Does this surface contain all perfect cuboids?

Question

Probably this is wildly known, but the closest I found was a surface with additional restrictive conditions.

A perfect cuboid is a cuboid having integer side lengths, integer face diagonals and an integer space diagonal leading to positive integer solutions of:

$$a^2+b^2=s_1^2\qquad \qquad \rm(1)$$ $$a^2+c^2=s_2^2\qquad \qquad \rm(2)$$ $$b^2+c^2=s_3^2\qquad \qquad \rm(3)$$ $$a^2+b^2+c^2=s_1^2+c^2 =s_4^2\qquad \qquad \rm(4)$$

wlog one can work with rationals and dividing by $c^2$ one can assume $c=1$ making (2) and (3) $a^2+1=s_2^2$ and $b^2+1=s_3^2$ and they can be parametrized by the substitutions $a=\dfrac{u^2-1}{2u}$ and $b=\dfrac{v^2-1}{2v}$

This leaves only (1) and (4). $s_1^2+1=s_4^2$ leads to the next parametrization $s_4=\dfrac{s^2+1}{s^2-1}$ and $s_1=\dfrac{2s}{s^2-1}$.

This makes (1) and (4) equal and the denominator vanish for $u=0$,$v=0$ or $s= \pm 1$.

So the surface comes from the numerator of (1):

$u^{4} v^{2} s^{4} + u^{2} v^{4} s^{4} - 2 u^{4} v^{2} s^{2} - 2 u^{2} v^{4} s^{2} - 4 u^{2} v^{2} s^{4} + u^{4} v^{2} + u^{2} v^{4} - 8 u^{2} v^{2} s^{2} + u^{2} s^{4} + v^{2} s^{4} - 4 u^{2} v^{2} - 2 u^{2} s^{2} - 2 v^{2} s^{2} + u^{2} + v^{2} = 0$

The trivial rational points contain $0$ and $\pm 1$.

(A) Does the surface contain all perfect cuboids?

(B) Is there a reason/heuristic to believe the surface might have non trivial rational points?

EDIT Per John's answer some congruences were found on the homogenized surface

$u^{4} v^{2} s^{4} + u^{2} v^{4} s^{4} - 2 u^{4} v^{2} s^{2} h^{2} - 2 u^{2} v^{4} s^{2} h^{2} - 4 u^{2} v^{2} s^{4} h^{2} + u^{4} v^{2} h^{4} + u^{2} v^{4} h^{4} - 8 u^{2} v^{2} s^{2} h^{4} + u^{2} s^{4} h^{4} + v^{2} s^{4} h^{4} - 4 u^{2} v^{2} h^{6} - 2 u^{2} s^{2} h^{6} - 2 v^{2} s^{2} h^{6} + u^{2} h^{8} + v^{2} h^{8}$

$$ v^{2} \cdot u^{2} \cdot s^{4} \cdot (u^{2} + v^{2}) \equiv 0 \mod {h}$$ $$ (s - h)^{2} \cdot (s + h)^{2} \cdot v^{2} \cdot h^{4} \equiv 0 \mod u$$ $$ (s - h)^{2} \cdot (s + h)^{2} \cdot u^{2} \cdot h^{4} \equiv 0 \mod v$$ $$ h^{4} \cdot (u^{4} v^{2} + u^{2} v^{4} - 4 u^{2} v^{2} h^{2} + u^{2} h^{4} + v^{2} h^{4}) \equiv 0 \mod s$$

The substitution $ s = u^{4} v^{2} + u^{2} v^{4} - 4 u^{2} v^{2} + u^{2} + v^{2}$ leads to a factor:

$$ u^{4} v^{2} + u^{2} v^{4} - 4 u^{2} v^{2} + u^{2} + v^{2} = 0$$

This is genus 1 curve with known trivial points, and it is of rank 0 (if I have done the computations right).

Answer 1

Looks OK to me (the approach anyway - I didn't check the numerator calculation). Ruslan Sharipov also found an explicit equation for the perfect cuboid surface, in a recent ArXiv paper at http://arxiv.org/abs/1104.1716. His derivation was much more intricate than yours, but the result looks very similar!

This surface is known to be a so-called surface of general type [ http://en.wikipedia.org/wiki/Surface_of_general_type ] and thus has only a finite number of rational points.

Most reckon it has no non-trivial (i.e. with all non-zero) rational points, and with equations like this it tends to be "small or nothing". Various people over the last century or so have claimed proofs of this; but I think the problem is still generally agreed to be open.

It would be interesting to look at congruence conditions on a homogenized version of your equation or Sharipov's. Maybe you would find the high degree strongly limited the number of solutions modulo smallish primes such as 17 and 23, although with four variables kicking around (in the homogenized equation) there are a lot of combinations!

Answer 2

One can obtain a similar parameterisation as follows: $$a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=a_1^2+a_2^2+a_3^2=c^2$$ in positive integers leads to rational numbers $d_k:=a_k/c$ whose squares are equal to $1-$ a rational square. This happens iff $d_k=(1-q_k^2)/(1+q_k^2)$ for some positive rational $q_k\neq 1$ for $k=1,2,3$. The only equation now is: $$d_1^2+d_2^2+d_3^2=1.$$ After getting rid of the denominators and passing back to integers, we obtain the equation:

$$s^{12}+s^8(u_1^4+u_2^4+u_3^4-4(u_1u_2)^2-4(u_1u_3)^2-4(u_2u_3)^2)- 16s^6(u_1u_2u_3)^2+s^4((u_1u_2)^4+(u_1u_3)^4+(u_2u_3)^4-4(u_1u_2u_3)^2(u_1^2+u_2^2+u_3^2))+(u_1u_2u_3)^4=0$$

to be satisfied by positive integers $s, u_1, u_2, u_3$ with a mild condition only:

$$u_k\neq s\text{ for }k=1,2,3$$

because we want nonzero solutions to the original problem.

There exists a near solution: $$a_1^2+b_1^2=a_2^2+b_2^2=a_3^2+b_3^2=c^2=a_1^2+a_2^2+a_3^2-2,$$ where $$a_1=1887,b_1=6016,a_2=2943,b_2=5576,a_3=5247,b_3=3496,c=6305.$$