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The surface $$ (x^{2} + y^{2} - 2 y + 1) \cdot (x^{2} + y^{2} + 2 y + 1) \cdot (x^{2} + y^{2} - 2 x + 1) \cdot (x^{2} + y^{2} + 2 x + 1) - z^2 = 0$$

has the points $\left(\dfrac{t}{s},\dfrac{t}{s},\dfrac{s^{4} + 4 t^{4}}{s^{4}}\right)$, $ \left(\dfrac{- s^{2} + t^{2}}{s^{2} + t^{2}},\dfrac{2 s t}{s^{2} + t^{2}} ,\dfrac{8 s^{3} t - 8 s t^{3}}{s^{4} + 2 s^{2} t^{2} + t^{4}}\right) $ (up to sign and swapping $x,y$)

Can the rest of the points be parameterized (or at least some of them)? (such points exist).

What software can do this?

The surface is related to Euler bricks.

EDIT

Some empirical observation about the points of interest. Of the 280 points found 246 have denominator of x that is sum of two squares, 250 have denominator of y that is sum of two squares and 252 have denominator of z that is a square.

Explaining the relation with Euler bricks per John's request (almost sure this have been done before).

So I naïvely wasted my time with perfect Euler bricks. $$\begin{align*}a^2+b^2&=s_1^2\qquad \qquad \rm(1)\\ a^2+c^2&=s_2^2\qquad \qquad \rm(2)\\ b^2+c^2&=s_3^2\qquad \qquad \rm(3)\\ a^2+b^2+c^2&=s_4^2\qquad \qquad \rm(4)\end{align*}$$

(4) can be parameterized by $(a,b,c) = (2s,2t,1-s^2-t^2)$ To eliminate (1) further parameterize (1) with $(s,t) = (x^2-y^2,2xy)$. Substituting leads to $a=2 x^{2} - 2 y^{2}, b=4 x y ,c= -x^{4} - 2 x^{2} y^{2} - y^{4} + 1$ and leaves only equations (2) and (3). (3) is the surface in question with $z=s_3$. The known parameterization make one of $a,b,c$ $0$.

Without the restricting factor of $2$ in $2s,2t$ the surface becomes

$$ (x^{2} + y^{2} - 2 x - 2 y + 2) \cdot (x^{2} + y^{2} - 2 x + 2 y + 2) \cdot (x^{2} + y^{2} + 2 x - 2 y + 2) \cdot (x^{2} + y^{2} + 2 x + 2 y + 2) - z^2=0$$

with known solutions $(t,t)$ and $x^2+y^2=2$ (up to sign)

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There's no surface there -- did you forget "$=0$"? –  joriki Aug 12 '11 at 21:21
    
@joriki yes, I forgot =0. Thank you. Fixed now. –  jorox Aug 13 '11 at 7:07
    
Have you tried switching to cylindrical coordinates, and then undoing the trigonometric portions with the Weierstrass substitution? –  J. M. Aug 13 '11 at 7:12
    
Something like this: do a conversion to either cylindrical or spherical coordinates, and let $\cos\,\theta=\frac{1-u^2}{1+u^2}, \quad \sin\,\theta=\frac{2u}{1+u^2}$... –  J. M. Aug 13 '11 at 8:25
    
@J.M Thank you very much! I will try this. –  jorox Aug 13 '11 at 8:51
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1 Answer

up vote 1 down vote accepted

I may have some results on this in due course. So bookmark this page and check back in a week or two.

But meanwhile, I'm curious to know how this relates to Euler bricks, unless you'd rather keep that under your hat.

Your equation is clearly equivalent over $\mathbb{Q}$ to the pair:

$ p^2 + q^2 + r^2 = 1 $

$ (1 - q^2) (1 - r^2) = t^2 $

and also (with some more work) to :

$ 1 + X^2 = Y^2 = (1 + Z^2) (1 + T^2) $

which mocks one by its apparent simplicity, in particular the seeming flexibility in choosing $Z$ and $T$, but which in truth is as intractable as the previous forms (and many others) !

These, especially the first, suggest you are investigating the "two face diagonals and body diagonal" case, or perhaps a slightly weaker verstion of that with the product of two face diagonals an integer. Not a bad idea.

Aside from one-parameter solutions, such as the ones you gave (in homogenous form), I think I've shown that this equation has a set of extra rational points derived from the Weierstrass curve $ X^3 - 13 X - 18 = Y^2 $, which has Mordell-Weil rank 1 with torsion point $X, Y = -2, 0$ and generator:

$X, Y = \dfrac{17}{4}, \dfrac{15}{8}$

But the proof is quite intricate, and needs careful checking, and also of course typing here! So it probably won't appear (if at all) before next week. I'll post again whether or not it holds up.

I am fairly certain there is no general, two-parameter solution. In other words this is not a rational surface. It's most likely what is called a K3 surface.

EDIT a week later:

On checking I found there was an error in what led to that Weierstrass equation. Why is it that every damned error in dealing with this brute of a problem seems to yield an intriguing or even definitive result?! It's not wishful thinking either, because one can rarely see the solution at the point where the error is made.

jojo, if you ping me your email, to jhnrmsdn at yahoo dot co dot uk, I can email you some more results and see if I can get you on the private Euler Brick Yahoo group (no promises, because I don't own the group).

Meanwhile here is an observation relating to your result. With suitable scaling, the two face diagonal and body diagonal case is equivalent to the set:

$x^2 + u^2 = 1$

$y^2 + v^2 = 1$

$x^2 + y^2 + z^2 = 1$

by the simple observation that the first two with the third imply respectively :

$u^2 = 1 - x^2 = y^2 + z^2$

$v^2 = 1 - y^2 = x^2 + z^2$

So plugging $x$ and $y$ of the following general parametrization of the third:

$x = \dfrac{2 a}{a^2 + b^2 + 1}$

$y = \dfrac{2 b}{a^2 + b^2 + 1}$

$z = \dfrac{a^2 + b^2 -1}{a^2 + b^2 + 1}$

into the first two and clearing denominators gives :

$(a^2 + b^2 + 1)^2 - 4 a^2 = U^2$

$(a^2 + b^2 + 1)^2 - 4 b^2 = V^2$

which is a stronger version of your result, split into two factors which must each be square rather than simply their product being a square.

Of course that isn't to say the corresponding factors in your result, which is derived in a different way, need be square. But it illustrates a characteristic feature of this problem - It's like a vast algebraic hall of mirrors, where similar forms show up, which may be weaker or stronger or even identical (the latter amounting to addition formulae which allow one to leapfrog between solutions a la chord-tangent process).

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Thank you! I am investigating exactly "two face diagonals and body diagonal" (almost sure done before) and will explain the question soon (with some empirical results about the points). Btw, do you have bounds of the degrees of the possible parameterizations? –  jorox Aug 14 '11 at 9:51
    
edited the question. Had a mistake with a factor of 2, gave another similar surface without it. –  jorox Aug 14 '11 at 13:28
    
Thanks for your explanation. But I'm not sure how you get your surface equation from it, or even which is the right equation to be using now! You're correct that $a = 2(v^2 - u^2)$, $b = 4uv, $c = 1 - (v^2 - u^2)^2 - (2 u v)^2$ in your notation satisfies your (1) and (4); but how does plugging these into your (3) lead to either surface equation? Maybe I'm being thick, and it just needs some extra manipulation. –  John R Ramsden Aug 14 '11 at 14:01
    
Oh and by the way, welcome to the Euler Brick addicts' club ;-) –  John R Ramsden Aug 14 '11 at 14:04
    
My mess needs editing. First $(a,b,c) = (2s,2t,1-s^2-t^2)$. Then $(s,t) = (v^2-u^2,2uv)$. So $(a,b,c)=(f(u,v),g(u,v),h(u,v))$. (3) is $F(u,v)$ and $u,v$ correspond to $x,y$ in the surface. The second surface is after removing the restricting factor of $2$. Thank you again. –  jorox Aug 14 '11 at 14:37
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