Newsgroups: sci.math From: iain@stt.win-uk.net (Iain Davidson) Date: Thu, 08 Feb 1996 12:48:14 GMT Subject: soln. to set Diophantine Eqns I was wondering if it was possible to find a formula for non-trivial solutions to this set of equations qq + 2pr + 2kst =0 ktt + 2ps + 2qr = 0 rr + 2pt + 2qs =0 where k,p,q,r,s are all integers or can it be shown that no non-trivial solutions exist. Any help would be welcome. I. Davidson Iain Davidson Tel : +44 1228 49944 4 Carliol Close Fax : +44 1228 810183 Carlisle Email : iain@stt.win-uk.net England CA1 2QP ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: soln. to set Diophantine Eqns Date: 12 Feb 1996 08:31:14 GMT Summary: You have to look for solutions in the right way. Keywords: elliptic curves, branched covers, projective curves, polar forms In article <75@stt.win-uk.net>, Iain Davidson wrote: > >I was wondering if it was possible to find a formula for non-trivial >solutions to this set of equations > >qq + 2pr + 2kst =0 > >ktt + 2ps + 2qr = 0 > >rr + 2pt + 2qs =0 > > >where k,p,q,r,s are all integers or can it be shown that no >non-trivial solutions exist. I found this to be a highly entertaining problem. I would like to know of its origin. There are many solutions to this system of equations. Perhaps the simplest non-trivial one is (k,p,q,r,s,t)=(24,4,-4,4,1,-1). I can describe how to find them all, although it is not so much a formula as a recursive procedure. After this discussion, I will close with some general comments which arose while I was attacking this problem in other ways; in particular, I will show that the problem is a little more natural than it might at first appear, and that it even admits a "polar form" akin to the statement of the "integral brick" problem. The resulting post is rather lengthy (~350 lines). I'm sorry; these things happen when I get carried away. **************** Rationally, the 3-fold described by the above equations is essentially the product of the affine plane and the elliptic curve y^2=x^3+x^2-28x+48. Here's the correspondence. Suppose you have integers (or rationals) k, p, q, r, s, t satisfying q^2 + 2pr + 2kst =0.........(1) kt^2 + 2ps + 2qr = 0........(2) r^2 + 2pt + 2qs =0..........(3) If r=0, then it is easy to check that (p,q,r,s,t,k)= (p,0,0,0,0,k), (0,0,0,s,0,k), (0,0,0,s,t,0), or (-4e^4f, -4e^3f, 0, ef, f, 8e^5) for some integral values of the remaining variables (which are otherwise unrestricted). These are solutions which are not picked up by the mechanism I am about to describe. These and other similar exceptions I will consider "trivial". Thus I will assume r (and various other quantities) are nonzero. Multiply the three equations (1)-(3) by -64s^2pt/r^6 128s^3p/r^6 and 32s^2(q^2+2pr)/r^6 respectively and add. If you define rational numbers x and y by x = -4sq/r^2 + 2 y = (16s^2p + 12 qsr )/r^3 + 2 the the sum becomes (after a bit of simplification) y^2 = x^3 + x^2 - 28x + 48 that is, this pair (x,y) defines a rational point on the given elliptic curve. So to the solution (k, p, q, r, s, t) we associate the point (r, s/r, x, y) in A^2 x E (A is affine space, E is the elliptic curve). Note that we have assumed r<>0; we might as well assume s<>0 too, since the above procedure takes any point with s=0 to a point (r, 0, 2, 2). No other point with r<>0 maps to (2,2) or (2, -2). The formulas above are invertible: given such a point (r,u,x,y), we compute p = (y+3x-8)/16 ru^(-2) q = -(x-2)/4 ru^(-1) r = r s = ru t= 4(x-4)/(y+3x-8) ru^2 k = (y+3x-8)(x^2+2x+2y-12)/(128(x-4)) u^(-5) and then discover that (i) this map is the inverse to the previous one (ii) if (x,y) is in E, then (k,p,q,r,s,t) is a point on the 3-fold. As remarked in the previous paragraph, we have here assumed u<>0. In the description of t and k, note that for (x,y) in E we have (y+3x-8)/(x-4) = -(x-2)^2/(y-3x+8), so that these rational functions are regular at x=4 (where y=+-4). We avoid the points with x=2 (as noted above, these are not in the image of the 3-fold anyway). It is now easy to compute the integral solutions as well. We may compute all rational solutions by taking each rational point (x,y) on E, then allowing u and r to vary over the (nonzero) rationals. With x and y fixed, we see there is a maximal positive rational u=u0 making k integral; the other choices for u yielding an integral k are u=u0/u1 for any integer u1. With u then fixed as well, there is a minimal positive r = r0 making p, q, r, s, and t all integral; the other choices are of the form r=r0*r1 with r1 an arbitrary integer. This gives a collection of solutions of the general form (p,q,r,s,t,k) = (p' r1, q' r1 u1, r' r1 u1^2, s' r1 u1^3, t' r1 u1^4, k' u1^5). Thus the determination of all integral points amounts to a determination of all rational points on the elliptic curve. I used John Cremona's mwrank program to determine that this curve has a single torsion element of order 2 (namely T = (x,y) = (3,0) ) modulo which it is infinite cyclic, with the point P = (2, -2) as generator. More elements of the elliptic curve are easily generated by the chord-and-tangent process; here are the first few (note that if Q=(x,y), -Q = (x, -y). ) P=(2,-2) P+T = (-2,10) 2P=(4,-4) 2P+T = ( 8,20) 3P=(-6,-6) 3P+T = (66/27, 10/27) 4P=(26/8, 11/8) 4P+T = (23, -110) (assuming my arithmetic is correct). There is no "formula" which constructs all these points, that is, this curve is not birationally equivalent to P^1, so there is no nonconstant rational function whose image lies in the curve E. Using our earlier formulas, we now compute the smallest "integral" points on the original curve. We avoid the points with x=2 as noted above. SO starting with T=(3,0), for example, we compute p = (1/16) ru^(-2) q = (-1/4) ru^(-1) r = r s = ru t = (-4) ru^2 k = (3/128) u^(-5) so that u0=1/4, k=24 u1^5. Then we need r0=4u1^2 in order to make t integral, so the general solution corresponding to this point on E is p = r1 q = -4 r1 u1 r = 64 r1 u1^2 s = 256 r1 u1^3 t = -4096 r1 u1^4 k = 24 u1^5 **************** I will close with some other observations about the original system of equations. It seemed natural to think of k as a "parameter", leaving this (for each k) as a system of 3 homogeneous quadratics in affine 5-dimensional space; apart from the trivial solution (p,q,r,s,t)=(0,0,0,0,0), the rest of the solution set may then be viewed as a curve in projective space P^4. The poster asked if any of these curves (for various integral k) have any rational points. Faltings has shown that a rational curve of genus g > 1 can have only finitely many rational points. Thus for each k there are at most a finite number of solutions to this set of equations. Faltings' result does not enable us to tell whether that number is zero or not. We investigate each of the different curves (with different k's). Let us consider the case k=0. In that case, the original equations reduce to qq + 2pr =0 2ps + 2qr = 0 rr + 2pt + 2qs =0 If p=0, then we must have q=0 and then r=0 leaving s and t arbitrary. If p <> 0, then we use the three equations in order to compute r = -q^2/(2p), s = -qr/p = q^3/(2p^2), and t = -(r^2+2qs)/(2p) = -(5/8) q^4/p^3 so that the curve is the really union of two irreducible curves, parameterized by P^1(Q) by f([u:v]) = [0:0:0:u:v] and g([u:v]) = [8u^4 : 8vu^3 : -4v^2u^2 : 4v^3u : -5v^4 ] (Here the coordinates in the image give p, q, r, s, and t in that order.) These two curves meet at [0:0:0:0:1] only. Now, the curves with the various nonzero values of k are distinct over Q (e.g., some may be empty, other may have infinitely many solutions), but let us consider solutions in any field K which contains an element j with j^5=k. Over such a field, _all_ these curves are isomorphic. Indeed, if [p:q:r:s:t] is a solution to the original curve, then let p'=p/j^4, q'=q/j^3, r'=r/j^2, s'=s/j, and t'=t. Then we may compute that the new point [p':q':r':s':t'] is a point on the curve corresponding to k=1; moreover, this transformation is invertible. So over any field with such an element j, the search for solutions may be restricted to the case k=1. (To find rational solutions to any of the other curves, "simply" enumerate all solutions to this basic curve in Q(k^(1/5)) , transform back to solutions of the more general curve, and then check to see which transformed solutions lie in Q.) **************** So we may focus on the basic curve in P^4: qq + 2pr + 2st =0 tt + 2ps + 2qr= 0 rr + 2pt + 2qs =0 There is a rational map P^4 --> P^2 defined by projection, that is, [p:q:r:s:t] -> [q:r:t]. (Notice that the domain of this function is all of P^4 except the points [u:0:0:v:0] as [u:v] ranges over P^1.) If [p:q:r:s:t] is a point on our curve, so that the equations are satisfied, then multiplying those equations respectively by (2pqrt-2pt^3-q^3t+qr^3); 2(t^2-qr)^2; (q^2t^2-r^3t-2pqr^2-2prt^2): and adding, we get 2(t^2-qr)^2(t^2 + 2qr) + (r^3-q^2t)(q^3-r^2t) = 0 which simplifies a little when you expand it, to: 5 5 3 3 2 2 2 6 q t + r t - 5 q r + 5 q r t - 2 t = 0 Thus, the image of our curve is contained in the curve this describes in P^2. This is almost isomorphic to the original curve. Over points where t^2-qr <> 0, one can use two of the equations above to solve for p and s; the statement that [q:r:t] satisfies this last equation implies the remaining equation of the original set is also satisfied; moreover, for such [q:r:t], these are the only points [p:q:r:s:t] of the original curve which lie in the inverse image. So the projection is an isomorphism here. If t^2=qr, then it is not hard to work with the above equation and conclude that q^5=r^5 and then that the only such points on the curve are of the form [w: w^(-1): 1] where w is a fith root of 1. (w=1 always gives one such point; if there is a fifth root of unity in K, then there are 5 such points.) For these point we may return to the original curve in P^4 and conclude there are exactly two choices for (p, s): the points of the original curve lying over [w: w^(-1) : 1] are [w^3: w : w^(-1) : (-3/2)w^(-3) : 1] and [(-3/2)w^3 : w : w^(-1) : w^(-3) : 1]. Also note that the original curve in P^4 includes points not in the domain of the projection function -- points with q=r=t=0. But in such cases, ps=0 as well, so there are only two additional points, [1:0:0:0:0] and [0:0:0:1:0] So we conclude that the projection map P^4 -> P^2 induces the following: on the curve in P^4, the map is defined at all but two points [1:0:0:0:0] and [0:0:0:1:0]. It identifies the (up to 5) pairs of points [w^3: w : w^(-1) : (-3/2)w^(-3) : 1] and [(-3/2)w^3 : w : w^(-1) : w^(-3) : 1]. Otherwise it is one-to-one and onto the given curve in P^2. The two curves are birationally equivalent. We conclude that over sufficiently large fields, the study of our curve in P^4 is equivalent to the study of a curve in P^2. **************** While we are discussing branched coverings, let me mention two or three more. Assume the field K contains a fifth root of unity w. Then the cyclic group ~= Z/5Z acts on P^4 as follows: w^(i)[ q : r : t ] = [ w^(i)q : w^(-i)r : t ]. This action preserves our curve setwise. The equivalence classes are the inverse images of points under the map P^2 -> P^3 given by f([q:r:t]) = [q^5 : r^5 : qrt^3 : t^5 ]. Writing f([q:r:t])=[x:y:u:v], we see that the image of each point in our curve will lie on the surface 5 u^2 - 2 v^2 + xv + yv = 0 and, like all of f(P^2), on the surface xyv^3=u^5 The branch points of this covering are of course the points where the map f fails to be 5-to-1, i.e., the points fixed by Z/5Z. The only such point is [0:0:1], but this point of P^2 isn't on our curve. Thus this f is an 5-fold unbranched covering of a curve in P^3 by the one in P^2. The cyclic group Z/2Z in turn acts on P^3 by having the non-identity element send [x:y:u:v] to [y:x:u:v]. This action also preserves our curve (setwise). The orbits under the Z/2Z action are the inverse images of points in the map [x:y:u:v] -> [(x+y)v:xy:uv:v^2] except on the subset where v=0. We can combine this with the previous paragraph: the dihedral group D_10 acts on P^2 via the preceding two actions; the equivalence classes are precisely the inverse images of points under the map g: P^2 -> P^2 given by g([q:r:t]) = [(q^5+r^5):(qr)t^3:t^3], except that as usual we have collapse many points to few on the set t=0. This last map sends points of our curve to the curve zv + 5 u^2 - 2 v^2 = 0 This is clearly a rational curve in P^3 (i.e. it is birationally equivalent to P^1, under the maps [z:u:v] -> [u:v] and [u:v] -> [5u^2-2v^2:uv:v^2]; the point [1:0:0] is not in the domain. **************** Thinking to look at the p-adic solutions via Newton's method, I computed the derivative of the 3 defining polynomials (with respect to p,q,r,s,t). Do it yourself: you'll see a clear pattern which I can use to rewrite the equations in a way which reveals more structure. Indeed, the equations may be written in matrix form as (p,q,r,s,t) . A^n . B . (p,q,r,s,t)^t = 0 (the "^t" is for transpose) where [ 0 1 0 0 0 ] [ 0 0 0 0 1 ] [ 0 0 1 0 0 ] [ 0 0 0 1 0 ] A = [ 0 0 0 1 0 ] , B = [ 0 0 1 0 0 ] , n = 0, 1, 2 [ 0 0 0 0 1 ] [ 0 1 0 0 0 ] [ k 0 0 0 0 ] [ 1 0 0 0 0 ] So we are looking for integral isotropic vectors of these three (clearly linked) quadratic forms. Now, how about a change of basis? If you've ever seen circulant matrices, you know what to try. As in some of the previous sections, we will have to forego rationality. If w = one fixed primitive 5th root of 1, then we can use a basis consisting of the vectors e_i = (1, w^i, w^(2i), w^(3i), w^(4i) ) for i = 0, 1, 2, 3, 4. Note that the change of basis matrix has an inverse which is integral after multiplication by 5. (Keyword: Vandermonde). Thus vectors of algebraic integers (p,q,r,s,t) may be expressed as sums Sum( x_i e_i ) with each 5*x_i an (algebraic) integer; conversely, any Sum( x_i e_i ) with x_i integral gives an integral vector (p,q,r,s,t). (The factor of 5 will not be a concern since the equations are _homogeneous_ quadratics.) How are the quadratic forms expressed in this new basis? We need to compute Q_{ij}^{(n)} := e_i A^n B e_j^t, for which I get [ 1 0 0 0 0 ] [ 0 w^4 0 0 0 ] Q^(0) = 5 [ 0 0 w^3 0 0 ] [ 0 0 0 w^2 0 ] [ 0 0 0 0 w ] that is, Q^(0)_ij = 5 w^(4i) delta(i,j), where we are indexing our rows and columns with {0,1,2,3,4}. The others are a bit messier: Q^(1)_ij = w^(3i) [ 5 delta(i,j) + (k-1) w^{4(i-j)} ] Q^(2)_ij = w^(2i) [ 5 delta(i,j) + (k-1) (w^{4(i-j)}+ w^{3(i-j)}) ] However, if we make the assumption of previous sections that k=1, then all these quadratic forms are diagonalizable; that is, the original equations amount to this: we are trying to find algebraic integers x_0, x_1, x_2, x_3, x_4 so that Sum w^(4i) x_i^2 = 0 Sum w^(3i) x_i^2 = 0 Sum w^(2i) x_i^2 = 0 Thus the problem at hand may be described in several equivalent ways: (a) given a linear plane through A^5, find the points on it all of whose coordinates are squares (b) find two squares such that each of three given linear combinations of them are also squares (c) find two field elements such that each of five given linear combinations of them are squares I can't decide which one sounds easiest, or most impressive. Just for comparison, I mention the (still unsolved) question of the "integral brick". This asks for the dimensions a, b, c of a rectangular brick such that these three lengths, the lengths d, e, and f of the surface diagonals, and the length g of the main diagonal, are all integers (or rational). It may be expressed, for example, as the problem "Solve 0=A+B-D=A+C-E=B+C-F=A+B+C-G in squares". (The brick problem looks for a point on a _surface_ in P^6.) **************** SUMMARY: Viewing k as a separate parameter introduces a natural family of interesting curves in P^4, the existence of rational points on which is difficult to determine. Viewing the bundle of curves as a surface in A^6 allows an enumeration of all points, without necessarily answering the question of the existence of solutions on any individual curve in the family! dave ============================================================================== To: rusin@vesuvius.math.niu.edu Date: Tue, 13 Feb 1996 19:23:53 Subject: Re: soln. to set Diophantine Eqns From: Iain Davidson Dear Dave, Thanks for your reply, >I found this to be a highly entertaining problem. I would like to >know of its origin. Equations were an attempt to solve X^5 + kY^5 = Z^2 GCD(X,Y,Z) =1 I took the norm of (p + q*alpha + r*alpha^2 + s*alpha^3 + t*alpha^4)^2, where alpha is the real fifth root of K. Setting the three equations to zero means that the only terms left in the form are X^5 + kY^5 which is a square. Unfortunately, I don't have the mathematical background to follow through your reply. Could you suggest some background reading as I would like to better understand what you were saying. You could obviously get larger sets of equations by considering higher powers. 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 ============================================================================== [Remark 1998/11/19: The curve 5 5 3 3 2 2 2 6 q t + r t - 5 q r + 5 q r t - 2 t = 0 has genus 5.]