1/e = 0 We look for some embedded curves of genus 0. How should we find them? Your method worked, and seemed easy: let C be the intersection of S with a linear hyperplane H = 0 in P^5. Very well: suppose H = alpha a + ... + delta d. (We needn't include a multiple of e since we may always add a multiple of (a+b+c+d+e) to H without affecting the definition of C.) Since H must be nonzero, permute variables to assume a has a nonzero coefficient. Eliminate a and e to project S isomorphically to the curve S' in P^3 given by a single quartic in {b,c,d}. Now, it's hard in this presentation to decide whether or not C has genus zero. But this equation is a cubic in b (say) with lead coefficient beta*(beta+1)*(c+d) (I am assuming, as I may, that alpha=-1). So we consider only the cases in which beta=0 or -1. Then our one equation is quadratic in b, and so is solvable for b iff its discriminant is a square. Thus our curve (with beta=0 or -1) is isomorphic to one presented by an equation Y^2 = homogeneous polynomial of degree 6 in c and d. Now, as noted before, this curve is parameterizable iff of genus zero, and so definitely not unless singular. This requires the sextic have repeated roots, and so its descriminant must be zero. Maple can factor this discriminant; for both beta=0 and beta=-1, it's a product of powers of gamma, delta, gamma+1, delta+1, gamma-delta, and polynomials of higher degree in those variables. Playing with the possibilities for a minute shows that the only candidates for a genus-zero curve are the intersections of the surface S with hyperplanes H obtained by: permuting the variables in, or scaling, one of these: a = beta b a+b = gamma c a+b+c = delta d a+b = gamma (c+d) Taking in addition linear combinations with the other linear form a+b+c+d+e=0 and allowing also permutations which involve e, we see the last two types are equivalent to the first two. Let me start afresh with these. From a+b+c+d+e=0, 1/a+1/b+1/c+1/d+1/e=0 and a=beta b, we get a curve which can be described with just coordinates b,c,d as a quadratic in b: 2 2 2 c d (beta + 1) (d + c) + (c d beta + 3 c d beta + c d + d beta + c beta) b 2 + beta (beta + 1) (d + c) b which has a rational point iff discrim(",b) is a square. That discriminant is a quartic in c, so we get a curve of genus 1 unless _its_ discriminant is zero. That happens only for beta=0, beta=-1, and some irrational beta. Taking beta=0 would lead to a=0, which is invalid for our original problem; taking beta=-1 leads to the condition c^2+cd+d^2=0, which has no rational solutions. Starting instead with a+b+c+d+e=0, 1/a+1/b+1/c+1/d+1/e=0 and now a+b=gamma c, we get a curve described in b,c,d-coordinates as 2 2 2 2 c d gamma (d + gamma c + c) + gamma c (gamma c d + d + c d + c + gamma c ) b 2 2 2 2 + (-c d - gamma c d - d - c - gamma c ) b . Playing the double-discriminant game again leads to the conclusion that this curve can only have genus zero if gamma=0, -1, 3, or -4. The first two cases can be converted into the case a=beta b, and so will lead to no nonzero rational solutions. The case gamma=-4 gives rational solutions iff 3 c^2 + 3 c d - d^2 = 3*square. This is indeed a curve of genus zero now, and most of its infinitely many points have abcde <> 0. (The case gamma=3 is really the same: a+b=3c and a+b+c+d+e=0 imply d+e=-4c.) It looks like, up to permutation of the five variables, you have found the only embedded curve of genus zero (apart from those in the hyperplanes a=0, etc.) That is, you may have the only parameterized solution to the original problem. At one point in the analysis I assumed ("beta=0 or -1") that a cubic in b could not lead to a rational curve, and although I know that's not really true, I doubt any more examples could be found in this way. It seems to me the only real possibility for finding other rational curves would be to intersect the surface S with hypersurfaces H of higher degree than 1. I haven't tried to do so. * * * * * * * * * * * * * I should also comment on your idea of solving Y^2 = quartic in b. You wrote: >First and third factors are quadratic in b. >Each can be made square (times a const.) by finding k so discriminant is 0 In other words, you try to solve the problem by arranging the quartic so that one quadratic factor is _identically_ a square (times a constant). You can also do this for any particular (factored) quartic in which the quadratic factors are not identically squares. The idea is to choose b to make one factor be a rational square (here you're reduced to parameterizing the points on a conic); substitute in your solutions for b, and then the condition that (quartic in b) be a square is equivalent to the condition (quartic in that rational parameter) be a square. This new quartic defines a different homogeneous space (in general); what it is is the homogeneous space which is 2-isogenous to the original one (when the quartic factors into two quadratic factors, you're defining an elliptic curve with 2-torsion). This finds for you all the points on the original curve which are in the image of the 2-isogeny, which is about halfway towards saying you're finding points on the original curve which are squares (doubles) in the _group_ of points in the elliptic curve. You can get the rest of the points on the curve by a similar procedure: rather than ask that both quadratic factors be squares, ask that each be d*(a square) for the same d. It turns out that the only d which will lead to points on the elliptic curve are those dividing 4*discriminant, so this is a finite process. This is precisely what "descent" on the elliptic curve is about. * * * * * * * * * * * * * >> (I've been mulling over the general situation of arithmetic on algebraic >> surfaces for a while, and I'm anxious to collect all possible tricks.) > >I hope you'll share all these tricks, perhaps at your web site. My only other real trick has been to look for presentations of the surface as an elliptic surface in such a way that the generic rank is positive, and then finding an expression for a generator as a function of the parameter separating the fibres. This is an idea orthogonal to the one you used: rather than hoping there is one special fibre (of genus zero), I hope I can pick out one special point in each fibre. This is the method I used to find a rational solution to that "spider on the box" problem. In general, it's tricky to decide whether or not parameterized solutions exist. For example, I once asked Elkies if there were rational curves in the surface x^4+y^4+z^4=1 (in which he found infinitely many rational points). His answer --- "It might be known that there are no *smooth* such curves, because once they are smooth one can determine their self- intersection from the adjunction formula, use the Galois structure of the Neron-Severi group, yada^3. But proving that there are no rational curves at all seems much harder." --- did not lead me to think I could answer this question with mindless playing. (It turns out Bremner also addressed this particular question in a 1987 paper.) Thank you for remembering my web site. If it's OK with you I'll probably add your correspondence on this topic somewhere under elliptic curves. By the way, since a lot of Diophantine questions have arisen lately, I'm starting to prepare a FAQ which can provide pointers to classes of Diophantine problems which have fairly well-understood methods of solutions. I don't think it will be ready for awhile, but you can take a look at a current working copy at 98/diophantine which I expect to change a lot before I start mentioning it in newsgroup posts. dave ============================================================================== From: Dave Rusin Date: Fri, 18 Sep 1998 00:14:42 -0500 (CDT) To: rusin@math.niu.edu Subject: both solutions are the same Note that one denominator in each formula is a product of two quadratics. Can we express n in terms of m? Not quite: by scaling we can at best assume lhs=(common factor over a,b,c,d,e,)*rhs. Still, compare roots of both products-of-quadratics and look for a rational map sending one to other: m=(5*n+18)/(n+4) works. ============================================================================== Subject: Re: 1/a + 1/b + 1/c + 1/d = 1/(a+b+c+d) another family To: rusin@math.niu.edu (Dave Rusin) Date: Fri, 18 Sep 98 19:37:02 CDT From: Jim Buddenhagen Thanks for your detailed and interesting comments. In due course I'll try to respond. In the mean time, by a different method I found new solutions: # some solutions to Diophantine eqn ee (for k not -1,0,-1/2) ee := 1/a+1/b+1/c+1/d-1/(a+b+c+d); abcd := [-k*(k+1)*(k^2+k+2)*(k^3+2*k^2+k+1), -k*(2*k+1)*(k^3+k^2-1), k*(k+1)* (k^2+k+2)*(k^3+k^2-1), (k^3+k^2-1)*(k^3+2*k^2+k+1)]; abcd2 := [k*(k+1)*(k^2+k+2)*(k^3+k^2-1), (k+1)*(2*k+1)*(k^3+2*k^2+k+1), -k*(k +1)*(k^2+k+2)*(k^3+2*k^2+k+1), (k^3+k^2-1)*(k^3+2*k^2+k+1)]; I have to give more thought to how these relate to the others. -- Jim Buddenhagen jb1556@daditz.sbc.com ============================================================================== From: Dave Rusin Date: Mon, 21 Sep 1998 10:19:05 -0500 (CDT) To: jb1556@daditz.sbc.com Subject: Re: 1/a + 1/b + 1/c + 1/d = 1/(a+b+c+d) another family Jim, > In the mean time, by a different method I found new solutions: These are great; they allow us to construct an infinite family of parameterized solutions! For example, try abcd3:= [k*(k+1)*(k^2+k+2)*(k^3+k^2-1)*(k^9+5*k^8+18*k^7+39*k^6+61*k^5+66*k^4+41*k^3+ 11*k^2-1)*(k^3+k^2+4*k+1), k*(2*k+1)*(k^3+2*k^2+5*k+3)*(k^3+2*k^2+k+1)*(k^9+5* k^8+18*k^7+39*k^6+61*k^5+66*k^4+41*k^3+11*k^2-1), (k^9+4*k^8+14*k^7+31*k^6+51* k^5+60*k^4+41*k^3+15*k^2+3*k+1)*(k^9+5*k^8+18*k^7+39*k^6+61*k^5+66*k^4+41*k^3+ 11*k^2-1), -k*(k^3+2*k^2+5*k+3)*(k^3+2*k^2+k+1)*(k+1)*(k^2+k+2)*(k^9+4*k^8+14* k^7+31*k^6+51*k^5+60*k^4+41*k^3+15*k^2+3*k+1)]; Of course I am curious once again to know how you found your solution. > I have to give more thought to how these relate to the others. Here's what you have done. Recall you viewed the problem as an algebraic surface defined in variables a,b,c (you set d=1 at one point, I think). You can look at the surface like this: here I'll mark some rational points "*". (These coordinates are fictitious but they give the idea). -+-------------- b=4 | -+-*-*-*-------- b=3 | -+-------------- b=2 | -+-*-*---------- b=1 | -+-------------- b=0 | -+-*-*---------- b=-1 | -+-------------- b=-2 | c=0 c=1 etc The surface floats above the b-c plane somehow (a depends quadratically on b and c) but I'll just focus on its projection to the b-c plane. On each horizontal line we fix say the b-coordinate. For some b, we find we get an elliptic curve of positive rank, which is enough to get infinitely many rational points on the surface, but not a parameterization. Other b give no rational points. No b gives a singular (genus-0) curve. What you did instead was to focus on k=b+c. Let's look at the lines k=const: -+-\-\---------- b=4 |\ \ \ -+-*-*-*-------- b=3 |. \ \ \ -+-.-\-\-\------ b=2 |\ . \ \ \ -+-*-*-\-\-\---- b=1 |\ \ . \ \ \ -+-\-\-.-\-\-\-- b=0 |\ \ \ . \ \ \ -+-*-*-\-.-\-\-\ b=-1 | \ \ \ . \ \ -+---\-\-\-.-\-\ b=-2 | .............singular curve c=0 c=1 etc This time the picture is different: some of those lines give rise to singular curves, which by themselves give parameterized curves in the surface. Those are the ones you found before. In addition, all the other slanted lines have rational points on them -- that is, these are all elliptic curves of _positive_ rank. Moreover, they have positive rank "generically", that is, we can locate a rational point "*" on each of them using rational functions of k to give the coordinates a,b,c. You can get a parameterized curve in the surface, if you like, by "connecting the dots", in this picture giving a zig-zag curve which is mostly vertical. This now lets me get infinitely many other parameterizations: if you have given a curve joining one point P_k in each elliptic curve E_k, I simply draw the curve joinging -P_k. I can get another curve joining each 2P_k, or each 3P_k, ... Indeed, this is precisely the procedure I alluded to in my last letter: >My only other real trick has been to look for presentations of the surface >as an elliptic surface in such a way that the generic rank is positive, >and then finding an expression for a generator as a function of the >parameter separating the fibres. This is an idea orthogonal to the one you >used: rather than hoping there is one special fibre (of genus zero), I hope >I can pick out one special point in each fibre. You can also put this into the context of the main discussion in my previous letter: >Let me start afresh with these. From a+b+c+d+e=0, 1/a+1/b+1/c+1/d+1/e=0 and >a=beta b, we get a curve which can be described with just coordinates b,c,d as >a quadratic in b: [...] Here I was looking at the horizontal lines. In my last letter I showed none was singular; I already knew the generic rank was zero, too. Then I tried the slanted lines, since you had been successful there: >Starting instead with a+b+c+d+e=0, 1/a+1/b+1/c+1/d+1/e=0 and now >a+b=gamma c, we get a curve described in b,c,d-coordinates as [another quadratic in b]. Taking discriminant w.r.t. b gave me a curve of the form y^2=quartic in d/c. In my last letter I found the curves in this family which were of genus 0; your present example prompted me to observe the others all have rank at least 1. Indeed, with a bit of algebra these curves can be rendered into Weierstrass normal form as 2 2 2 2 2 3 V1 = 16 k (k + 1) U2 + k (k + 1) (k + k - 4) U2 + U2 (this is your k, = my -gamma); your parameterization corresponds to the observation that there is a point on this curve with U2=4*(1+k)^2 So, I'm not really sure how you're doing it, but you're finding all the rational points on this surface I would ever expect to find! dave ============================================================================== Subject: Re: 1/a + 1/b + 1/c + 1/d = 1/(a+b+c+d) another family To: rusin@math.niu.edu (Dave Rusin) Date: Mon, 21 Sep 98 15:24:25 CDT From: Jim Buddenhagen Dave, > abcd3:= > [k*(k+1)*(k^2+k+2)*(k^3+k^2-1)*(k^9+5*k^8+18*k^7+39*k^6+61*k^5+66*k^4+41*k^3+ > 11*k^2-1)*(k^3+k^2+4*k+1), k*(2*k+1)*(k^3+2*k^2+5*k+3)*(k^3+2*k^2+k+1)*(k^9+5* > k^8+18*k^7+39*k^6+61*k^5+66*k^4+41*k^3+11*k^2-1), (k^9+4*k^8+14*k^7+31*k^6+51* > k^5+60*k^4+41*k^3+15*k^2+3*k+1)*(k^9+5*k^8+18*k^7+39*k^6+61*k^5+66*k^4+41*k^3+ > 11*k^2-1), -k*(k^3+2*k^2+5*k+3)*(k^3+2*k^2+k+1)*(k+1)*(k^2+k+2)*(k^9+4*k^8+14* > k^7+31*k^6+51*k^5+60*k^4+41*k^3+15*k^2+3*k+1)]; Yes. I found it and an infinite seq of others also (in fact I sent email to the original nmbrthry list poster a few days ago about this). > Of course I am curious once again to know how you found your solution. I did it two different ways. The first was: ee:=1/a+1/b+1/c+1/d-1/(a+b+c+d): eee:=numer(simplify(")): readlib(discrim): subs(d=1,eee): # Y2 must be a square Y2:=factor(discrim(",a)): Y2k:=factor(subs(c=-b+k,Y2)); # note b=0 and b=k give points on y^2=Y2k with y coord k*(k+1) # do arith. on y^2=Y2k directly, start with P0=[0,k*(k+1)] b0:=0: y0:=k*(k+1): P0=[b0,y0]; r:=convert(taylor(sqrt(Y2k),b=0,3),polynom): factor(Y2k-r^2); solve(",b); # maple may not put non-zero term 1st so adjust next line if necessary b1:=factor("[1]); y1:=sqrt(factor(subs(b=b1,Y2k))); P1:=[b1,y1]; # next point of seq on y^2=Y2k # here is another b-term in seq subs(k=b1,b1):b2:=factor("); The second method was to use Fibonacci-like recursion relations I had developed a couple of years ago for y^2=even_quartic(x). Must start with a non-torsion point. Here is an example to show what such recursions look like. Suppose we want values of x such that 2*x^4-2*x^2+1 is a rational square. Start with x_0=1 and x_1=1. Let x_(i+1)=(((x_i)^2 - 4)/(1-8*(x_i)^2)) / x_(i-1) Then x_0, x_1, x_2, ... will make 2*x^4-2*x^2+1 a rational square. I transformed (by shifting b) the eqn y^2=Y2k (above) to y^2=even_quartic(b') and used this method starting with the point corresp to [0,k*(k+1)]. The next point is the abcd of my previous email to you, and the next is the abcd3 point you mention above. Of course both of these methods amount to arith on the quartic-form elliptic curve directly. > > I have to give more thought to how these relate to the others. [thanks for all your comments on this-- your ascii art rivals John Baez's :-] > > So, I'm not really sure how you're doing it, but you're finding all the > rational points on this surface I would ever expect to find! See above. --Jim Buddenhagen jb1556@daditz.sbc.com ============================================================================== From: Dave Rusin Date: Mon, 21 Sep 1998 15:55:40 -0500 (CDT) To: jb1556@daditz.sbc.com Subject: Re: 1/a + 1/b + 1/c + 1/d = 1/(a+b+c+d) another family Thanks! I decided to go whole-hog with the change of variables. If we substitute vars:={ a=(2*k*c*(1+k)*(32*k^5+4*k^5*U+96*k^4+8*k^4*U+96*k^3+32*k^2+k^2*U^2-16*k^2* U-12*U*k+3*U^2*k+4*U^2)-2*(1+k)*(4*k^3+4*k^2+4*k+U+4)*k*c*V) /(256*k^5+64*k^2+256*k^3+8*V*k^2-8*V*k^3+8*V*k+64*k^6-8*V*k^4+V^2+ 384*k^4-64*k^3*U-56*U*k^2+7*U^2*k^2-k^4*U^2+16*k^5*U+2*k^3*U^2-4*V*U -16*k^4*U+8*k^6*U+4*U^2+8*k*U^2-16*k*U-4*V*k*U), b=-(-2*V*k-2*V+2*k^3*U+U^2-2*k*U)*(-U+8*k+4+4*k^2)*k*c /(256*k^5+64*k^2+256*k^3+8*V*k^2-8*V*k^3+8*V*k+64*k^6-8*V*k^4+V^2+ 384*k^4-64*k^3*U-56*U*k^2+7*U^2*k^2-k^4*U^2+16*k^5*U+2*k^3*U^2-4*V*U -16*k^4*U+8*k^6*U+4*U^2+8*k*U^2-16*k*U-4*V*k*U), #c= c, d=-c*2*(k+1)*k*(-U+8*k+4+4*k^2)/(8*k^3-U*k^2+16*k^2-3*k*U+8*k-2*U+V), e= c*(k+1)*(U*k^2+k*U+2*U-V)/(8*k^3-U*k^2+16*k^2-3*k*U+8*k-2*U+V) }; Then the original equation 1/a+1/b+1/c+1/d+1/e=0 becomes 2 2 2 2 2 3 V = 16 k (k + 1) U + k (k + 1) (k + k - 4) U + U # eq5:=V^2=16*k^2*(k+1)^2*U+k*(k+1)*(k^2+k-4)*U^2+U^3; which is clearly now an elliptic surface with some singular fibres and the others possessing an element of order 2 (U=0) and an element of infinite order (U=(2-2*k)^2). This surface looks just a wee bit too complicated to have been studied before, but it doesn't seem out of the ordinary in any way. The inverse map is inv:= {k=(b+a)/c, V=4*b*(b+a)*(b+a+c)^2*(2*c*b+2*c*d+d*b)*(c*b+a*c+d*b+a*d+d^2+c*d+c^2)/c^5/d^3, U=4*(b+a)*(b+a+c)*(a*c*d+a*d*b+c*b*a+d^2*b+2*d*c*b+c*b^2+c^2*b+c*d^2+c^2*d+ d*b^2)/c^3/d^2}; dave ============================================================================== From: Dave Rusin Date: Mon, 21 Sep 1998 16:25:24 -0500 (CDT) To: rusin@math.niu.edu Subject: reciprocals: grand xforms eq:=1/a+1/b+1/c+1/d+1/e; vars:=[a,b,c,d,e]: subs(e=-a-b-c-d,vars); subs(a=-b+k*c,"); subs(b=b1+k*c/2,"); subs(d=dd*c,"); factor( subs(b1=b2*c,")); vars:=": numer(factor(subs({a=vars[1],b=vars[2],c=vars[3],d=vars[4],e=vars[5]},eq))); coeff(",b2,2); -"/2; factor(subs(b2=b3/",vars)); vars:=": numer(factor(subs({a=vars[1],b=vars[2],c=vars[3],d=vars[4],e=vars[5]},eq))); factors(")[2]; "[2][1]; eq2:=": fixquart("+b3^2,dd,0); subs({dd="[1],b3="[2]},vars); vars:=factor("): numer(factor(subs({a=vars[1],b=vars[2],c=vars[3],d=vars[4],e=vars[5]},eq))); factors(")[2]; col("[3][1],V); col("/(k+1)^2/k^6,V); eq3:=-V^2+col(V^2+",U); subs({U=U1/4/k^2/(k+1)^2,V=V1/8/k^3/(k+1)^3},vars); vars:=factor("); numer(factor(subs({a=vars[1],b=vars[2],c=vars[3],d=vars[4],e=vars[5]},eq))); factors(")[2]; col("[2][1],V1); subs(U1=U2-8*k-4-4*k^2,vars); vars:=factor("); # Result: vars:= [(256*k^5+64*k^2+256*k^3-16*V1*k^2-16*V1*k^3-16*V1*k+64*k^6-8*V1*k^4+V1^2+384* k^4-8*V1-64*k^3*U2-72*U2*k^2+11*U2^2*k^2-k^4*U2^2+24*k^5*U2-2*V1*U2+8*k^6*U2+8 *U2^2-U2^3+18*k*U2^2-24*k*U2-2*V1*k*U2)*k*c /(256*k^5+64*k^2+256*k^3+8*V1*k^2-8*V1*k^3+8*V1*k+64*k^6-8*V1*k^4+V1^2+ 384*k^4-64*k^3*U2-56*U2*k^2+7*U2^2*k^2-k^4*U2^2+16*k^5*U2+2*k^3*U2^2-4*V1*U2 -16*k^4*U2+8*k^6*U2+4*U2^2+8*k*U2^2-16*k*U2-4*V1*k*U2), # -(-2*V1*k-2*V1+2*k^3*U2+U2^2-2*k*U2)*(-U2+8*k+4+4*k^2)*k*c /(256*k^5+64*k^2+256*k^3+8*V1*k^2-8*V1*k^3+8*V1*k+64*k^6-8*V1*k^4+V1^2+ 384*k^4-64*k^3*U2-56*U2*k^2+7*U2^2*k^2-k^4*U2^2+16*k^5*U2+2*k^3*U2^2-4*V1*U2 -16*k^4*U2+8*k^6*U2+4*U2^2+8*k*U2^2-16*k*U2-4*V1*k*U2), c, -c*2*(k+1)*k*(-U2+8*k+4+4*k^2)/(8*k^3-U2*k^2+16*k^2-3*k*U2+8*k-2*U2+V1), c*(k+1)*(U2*k^2+k*U2+2*U2-V1)/(8*k^3-U2*k^2+16*k^2-3*k*U2+8*k-2*U2+V1)]; numer(factor(subs({a=vars[1],b=vars[2],c=vars[3],d=vars[4],e=vars[5]},eq))); factors(")[2]; col("[1][1],V1); eq5:=-V1^2+col(V1^2+",U2); #Result: # eq5:=-V1^2+16*k^2*(k+1)^2*U2+k*(k+1)*(k^2+k-4)*U2^2+U2^3 inv:=solve({a=vars[1],b=vars[2],d=vars[4]},{k,V1,U2}); #Improve these a little: factor(simplify(",{numer(subs(e=-a-b-c-d,eq))})); inva:= {k=(b+a)/c, V1=4*b*(b+a)*(b+a+c)^2*(2*c*b+2*c*d+d*b)*(c*b+a*c+d*b+a*d+d^2+c*d+c^2)/c^5/d^3, U2=4*(b+a)*(b+a+c)*(a*c*d+a*d*b+c*b*a+d^2*b+2*d*c*b+c*b^2+c^2*b+c*d^2+c^2*d+d*b^2)/c^3/d^2} varsa:={ a=(2*k*c*(1+k)*(32*k^5+4*k^5*U2+96*k^4+8*k^4*U2+96*k^3+32*k^2+k^2*U2^2-16*k^2* U2-12*U2*k+3*U2^2*k+4*U2^2)-2*(1+k)*(4*k^3+4*k^2+4*k+U2+4)*k*c*V1) /(256*k^5+64*k^2+256*k^3+8*V1*k^2-8*V1*k^3+8*V1*k+64*k^6-8*V1*k^4+V1^2+ 384*k^4-64*k^3*U2-56*U2*k^2+7*U2^2*k^2-k^4*U2^2+16*k^5*U2+2*k^3*U2^2-4*V1*U2 -16*k^4*U2+8*k^6*U2+4*U2^2+8*k*U2^2-16*k*U2-4*V1*k*U2), b=-(-2*V1*k-2*V1+2*k^3*U2+U2^2-2*k*U2)*(-U2+8*k+4+4*k^2)*k*c /(256*k^5+64*k^2+256*k^3+8*V1*k^2-8*V1*k^3+8*V1*k+64*k^6-8*V1*k^4+V1^2+ 384*k^4-64*k^3*U2-56*U2*k^2+7*U2^2*k^2-k^4*U2^2+16*k^5*U2+2*k^3*U2^2-4*V1*U2 -16*k^4*U2+8*k^6*U2+4*U2^2+8*k*U2^2-16*k*U2-4*V1*k*U2), c= c, d=-c*2*(k+1)*k*(-U2+8*k+4+4*k^2)/(8*k^3-U2*k^2+16*k^2-3*k*U2+8*k-2*U2+V1), e= c*(k+1)*(U2*k^2+k*U2+2*U2-V1)/(8*k^3-U2*k^2+16*k^2-3*k*U2+8*k-2*U2+V1) }; #subs(varsa,inva): #implify(numer(factor({seq(lhs("[i])-rhs("[i]),i=1..nops("))})),{eq5}); #takes a long time to verify that we get: {0} ! #(much faster the other way). ============================================================================== Subject: Re: 1/a + 1/b + 1/c + 1/d = 1/(a+b+c+d) another family To: rusin@math.niu.edu (Dave Rusin) Date: Mon, 21 Sep 98 16:40:49 CDT From: Jim Buddenhagen Thanks for the transformation and inverse, which I checked and they work. But I think it was easier to stay on the quartic! > Then the original equation 1/a+1/b+1/c+1/d+1/e=0 becomes > > 2 2 2 2 2 3 > V = 16 k (k + 1) U + k (k + 1) (k + k - 4) U + U > > # eq5:=V^2=16*k^2*(k+1)^2*U+k*(k+1)*(k^2+k-4)*U^2+U^3; > > which is clearly now an elliptic surface with some singular fibres and > the others possessing an element of order 2 (U=0) and an element of > infinite order (U=(2-2*k)^2). This surface looks just a wee bit too ^^^^^^^^^^^^^ Don't you mean U=(2+2*k)^2 here? > complicated to have been studied before, but it doesn't seem out of > the ordinary in any way. Except for having the "generic" point of infinite order. [inverse map deleted] --Jim Buddenhagen jb1556@daditz.sbc.com ============================================================================== From: Dave Rusin Date: Mon, 21 Sep 1998 16:52:10 -0500 (CDT) To: jb1556@daditz.sbc.com Subject: Re: 1/a + 1/b + 1/c + 1/d = 1/(a+b+c+d) another family > Don't you mean U=(2+2*k)^2 here? Yes, sorry >>it doesn't seem out of the ordinary in any way. >Except for having the "generic" point of infinite order. I guess it depends on what you consider "ordinary"! Actually I just started experimenting and have found that for specific choices of k, the rank is _very_ frequently greater than 1 for k > 8 (integral) Now, _that_ would be out of the ordinary. dave ==============================================================================