From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: set of 2 equations Date: 17 Jun 1998 19:05:01 GMT In article <6m6l83$cll$1@nnrp1.dejanews.com>, wrote: >plz find all of the integer solutions a, b, c to this set : > >(1) a^2 = 4(b+c) >(2) a^3 - 2(b^3) -4(c^3) = 0.5 * abc Well, I can find all the rational solutions (there are infinitely many) and some integral ones -- probably all of the latter, but to know for sure would require some ad-hoc arguments, I suspect, and I am not willing to try to make them. Here are the rational points; you can look for yourself for integral ones among them. We may eliminate c=a^2/4 - b to see {a,b} must satisfy 3 3 6 4 2 2 3 2 16 a + 32 b - a + 12 a b - 48 a b - 2 a b + 8 a b = 0 (_Integral_ solutions will also require a to be even.) The substitution b = a*e/4 shows that there must exist rational points on the curve specified by 3 2 2 3 2 (*) 2 a - 6 a e + 6 a e - e + a e - e - 32 = 0 Since the degree is 3, we will have an elliptic curve once a distinguished point is chosen for the origin; [4/3, -4/3] will do. With some more substitutions, the curve is transformed to Weierstrass normal form 2 3 Y + XY = X - 27712 - 288 X This is a curve of rank 1 with a torsion subgroup of order 3. (The torsion points are those with X=52). A generator has X=874/9. Transforming back to the previous coordinates puts the torsion points at [2,8] and [4,3] and the generator at [218/51, 118/51]. You may now find all rational solutions [a,e] to equation (*) with the so-call "chord and tangent process": given any two rational points on the curve, draw the line through them; it will intersect the curve in another rational point. (Taking the two starting points to be equal replaces that line with the tangent line through the point; again another rational point will be the intersection of the line and the curve.) There will be infinitely many rational points. You may easily convert their coordinates back to the initial values of {a,b,c}. For example, the four points mentioned above give {a,b,c} = {4/3, -4/9, 8/9}, {2,4,-3}, {4, 3, 1}, and {218/51, 6431/2601, 5450/2601} respectively. It follows from general principles that there will only be a finite number of integer points on the curve, but it is not in general possible to know an effective upper bound on their coordinates, so I would be reluctant to state that all integer points are found. dave ============================================================================== [Remark: note that a,b integral only forces e to be rational, but (*) makes e an algebraic integer, hence an integer: a | 4b . Introducing the integer f=a-e at the expense of a gives the equation 3 3 2 f + e + f e = 32 Considering which pair of the four terms have the smallest 2-part, we find 2^r||f, 2^s||e with (r,s)=(0,0), (0,1), (1,3), (2,3), (3,2). Hmm... I don't know how to find a provably complete set of integer points. Integer solutions include (e,f) = (8,-6) and (3,1): (a,b)=(2,4), (4,-3) These are the torsion points on the EC when taking (-4/3,8/3) as O. A short walk around the group shows no other points (e,f) are even _close_ to integral! ]