Thoughts about perfect cuboids.
Alireza Nejati
Consider the vector (a,b,c) where a,b, and c are integers. If
g=sqrt(a^2+b^2 +c^2) is integer then (a,b,c,g) is called a pythagorean
quadruple, or PQ. It's called a primitive pythagorean quadruple (PPQ)
if gcd(a,b,c)=1. Similarly, if f=sqrt(a^2 + b^2) is an integer then
(a,b,f) is called a pythagorean triple (PT).
Now consider the following transformations:
(a,b,c,g) --> (b,a,c,g)
(a,b,c,g) --> (a,c,b,g)
(a,b,c,g) --> (-a,b,c,g)
(a,b,c,g) --> (a+b+g,a+c+g,b+c+g,a+b+c+2g)
All 4 of these transformations preserve the property of being a PQ.
For the first 3 it's obvious, it's just the fourth that needs some
proof (and the proof isn't that hard, just some basic algebra). It can
also be shown that for *any* two PQs where the gcd of their first 3
elements is equal, there exists a transformation that maps them to
each other, and that transformation is either one of these 4
transformations, their inverses, or some combination of them (from
this article: http://webbox.lafayette.edu/~reiterc/nt/ppllpd/ppllpdpp.pdf).
We have the same situation for PTs (except there are 3 basic
transformations instead of 4; I'm not going to write them down but you
can search for Berggren trees).
Now, WLOG, assume gcd(a,b,c)=1. So, if (a,b,c,g) is a PPQ then there
must be some map that transforms it to (0,0,1,1), since that is also a
PPQ. Similarly, if (a,b,sqrt(a^2 + b^2)) is a PT there must be a
transformation (using the 3 ones for PTs) that maps it to (0,k,k) i.e.
a=0 and b=k, (and another which maps it to (k,0,k) i.e. a=k and b=0)
where k=gcd(a,b). A similar argument holds for (a,c,sqrt(a^2 + c^2))
and (b,c,sqrt(b^2 + c^2)). Now collect the a,b, and c components (i.e.
discard g,f, etc.) of the 'target' vectors in a list, and let's see
what we can observe:
(all the below vectors are (a,b,c))
(0,0,1) or (0,1,0) or (1,0,0)
(0,k,0) or (k,0,0)
(j,0,0) or (0,0,j)
(0,0,l) or (0,l,0)
Where j=gcd(a,c) and l = gcd(b,c).
Now note that any 3 of these can be chosen to be linearly dependent
from each other; however, once the 4th enters the picture, it must be
a linear combination of the others.
I think this somehow relates to the fact that, so far, we have found
euler bricks with 3 of the four diagonals integer, but we have never
found any with all four diagonals integer. I am very confident that by
studying this structure the problem could be chipped at. Discuss.
Bart Goddard
9295-6f0...@o18g2000prh.googlegroups.com:
> I think this somehow relates to the fact that, so far, we have found
> euler bricks with 3 of the four diagonals integer, but we have never
> found any with all four diagonals integer. I am very confident that by
> studying this structure the problem could be chipped at. Discuss.
You misspelled "disgust" as "discuss" and "they" as "we".
--
Cheerfully resisting change since 1959.
rasterspace
I haven't seen it. actually, there is no reason
to suppose that a perfect cuboid exists, although
it's easy to make perfect tetrahedra (any
of the seven combinations .-)
Alireza Nejati
It's also possible to construct perfect parallelopipeds, even if you
restrict it to just one obtuse angle.
If the 'hunch' I outlined in the original post is true no perfect
cuboid exists. I don't really think this is such a hard problem; it's
just that it's received very little attention. In the past 30 years or
so there have only been 3 papers (that I could find) that talked about
it at all, and all 3 only analyzed it using very basic mathematics.
rasterspace
good insight, that the parallepiped need only
be skew in ... two facets?
Alireza Nejati
I don't understand what you mean.
rasterspace
It's also possible to construct perfect parallelopipeds; even if you
restrict it to just one obtuse angle?
that'd be a corallary to the fact that
any six of the rectangular parallelopiped measures can be integer;
eh?
Gerry Myerson
<2d855258-37ed-4fde...@r19g2000prm.googlegroups.com>,
Alireza Nejati <alire...@gmail.com> wrote:
Have you seen Ronald van Luijk's undergraduate thesis,
On Perfect Cuboids? It uses some pretty advanced stuff.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Alireza Nejati
wrote:
> In article
> <2d855258-37ed-4fde-9a81-3b8c0e745...@r19g2000prm.googlegroups.com>,
Thanks, I appreciate the pointer.
rasterspace
I guess they can have six different facial diagonals ... and
four different body diagonals?
OwlHoot
> Some background:http://en.wikipedia.org/wiki/Euler_brick
>
> Consider the vector (a,b,c) where a,b, and c are integers. If
>
I can't help feeling a bit skeptical that just studying GCDs
of the sides will give enough information to find or exclude
perfect cuboid solutions.
Also, this problem has a nasty trait that if you "simplify"
it by (as it seems you're trying by throwing parallelepipeds
into the pot) equal complexities will pop up somewhere else!
Every now and then I have a crack at the "two face diagonals
and body diagonal all rational" case, with all sides rational
of course, which can be expressed as:
1 + x^2 = p^2
1 + y^2 = q^2
1 + x^2 + y^2 = r^2
If this was a rational variety, there'd be a two-parameter
solution, with x, y, p, q, r each a rational function of
two parameters, say u, v.
But I recently found that this is birationally equivalent
to the system:
k^2 + K^2 = 1
c^2 + s^2 = 1
d^2 + k^2 s^2 = 1
which of course for any given "modulus" k are the equations
defining the Jacobi Elliptic functions.
For that reason it seems highly unlikely that the variety
is rational. But that is only obvious for any given k, i.e.
a "slice" through the surface, and I suppose there's a small
chance it could be rational if k, s, c, d are all variables
on the same footing.
Anyway, following is a copy and paste sketch of how to
go from the diagonal conditions above to the Elliptic
function set:
Starting with a third system:
1 + x^2 = u^2 [1.1]
1 + y^2 = v^2 [1.2]
1 + z^2 = u^2 v^2 [1.3]
[1.1] * [1.2] with [1.3] =>
x^2 y^2 + x^2 + y^2 = z^2
2 p T
<=> x y = -------------
p^2 + q^2 + 1
2 q T
x = -------------
p^2 + q^2 + 1
(p^2 + q^2 - 1) T
y = -----------------
p^2 + q^2 + 1
p (p^2 + q^2 + 1)
=> T = -----------------
q (p^2 + q^2 - 1)
2 p p
=> x, y = -------------, -
p^2 + q^2 - 1 q
In [1.1], [1.2] these give resp:
4 p^2 + (p^2 + q^2 - 1)^2 = U^2 [2.1]
p^2 + q^2 = V^2 [2.2]
2 L V (L^2 - 1) V
[2.2] <=> p, q = -------, -----------
L^2 + 1 L^2 + 1
which in [2.1] =>
2 L V^2 - 1
(-------)^2 + (-------)^2 = <sq>
L^2 + 1 2 V
<=> x^2 + y^2 = 1
1 + z^2 = t^2
x^2 + z^2 = w^2
<=> x^2 + y^2 = 1
x^2 + z^2 = w^2
x^2 + y^2 + z^2 = t^2
But:
1 + x^2 = u^2 [1.1]
1 + y^2 = v^2 [1.2]
1 + z^2 = u^2 v^2 [1.3]
are equivalent to:
(1/u)^2 + (x/u)^2 = 1
(y/v)^2 + (1/v)^2 = 1
(z / uv)^2 + (1/u)^2 (1/v)^2 = 1
So we can take:
k := 1/u (so that 0 < |k| < 1)
1/v = sn(k, t)
y/v = cn(k, t)
z / uv = dn(k, t)
<=>
1
u = -
k
1
v = --------
sn(k, t)
k'
x = -
k
cn(k, t)
y = --------
sn(k, t)
k
z = --------
sn(k, t)
where:
k^2 + k'^2 = 1
Cheers
John R Ramsden
P.S. If nothing else, this shows how from any two solutions
to the rational box problem one can leapfrog to a third by
using the addition theorems for Jacobian Elliptic functions.
OwlHoot
>
> [..]
>
> which in [2.1] =>
> 2 L V^2 - 1
> (-------)^2 + (-------)^2 = <sq>
> L^2 + 1 2 V
>
> <=> x^2 + y^2 = 1
>
> 1 + z^2 = t^2
>
> x^2 + z^2 = w^2
I should have made it clear that the x, y, z in the latter
three are "new" and not the same as the starting ones (I did
say it was a sketch!)
rasterspace
Gerry Myerson
<ec5f2daa-7b72-43a8...@l22g2000pre.googlegroups.com>,
rasterspace <Spac...@hotmail.com> wrote:
> so'd any one see the proof of no PCs?
I can't find it anywhere, but I'm sure we had a discussion
here a few years ago where there was a news item about
a Russian student, maybe even a high school student,
entering a non-existence proof in some competition, but
then all efforts to track down the actual alleged proof
ended without success. In short, I don't think anyone
has seen the alleged proof, and I don't think there is
any proof to be seen.
Gerry Myerson
<ec5f2daa-7b72-43a8...@l22g2000pre.googlegroups.com>,
rasterspace <Spac...@hotmail.com> wrote:
> so'd any one see the proof of no PCs?
Found the discussion I was thinking of.
Started 8 November 2006 by Hauke Reddmann
with the Subject: Parametric solutions of Euler Brick?
Christopher Heckman (writing as Proginoskes) mentioned
the alleged proof, and much discussion ensued.
All available to you via the magic of Google.
OwlHoot
I've checked my algebra, and it seems OK. So in summary:
For each set of rational numbers x, y, z, u, v such that:
x^2 + y^2 = 1
x^2 + z^2 = u^2
x^2 + y^2 + z^2 = v^2
( what Tito Piezas calls a face cuboid on his page at
http://sites.google.com/site/tpiezas/0020 )
there exist rational numbers k, K, s, c, d such that:
k^2 + K^2 = 1
c^2 + s^2 = 1
d^2 + k^2 s^2 = 1
and conversely.
The explicit relations aren't nearly as bad as I feared:
x = c
y = s
z = K^2 - k^2(k - c)^2 / (2 k K (k - c))
u = (3 c - k) / K
v = K^2 + k^2(k - c)^2 / (2 k K (k - c))
So what happens if these are plugged into the remaining
face diagonal condition, I hear you ask - Well let's try:
Expressing this condition as y^2 + z^2 = w^2 we obtain,
after using c^2 + s^2 = 1 to simplify, and clearing
denominators:
K^4 + k^4(k - c)^4 = (2 k K c (k - c))^2 + W^2
I have a proof that this, combined with k^2 + K^2 = 1,
has a general two-parameter solution, which I'll post
after checking. (It is very intricate!)
So that should narrow down the choices when searching
for perfect cuboids (all faces an diagonals rational).
Cheers
John R Ramsden (jhnr...@yahooo.co.uk) remove o
rasterspace
I'm sure that I can prove it, using parity,
a la Fermat's results on right trigona.
:Scoll ye God-am palimpsesT:
"writing numbers backwards" was the problem
with Hensel's lemma for p-adic numbers;
there is really no sufficient reason to do that.
also, see Laurent sreies.
:Scoll ye God-am palimpsesT:
there is nothing weird about space being "curved,"
if interepreted in terms of refraction. however,
Minkowski's useless spacetime slogan seems
to have been used by the British Pyshcological Research Soc.
to create a popular misconception, known as Flatland.
and Minkowski was a great "N-d" geometer,
viz his generalization of Pick's theorem.
:Scoll ye God-am palimpsesT:
to say that Newton's corpuscle was
in any way a classical theory, belies the fact that
it was not a theory, at all, considering that
he got refraction completely wtong. which is hard to do,
if he really answered Liebniz's challenge
on the brachostochrone -- supposedly *the* classical problem
to initiate the caclulus by Bernoulli et al.
OwlHoot
>
> "writing numbers backwards" was the problem
> with Hensel's lemma for p-adic numbers;
Is that you, Nico? ;-)
I forgot to mention that the system:
k^2 + K^2 = 1
c^2 + s^2 = 1
d^2 + k^2 s^2 = 1
has an interesting "duality", in that one can
flip the roles of k, K, s, c to s', c', k', K'
So if you treat all terms as variables on the
same footing, there's a kind of two-way
intertwined leapfrogging process.
Oh and there was a slip in the two-parameter
solution I claimed to have found at the end
but which needed checking - A fourth power
accidently became a square at one point -
It's so easy to do.
Cheers
John R Ramsden