MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

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 exactly 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 will 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.

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.
p=11 q=13/9 r=-5/8
P=-958/589 Q=-533/357 R=-55/534 or
p=11 q=13/9 r=8/5
P=2 Q=1 R=Infinity

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?
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.)

Bonus Q5: Does someone has a list of all known parametric solutions? (I know at least 5 fundamental ones, including Eulers, of course.)

share|cite|improve this question
@Hauke: the arithmetic of these surfaces related to the perfect cuboid problem has been studied seriously by number theorists, a lot, using these sorts of methods, already. See for example the work of Ronald van Luijk. In some sense I suspect his work will both answer all your questions and also, implicitly, tell you the obstruction lying between these ideas and a resolution of the perfect cuboid problem. In some sense I know the obstruction already -- even if you could parametrise your solutions you're still going to be left with a hard diophantine problem. – Kevin Buzzard Jun 7 '11 at 18:04
THX for the reference! I would have been surprised if this approach hadn't had been tried yet, but it's not easy for an amateur to dig up the relevant primary literature. – Hauke Reddmann Jun 8 '11 at 10:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.