From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: The Diophantine system Date: 31 Jan 2000 05:56:25 GMT Newsgroups: sci.math Summary: concordant binary quadratic forms In article <89pn0ijwm08x@forum.swarthmore.edu>, Olga Rukosujeva wrote: >If n=2 system > x^2+y^2=z^2 x^2+(ny)^2=t^2 >has no solutions. > >Do you know other such n? Eh? For _any_ n, we have solutions (x,y,z,t)=(0,a,+-a, +-na) and (a, 0, +-a, +-a) for any a. So I suppose the statement means, "... no solutions other than these 8 families". But even then, I'm not sure why n=2 is singled out; the same result holds for many small n and indeed the challenge seems to be to find an n for which the family of equations _has_ other solutions. You didn't specify what "solution" means, but from the Subject line I gather you intend x,y,z,t to be rational or integral -- really these are the same questions since your equations are homogeneous. Well, a pair of homogeneous quadratic equations determines an elliptic curve, so (for each specific n ) this problem is likely to yield to known techniques. Specifically, since there are no nontrivial solutions with z=0, we may divide through by z and obtain the equations x^+y^2=1, x^2+(ny)^2=t^2. The rational solutions to x^2+y^2=1 are (-1,0) and all the points ((1-u^2)/(1+u^2), (2u)/(1+u^2)). Your second equation then requires that (1-u^2)^2 + (2nu)^2 be square, that is, we seek points on the curve v^2 = u^4 + (4n^2-2)u^2 + 1. We may even put this curve into Weierstrass form using the substitution X=(v+u^2-1)/2, Y=(vu+u^3+2n^2u-u)/2 to see we are looking for points on the curve Y^2=X(X+1)(X+n^2). The eight points we knew at the beginning have become the points (0,0), (-1,0), (-n^2,0) of order 2, and (n, +-(n^2+n)) and (-n, +-(n^2-n)) of order 4; these and the point at infinity form the torsion group (order 8) of these elliptic curves. So the question of whether or not there are other solutions is exactly the question of whether or not the rank is greater than zero. Well, I ran some calculations and found the rank indeed to be zero for all n between 2 and 20 inclusive except for the following, all of which have rank 1 ; shown are the coordinates for one of the generators: n X Y u v x y t (z=1) 10 -2 14 1/7 -148/49 24/25 7/25 74/25 11 4 50 2/5 221/25 21/29 20/29 221/29 12 3 42 2/7 339/49 45/53 28/53 339/53 14 -49 588 4 -113 -15/17 8/17 113/17 17 49 910 35/13 ... -528/697 455/697 7753/697 19 1/4 85/8 1/34 ... 1155/1157 68/1157 1733/1157 For each of these n you may construct all (infinitely many) solutions to your problem by means of the well-known (to those who know it) chord-and-tangent procedure. While I was at it, I took a look at the more inclusive family of elliptic curves of the form Y^2 = X(X+1)(X+N) where N is now any integer. Again it's true that the rank is zero for all N between -10 and 10 inclusive, except for N=0, N=1 (which give a singular curve) and these N, for which the rank is 1 and a generator [X,Y] is shown: -10: [98, 924] -6: [8,12] 7: [1,4] 10: [-2,4] The rule of thumb, when discussing a family of elliptic curves like this, is that apart from a finite number of exceptional N, all choices of N yield an elliptic curve with a rank at least equal to the "generic rank", in our examples zero. The rank will be frequently greater by 1, but also (with decreasing frequency) ever higher ranks also appear. (Sadly, it also begins to be true for large N that more and more curves in the family will have undetermined rank; the known algorithms for rank determination are not complete.) So we expect for example that there will be curves in these families of rank 2 or more. I did observe one such curve in the second family: when N=31, the curve Y^2 = X(X+1)(X+N) has rank 2 and independent points [X,Y]=[1,8] and [-4,18]. dave Elliptic curves: http://www.math-atlas.org/index/14H52.html ============================================================================== More generally, suppose we seek x,y for which both x^2+ny^2 and x^+my^2 are squares. Parameterizing x^2+ny^2=1 by x=(n-u^2)/(n+u^2), y=2u/(n+u^2) we find we must seek points on the curve v^2 = u^4 + (4m-2n)u^2 + n^2. Upon substituting with X=(v+u^2-n)/2, Y=u(v+u^2-n+2m)/2 we obtain the (symmetric) WNF Y^2=X(X+m)(X+n) . Of course there are three points of order 2 and some calculation with the doubling formulas show that the points with X=0, -m, -n are doubles iff m*n, m*(m-n), n*(n-m) are squares, respectivel; in the latter two cases one actually finds a half-point (i.e. X(X+m)(X+n)>0) iff m<0 resp n<0, and in the first case we similarly require both m,n positive (rather than both negative). These three cases may be obtained one from another by linear changes of variable anyway. Presumably situations like ax^2+ny^2 and bx^2+my^2 both being squares will lead only to homogeneous spaces and only with difficulty; can't even parameterize the initial quadratics, in general... ==============================================================================