From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Rational Cuboids - very special question Date: 21 Oct 1998 03:52:44 GMT Onno Garms wrote: >The rational cuboid problem is this: > >Is there a cuboid with sides, face diagonals and body >diagonal all rational? [According] to R.K. Guy, "Unsolved Problems >in Number Theory" this is an open question. >For further information check this book or the links on >index/11DXX.html > >If you leave away the condition that the body diagonal is >rational, there are solutions. (The smallest integral >solution being the cuboid with sides 44,117,240.) Before proceeding to the poster's question, let me point out the source of the difficulty with the "rational box" problem. The problem is homogeneous; that is, we could divide all lengths by the length of one of the sides (say) and express the problem as that of the existence of rational solutions to the equations 1 + y^2 = a^2 1 + z^2 = b^2 y^2 + z^2 = c^2 1 + y^2 + z^2 = d^2 (Here y and z are the other two sides. Note that the alternative problem suggested above requires the solution of just the first three equations.) Think about the _real_ solutions to this set of equations. These are the points (y,z,a,b,c,d) in R^6 which satisfy those equations. What does this solution set "look like"? Well, think: 4 equations in 6 unknowns, solution set has dimension 6-4=2. (Leave off the last equation and we still have a 5-3=2-dimensional solution set.) Whatever this mantra really means, the basic conclusion is valid: these equations describe an (algebraic) _surface_. As far as the number theory goes, that's a real problem: there is not really a good, complete, theory to which we can turn to answer this question. >Guy states that there is no such cuboid with ratio 3:4 >between two sides. J. Leech states the same thing in "The >rational cuboid revisited", Amer. Math. Monthly 84 (1977) >518-533. [I assume the "such" means, with no restriction on the body diagonal, that is, we seek rational solutions to the first three equations above.] Suppose we wanted to find such a box. Scaling by the shorter side in that 3:4 ratio, and letting y be the other side, we have y=4/3 (and so a=5/3). The remaining equations to be satisfied are 1 + z^2 = b^2 16/9 + z^2 = c^2 Now what happened is that we have added a constraint (y=4/3 x) and so reduced the dimension of the solution set by one: we expect a 3-2=1 dimensional solution set, i.e. a _curve_. This is a much happier situation: if the diagnosis is "curve", there is a course of treatment, and so while recovery is not guaranteed, the prognosis is excellent. In fact, these equations describe the intersection of two conic sections in 3-space, and so it is known that the intersection is an _elliptic_ curve, so it is nearly certain that a solution can be derived. >No one gives a proof. Though having searched literature >carefully, I cannot find one. Does anyone know where to find >a proof? A proof by infinite descent can be constructed, although frankly I am too lazy to create a concise one. The solutions to 1 + z^2 = b^2 can be parameterized with z = 2t/(t^2-1), t rational. Thus we seek rational solutions to the single equation 16/9 + 4t^2/(1+t^2)^2 = c^2; multiplying by 9/4 (1+t^2)^2, we see we are just asking that 4 t^4 + 17 t^2 + 4 be square. At this point we recognize an elliptic curve with 2-torsion, an upper bound on whose rank is obtained by descent (i.e. using 2-isogeny); I prefer to let software carry out the computations, and so learn that the rank is zero and the torsion has order 8. The only rational points are those with t= +- 1 or 0. You can provide details by mimicking the proof of Fermat's theorem ("All solutions to y^2 = t^4 + 1 are trivial") as found e.g. in the Cassel's little book on elliptic curves. dave