Boxing the Rational Box - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:59:31Z http://mathoverflow.net/feeds/question/67128 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67128/boxing-the-rational-box Boxing the Rational Box Hauke Reddmann 2011-06-07T13:08:41Z 2011-06-07T13:08:41Z <p>You surely all know the Perfect Cuboid problem. Here is a bit of pondering on the Euler box (space diagonal doesn't have to be rational).</p> <p>We start with the generators p,q,r and the surface (p^2-1)(q^2-1)(r^2-1)=8pqr. (A:B=(p^2-1)/2/p etc.) (Q1 - what generic name does such a surface have?) A key feature is that it contains rational lines like p=0,q=1,r=any. I call (0,1) a special point. Take any two different of them with p1!=p2,q1!=q2 and define s=(p-p1)*(q-q2)/(p-p2)/(q-q1). For example, s=(p-1)(q-1)/p/q. Eliminate p. We get a new surface in (q,r,s) which is <em>exactly</em> of the same type as the old - sextic, quadratic in each single variable. Rinse and repeat - you can do this substitution as long as you like provided there are two special points with say r1!=r2,s1!=s2 (and which are rational, I have to add - they <em>will</em> exist but might be real!). (Q2 - this is elliptic curve magic in disguise, right?) And since the surface stays quadratic in each variable, you can always jump from a solution e.g. p,q,r to -1/p,q,r (for qrs this already looks a bit more complicated). In effect, you have a chain pqr->qrs->rst->Rst->QRs->PQR and you can turn one solution pqr into another PQR this way. </p> <p>Examples (I don't give the intermediates): P=(p*(1+p-q-p*q+2*r+2*q*r+r^2-p*r^2-q*r^2+p*q*r^2))/(1+p+q+p*q+2*p*r-2*p*q*r-r^2+p*r^2-q*r^2+p*q*r^2), Q and R cyclic with p->q->r->p.<br> p=11 q=13/9 r=-5/8<br> P=-958/589 Q=-533/357 R=-55/534 or<br> p=11 q=13/9 r=8/5<br> P=2 Q=1 R=Infinity </p> <p>Note you can reverse the last example, start from P=any and regain the Euler parametric solution this way! Of course this provokes Q4: By the analogy to elliptic curves, is this process described by a group? Does the above chain end somewhere (or runs back into p,q,r eventually) and thus also forms a sort of group? (Which means there are only finitely of these transformations - in fact, I know of only 2 fundamental ones and the rest is concatenation.) And: Can all solutions p,q,r be derived from say, x,1,0 in finitely many steps?<br> My hope is of course that you can define an invariant for the transformations and use it to prove that the perfect cuboid doesn't exist. (I came pretty close but know zilch of elliptic curves and am stuck.)</p> <p>Bonus Q5: Does someone has a list of all known parametric solutions? (I know at least 5 fundamental ones, including Eulers, of course.)</p>