Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why hasn't Perfect Cuboid Problem been solved yet, whereas (possibly) more nontrivial ones such as FLT and Sphere packing have been solved?

I understand that calling some problems more nontrivial may be naive and seemingly trivial problems can be deceptively tricky, as with the FLT. All the same, FLT and to a lesser extent, Sphere Packing, garnered lots of attention by successive generations of mathematicians, until someone decided to finish it off and succeeded.

But, AFAIK, the Perfect Cuboid (PC) problem hasn't generated this kind of attention, perhaps because Fermat didn't leave a note about it. Is that the reason for PC remaining unsolved? One of the standard references for PC , Unsolved Problems in Number Theory , suggests several numerical results (p.178), but of course nothing like a proof, much like the status of FLT and Sphere Packing many decades ago.

share|improve this question
I think it is much harder to tell people about that problem, so it is going to get less attention. –  Douglas Zare Apr 8 '11 at 4:38
Another reason is that it is not a particular interesting problem from a mathematical perspective, its just a random diophantine system. Its truth value has probably no implications. –  user1708 Apr 8 '11 at 4:43
There aren't many implications of Fermat's Last Theorem, either. –  Douglas Zare Apr 8 '11 at 5:03
FLT is probably famous because of the little note, rather than its use. At the same time, the proof of FLT proved the Taniyama-Shimura conjecture. So it was useful –  picakhu Apr 8 '11 at 5:09
@Ganesh: I see the individual tallies (this is one of the reputation-dependent "privileges"). The two people who upvoted it must have retracted their upvotes. –  joriki Apr 8 '11 at 7:16

3 Answers 3

up vote 4 down vote accepted

If I were to take a guess, I'd suggest that the reason it hasn't been solved yet is because there's not any apparent practical application, and nobody's put a bounty on it that's large enough to make it worth anybody's trouble to solve.

share|improve this answer
Bounties and practical applications are not at all what motivate most mathematicians to attack a given problem. –  Chris Eagle May 28 '12 at 19:08
Hey, I just said it was a guess. Maybe I'm just being a bit cynical, but I sincerely think if a sizable bounty was put on this problem, I think it'd be solved (one way or the other) in under a year. –  Mark May 28 '12 at 21:07

The problem lies in the direction of search and the definition of the perfect cuboid. If one divides all the variables by the length of the internal diagonal, the result is a Rational Cuboid with an internal diagonal with a length of 1. This sets a limit on all values contained in the problem. The problem becomes a question of rational or irrational rather than integer or noninteger.

share|improve this answer
One could do the same thing with FLT: It can be rewritten as solving $x^n + y^n = 1$ in rationals $x,y$. –  Ted Nov 6 '11 at 21:04
Perhaps it is just me, but I am not sure how this answers the question, or why it was accepted. –  Eric Naslund May 28 '12 at 19:11

It seems that the perfect cuboid problem is one approaching the difficulty of, if not more difficult than solving Fermat's Last Theorem which took Andrew Wiles many years of his career and the use of extrememly complex mathematics to do.

However, if the problem is approached logically there seems to be a good reason why one can't exist. I was never one to believe that a seemingly impossible problem should be tackled by using computer programs to try as many combinations as possible just to see if a solution could be found. To me that admits defeat and actually proves nothing unless a set of figures that proves the case is found which for a perfect cuboid hasn't happend to date.

I believe there is an answer and it's been staring us in the face but been ignored. Take the three face diagonals of a cuboid with dimensions X, Y Z. The three face diagonals are given by:-

X^2 + Y^2 = D1^2;
X^2 + Z^2 = D2^2;
Y^2 + Z^2 = D3^2;

We need go no further because all we need do is look at Y^2 and think about symmetry.

share|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.