# Does this surface contain all perfect cuboids?

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).

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It should be widely instead of wildly. Funnier this way, of course. –  Will Jagy Sep 10 '11 at 2:37

Careful: it is not true that a surface of general type always has only finitely many rational points, because it can have rational or elliptic curves that have infinitely many rational points. (Indeed this happens here with the degenerate cuboids having a "side" of length zero.) The Lang-Bombieri conjecture is that all but finitely many points are on such curves, but nobody has any idea how to prove it. Also, reducing modulo $N$ can't work, at least not by itself, because that will not distinguish trivial points (with a zero side) from nontrivial ones. –  Noam D. Elkies Aug 27 '11 at 21:57
@John Thanks. I am familiar with Sharipov's paper. btw, rational points on his surface do not necessary produce perfect cuboids, there are additional restrictions. Check the conditions in Theorem 5.3 and "...within the open domain $D_{ab}$" Brute force search on Sharipov's surface produced many nontrivial points, but none satisfying the additional conditions. –  joro Aug 28 '11 at 6:08