Is the cube the only regular solid (including regular solids of
n-dimensions) that can be combined to form a larger regular solid?

Hi, noaks-ga:

The short answer is that yes, only n-cubes have the property of
packing together to form a larger regular "solid" (more properly,
regular polytope) for dimensions greater than 2.  I'll assume it is
obvious that n-cubes have this "self-similar dissection" property in
all dimensions, focusing attention on the elimination of all other
possibilities.

I will present various arguments below to deal with all specific
cases.  As a way of organizing the material, I'll explain for each
dimension:

1.  What the regular polytopes are.
2.  Which ones can be used to "tile" the plane or "fill space".
3.  How formulas for volume, hypervolume, etc. apply to each.

Because you've asked what "can be combined to form a larger regular
solid," I've assumed that overlapping is not allowed.  Throughout I'll
exclude from consideration the "nonconvex" regular solids (and their
higher dimensional analogs), because these are self-intersecting. 
[However see my Additional Interest link at the end for some Web pages
where these nonconvex regular "solids" are discussed.]

*  *  *  *  *  *  *  *  *  *  *  *  *  *  *

Just to cover all the bases, in two dimensions there are infinitely
many convex regular polygons, one similarity class for any number
greater than 2 of equal sides and equal angles.  Which of these can be
used to dissect a larger regular polygon?

In two dimensions we can:

 - Dissect an equilateral triangle into smaller equilateral triangles.

 - Dissect a square into smaller squares.

 - Dissect a regular hexagon into equilateral triangles.

and that's all for regular dissections of regular polygons.

To prove this one might want to separate those cases (a) where a
regular polygon can be dissected into self-similar polygons from those
(b) where a regular polygon can be dissected into congruent regular
polygons dissimilar to the larger one.

In particular cases (a) must involve regular polygons that can tile
the plane.  For if one such can be dissected into smaller "copies" of
itself, then turning the construction outword we could recursively
enlarge upon the original figure until the entire plane becomes filled
with tightly packed copies of itself.

It is known that the only regular polygons which tile the plane per
case (a) are equilateral triangles, squares, and regular hexagons.  I
won't belabor the point here, but it foreshadows an argument we will
find useful in higher dimensions.

In case (b) we would direct attention to how the interior angle of the
larger regular polygon could be subdivided into a whole number
multiple of an interior angle of a smaller regular polygon.  Since
already with a square the interior angle is pi/2 or 90 degrees, no
multiple of this (greater than 1) can be accomodated by the
(non-reflexive) interior angle of another regular polygon.

Having said as much (and having much more to cover about the higher
dimensional cases), I leave further details of the dissection of
regular polygons by regular polygons to the proverbial "interested
reader".

*  *  *  *  *  *  *  *  *  *  *  *  *  *  *

In three dimensions there are five convex regular polyhedra, all
widely known since the times of the ancient Greeks as Platonic Solids
because Plato enumerates them in his Timaeus, embuing them with
mystical significance:

  - regular tetrahedron

  - cube

  - regular octahedron

  - regular dodecahedron

  - regular icosahedron

[Since we deal here only with regular polyhedra and their higher
dimensional counterparts, I'll omit "regular" in association with the
specific geometric names from now on.]

Again it is helpful to know that of these, only the cube is "space
filling", in the sense of being able to tightly pack copies of a
regular polyhedron so that no gaps in space occur:

[Space-Filling Polyhedron]
(Eric W. Weisstein, From MathWorld--A Wolfram Web Resource)
http://mathworld.wolfram.com/Space-FillingPolyhedron.html

"A space-filling polyhedron is a polyhedron which can be used to
generate a tessellation of space. Although even Aristotle himself
proclaimed in his work On the Heavens that the tetrahedron fills
space, it in fact does not (Hilbert and Cohn-Vossen 1999, p. 45). The
cube is the only Platonic solid possessing this property (Gardner
1984, pp. 183-184)."

Since the cube is space-filling but none of the other regular
polyhedra are, we immediately have that the cube cannot be dissected
into any of the others, for that would imply they also were
space-filling.  Likewise, as in our suggested argument about regular
polygons, none of the regular polyhedra except the cube can be
dissected into smaller self-similar figures, for that too would imply
they can fill space.

To advance one other sort of general argument, we require formulas for
the volumes of these regular polyhedra, which can be found here:

[The Platonic Solids -- David A. Fontaine]
http://davidf.faricy.net/polyhedra/Platonic_Solids.html

Assuming edges of unit length, we have these values:

polyhedron        volume           dihedral angle
------------ ----------------- ----------------------
tetrahedron      SQRT(2)/12         arccos(1/3)
cube                  1                pi/2
octahedron       SQRT(2)/3          arccos(-1/3)
dodecahedron  (15+7SQRT(5))/4  2arctan((1+SQRT(5))/2)
icosahedron    5(3+SQRT(5)/12  2arctan((3+SQRT(5))/2)

From these volume formulas we can deduce the impossibility of
dissection for most combinations.  If a larger regular polyhedron is
to be dissected into a whole number of equal smaller regular
polyhedra, then the edge length of the larger polyhedron must be an
integer multiple of the edge length of the smaller polyhedra.  It
follows that the respective volumes of the two regular polyhedra for
unit edge lengths must be commensurate (have a rational ratio).

But among the distinct types of regular polyhedra, only the
tetrahedron and octahedron have a ratio of volumes that is rational.

So having dealt with the cases (a) of a regular polyhedron dissected
into self-similar copies by means of the space-filling argument, we
have most of the cases (b) of a regular polyhedron dissected into
copies of a different kind disposed of by these volume formulas.  I
summarize what the results are in a table with "footnotes" for the
tetrahedron/octahedron combinations that require still a special
argument:


 \     |                     Larger polyhedron
   \   |  tetrahedron    cube    octahedron  dodecahedron icosahedron
     \ |     (1)         (2)        (3)          (4)         (5)
----------------------------------------------------------------------
       |
S      |
m  (1) |     no          no        no[A]         no          no
a      | 
l      |
l      |
e  (2) |     no          yes        no           no          no
r      |
       | 
p      |
o  (3) |    no[B]        no         no           no          no
l      |
y      |
h      |
e  (4) |     no          no         no           no          no
d      |
r      |
o      |
n  (5) |     no          no         no           no          no
       |

[A]  If one tries to fit 4 tetrahedra into the corners of the
octahedron, one finds that they leave a gap, ie. they will not fill
out the octahedron's "solid angle" at a vertex.  Concretely, the
dihedral angle (between two faces) of a regular tetrahedron is
arccos(1/3) or approximately 70.53 degrees, and the dihedral angle of
a regular octahedron is arccos(-1/3) or approximately 109.47 degrees. 
Clearly one tetrahedron is too "narrow" to fit along the edge of an
octahedron, and two tetrahedra flush face-to-face will be too "wide"
to fit there.

[B]  It is also clear from the angular data just cited that a regular
octahedron cannot fit (edge down) within the dihedral angle of the
tetrahedron; its shape is too "blunt".

This completes our analysis of the "regular solids" cases.

*  *  *  *  *  *  *  *  *  *  *  *  *  *  *

The most interesting dimension is four, where there are six regular
polytopes.  Five of are analogs of the regular three-dimensional
figures, and one is unique:

 - hypertetrahedron (or 4-simplex)

 - hypercube (or 4-cube, aka "tesseract")

 - hyperoctahedron (or 4D cross-polytope, with 16 tetrahedral faces)

 - hyperdodecahedron (or 120-cell, with 120 dodecahedral faces)

 - hypericosahedron (or 600-cell, with 600 tetrahedral faces)

 - the "24-cell" (with 24 octahedral faces)


While some familiarity with the regular solids could be expected, the
generalizations to higher dimensions are not widely taught.  For a
starting point, one might enjoy the fruits of this person's labors to
visualize the unfamiliar:

[Polytopes]
http://home.inreach.com/rtowle/Polytopes/polytope.html

[Polytopes (continued)]
http://home.inreach.com/rtowle/Polytopes/Chapter2/Polytopes2.html

"The ability of equal hypercubes to close-pack and fill an n-space is
significant. It is not at all a commonplace, for a regular polytope to
fill space. Consider the plane: we may tile it with regular triangles,
with squares, or with hexagons. By mixing different types of regular
polygon, we may obtain a few more "Archimedean" tilings; however, our
options are rather limited. There is no tiling of regular polygons
which includes a pentagon, for instance, not to speak of a heptagon
(7-gon), or a 17-gon.

"In three dimensions, we have the space-filling of cubes, and that is
that. There are a few Archimedean space-fillings by mixtures of
regular and Archimedean polyhedra. For instance, one may fill 3-space
with Platonic tetrahedra and octahedra.

"In four dimensions, surprisingly, there are three space-fillings by
regular polytopes (the hypercube, the cross polytope, and the 24-cell
may all be used to fill the 4-space). But in all higher spaces there
is only the hypercube."

*  *  *  *  *  *  *  *  *  *  *  *  *  *  *

It is especially interesting that in four dimensions the 4-cube is
_not_ the only "space-filling" regular polytope.  It must share that
property with the 4-dimensional cross-polytope (the analog of the
octahedron) and with the "24-cell", so-called because it has 24
octahedral "faces".

It nevertheless remains true that only the 4-cube can be dissected
into regular polytopes, and only by using 4-cubes to do it.

These "polychoron" (a term for the 4-dimensional "solids" proposed by
Norman Johnson) have "hypervolumes" which are nicely summarized on a
Web page at this same site linked earlier:

[Regular Convex Four-Dimensional Polytopes]
http://davidf.faricy.net/polyhedra/Polytopes.html

Note that instead of "dihedral angles" (angles between faces sharing
an edge), we now have "dichoral angles" (angles between regular
polyhedron sharing a face).

Assuming edges of unit length, we have these values:

polychoron           hypervolume          dichoral angle
--------------- ------------------- --------------------------
4-simplex           SQRT(5)/96             arccos(1/4)
4-cube                   1                    pi/2
cross-polytope          1/6                  2pi/3
"120-cell"      15(105+47SQRT(5))/4          4pi/5
"600-cell"        25(2+SQRT(5))/4   2arctan((2+SQRT(5))SQRT(3))
"24-cell"                2                   2pi/3

With three of these polychoron able to "fill space" and three unable
to do so, and with irrational ratios of the (normalized) volumes for
all combinations except among those which "fill space" (the 4-cube,
the cross-polytope (aka hyperoctahedron), and the 24-cell), we can
again summarize the "dissection" results, using "footnotes" where more
argument is needed:


 \     |                     Larger polychoron
   \   |  4-simplex  4-cube  cross-polytope "120-cell" "600-cell" "24-cell"
     \ |     (1)       (2)         (3)         (4)        (5)        (6)
----------------------------------------------------------------------------
       |
S      |
m  (1) |     no        no          no          no         no         no
a      | 
l      |
l      |
e  (2) |     no        yes        no[C]        no         no        no[D]
r      |
       | 
p      |
o  (3) |     no       no[E]       no[F]        no         no        no[G]
l      |
y      |
c      |
h  (4) |     no        no          no          no         no         no
o      |
r      |
o      |
n  (5) |     no        no          no          no         no         no
       |
       |
       |
   (6) |     no       no[H]       no[I]        no         no        no[J]
       |


Notes:  See earlier results on dissection of regular polyhedra as needed.

[C]  If the cross-polytope were dissected into 4-cubes, its tetrahedral
     hyperfaces would have to be dissected into cubes but cannot be.

[D]  If the 24-cell were dissected into 4-cubes, its octahedral
     hyperfaces would have to be dissected into cubes but cannot be.

[E]  If the 4-cube were dissected into cross-polytopes, its cubic
     hyperfaces would have to be dissected into tetrahedra but cannot be.

[F]  If the cross-polytope were self-similarly dissected, its tetrahedral
     hyperfaces would have to be self-similarly dissected but cannot be.

[G]  If the 24-cell were dissected into cross-polytopes, its octahedral
     hyperfaces would have to be dissected into tetrahedra but cannot be.

[H]  If the 4-cube were dissected into 24-cells, its cubic hyperfaces
     would have to be dissected into octahedra but cannot be.

[I]  If the cross-polytope were dissected into 24-cells, its tetrahedral
     hyperfaces would have to be dissected into octahedra but cannot be.

[J]  If the 24-cell were self-similarly dissected, its octahedral
     hyperfaces would have to be self-similarly dissected but cannot be.

This completes our analysis of the regular polychora cases.

*  *  *  *  *  *  *  *  *  *  *  *  *  *  *

All convex regular polytopes ("solids" in possibly higher dimensions)
are described here:

[Platonic Solids in All Dimensions -- John Baez]
http://math.ucr.edu/home/baez/platonic.html

However in dimension five and higher, the regular polytopes reduce to
only three families:

 - n-simplexes (generalizing the triangle/tetrahedron)

 - n-cubes 

 - n-dimensional cross-polytope (2^n "faces" which are (n-1)-simplexes)

and we recall from a previously quoted link that of these, only the
n-cube is "space filling".

For the (hyper)volume V_n of a regular n-simplex with unit length edges, see:

[Regular Polytopes]
http://www.math.rutgers.edu/~erowland/polytopes.html

  V_n = SQRT(n+1)/(2^(n/2) * n!)

I could not find a ready online reference for the (hyper)volume U_n of
a regular n-dimensional cross-polytope with unit length edges, so I
worked it out:

  U_n = (2^(n/2))/n!

Of course the (hyper)volume of the n-cube with unit length edges is 1.

We are set to polish off all the higher dimensional regular dissection cases:


 \     |                Larger polytope
   \   |  n-simplex        n-cube           cross-polytope
     \ |     (1)             (2)                 (3)
-----------------------------------------------------------
S      |
m      |
a  (1) |     no              no                 no[K]
l      |
l      |
e      |
r      |
   (2) |    no[L]            yes                no[M]
p      |
o      |
l      |
y      |
t      |
o  (3) |    no[N]            no                 no
p      |
e      |

Notes:

[K]  The 2^n "faces" of the n-dimensional cross-polytope are
(n-1)-simplexes, as are the faces of the n-simplex.  A dissection of
the larger polytope then requires either that (1) we have one
n-simplex covering each face of the cross-polytope, or else that (2)
the faces of the cross-polytope are themselves "self-similarly"
dissected into (n-1)-simplexes.  We have already ruled out the latter
possibility (for n-1 > 2, so it applies here).

Nor is it possible to cover each "face" of the cross-polytope with an
equal "face" of a different n-simplex, because the collective size of
those 2^n n-simplexes is greater than that of the cross-polytope. 
That is:

  U_n = (2^(n/2)/n! < SQRT(n+1)*(2^(n/2)/n! = (2^n)*V_n

[L]  Both n-simplex and n-dimensional cross-polytope have hyperfaces
that are (n-1)-simplexes, and the (n-1)-simplex cannot be dissected by
the (n-1)-cubes that are hyperfaces of the n-cube.

[M]  Similar to [K], we would either have to cover each (n-1)-simplex
"face" by an (n-1)-simplex "face" of a cross-polytope, or else dissect
a regular (n-1)-simplex into smaller regular (n-1)-simplexes (which is
not possible for n-1 > 2).  But the former option is impossible as
well because a single cross-polytope is already bigger than the
n-simplex we're trying to dissect:

  V_n = SQRT(n+1)/(2^(n/2) * n!) < (2^n)/(2^(n/2) * n!) = U_n

for all n > 0.

This completes our analysis of the higher dimensional regular polytope cases.

*  *  *  *  *  *  *  *  *  *  *  *  *  *  *

Additional Interest:

As promised here's a link to a discussion/visualization of some
self-intersecting "regular" polyhedra, otherwise excluded from my
analysis but mentioned in some the Web pages linked to above:

[The Kepler-Poinsot Polyhedra]
http://www.georgehart.com/virtual-polyhedra/kepler-poinsot-info.html

Such nonconvex constructions also have higher dimensional counterparts.

*  *  *  *  *  *  *  *  *  *  *  *  *  *  *

If Clarification for some part of my Answer would be helpful, please let me know.

regards, mathtalk-ga

Very complete, well cited, and accessible (which is important to me). 
It is sort of like music for the eyes.

Let me poke my nose into this. I don't know what branch of Math are we
talking about, or why is this question asked, but it invites some
comments. Here is one, based on simple thinking only.

1.) What do you mean by regular solids? In 2D, a regular polygon is
one that is equilateral and equingular. So a rhombus, which is an
equilateral parallelogram, but not equiangular, is not considered a
regular polygon. Do I take it then that, in 3D, a regular solid is one
that is of equal faces and of equal included angles/spaces at the
vertices of this polyhedron by these faces? So with convex polyhedra
of 6 faces, only the cube is considered a regular solid?
Suppose we build up a rhombus to become a prism such that all the 6
faces are squares, like a cube, but a "cube" leaning to one
side---could that be considered a regular solid? The faces are equal
but the included spaces by these faces at the 8 vertices of this
hexahedron are not equal.

2.) Also, what do you mean by "...a larger regular solid?" Would that
be of the same shape? Cubes for a larger cube? Or a set of one regular
solid for even another, but larger, regular solid of different shape?

------------
If the cube is the only regular hexahedron, and whatever is the answer
for (2.), then, yes, only the cube is the regular solid to form a
larger regular solid.

If the "rhombic" hexahedron mentioned above could be considered as a
regular solid, and only a larger "rhombic" hexadron should be formed,
then, no. The cube and this "rhombic" hexahedron would both form
larger regular solids of their same shape.

Let "solid" be the unit size of 1 cube or 1 "rhombic" hexahedron.
>>>1 solid
>>>Next larger shape, 8 solids 
>>>Next, 27 solids
>>>Next, 64 solids
...etc.

That means the sequence of the number of unit solids to make the
larger shape is 1^3, then 2^3, then 3^3, then 4^3,....etc.

----------
What about the other regular polyhedra?

a) n=4, Four-faced polyhedron, or a tetrahedron. Any number of this
regular tetrahedron cannot form a larger regular terahedron. See one
of the slanting face of the solid. Fix one the 4 vertices as the apex
or tip of the solid. The next larger regular tetrahedron would have
only one unit solid at the apex. In the next layer below this apex,
how many unit solids could be tightly fitted in? You cannot make the 3
corners of the base of this layer. So, no way.

b) n > 6. The regular polyhedra will tend to become spheres as n, or
the number of faces, increases. So, for simplicity, could you form a
larger sphere out of a set of smaller spheres with no gap
between/among the smaller spheres? No. So, no way also.

tichol-ga,
sorry, but in your first point there is a contradiction.  A skewed
six-sided figure, a cube shoved to the side, no longer has six square
faces.  Full stop!

I wanted to propose that tetrahedrons fit the bill, but I was glad
that mathtalk-ga got his answer in before I embarrassed myself.

Hello, myoarin-ga.

Oh yes, the 6 faces of the "rhombic" solid are not all squares. My
mind was keyed to a solid that could be stacked up to create a larger
same shape. That's okay. Anyway that "rhombic" solid is not really a
regular solid because the cube is the unique regular hexahedron. Let
us just say I have a playful, sometimes curious, sometimes
inquisitive, mind.

About your "...embarrassed...", nobody is perfect, myoarin-ga, so go
ahead and "experiment". If what you said is not correct or not
precise, then somebody who cares would correct you. This is a
usergroup site, remember.
That is unless you do not want to make mistakes. 
I will not say you learn by your mistakes. That might not apply to you.

So you thought about the tetrahedron but you held on. So if
mathtalk-ga did not answer the question, you will hold on till
forever? I see. But that is not me. I am not afraid to make mistakes.
I welcome other people comments on things I am interested on. I want
other people's opinions different from mine. That is why I like to
solve problems in different ways. I am not happy when a problem can be
solved in one way only.

Thank you for you comment on my comment.

Regards,
ticbol

Thanks Ticbol-ga.
Maybe I'll go back to the site that shows how to make Tetra-s and play with a few.
Those kites just seem to suggest that a larger "regular solid" can be
made with tetra-s, but maybe there is a space inside them that is not
one.
Besides, Mathtalk-ga is a H... of a lot more knowledgeable on the subject.

tichol-ga,
I made 6 tetrahedrons  - with lots of paper strips to make more - but
it doesn't work.  One can stack them neatly, point to point, making a
larger tetra, but the emply space can't be filled .... "Of course,
silly," mathtalk-ga is muttering ...
But I'm real adept at folding them now ...

myoarin-ga,

You are still at this?

I thought you accepted mathtalk-ga's answer as correct? If his answer
covered everything, why even try to see if so many tetras can do the
work?
You don't understand mathtalk-ga's answer, do you? The "massiveness"
of his answer was enough for you to accept it as correct, don't you?

I don't follow his answer. Too complicated for me. But that does not
mean he is wrong or not. I just don't understand it.

What if I say also "Of course, silly...", would you stop trying now
with your tetras?

I'd feel terrible if anyone stopped trying to understand this Q for
themselves because of my "massive" approach to an answer.  The topic
is an interesting challenge for visualization even in three
dimensions!

For one thing I simply reference other sources for the fact that
tetrahedrons don't "tile" space.  It's a tempting mistake to think
they would, one I nearly gave into when preparing the Answer myself!

It may be helpful to thing about cutting off the four corners of a
tetrahedron at the midpoints of the edges.  The four removed corners
are regular tetrahedrons at half the scale of the original.  What is
left of the original tetrahedron is a regular octahedron, with one
face for each removed corner and one from the center triangle
remaining on each original face.

This shows that one can cover space with a mix of regular tetrahedra
and octahedra (having congruent triangular faces), but the regular
octahedron cannot be dissected into regular tetrahedron.  I gave an
argument above to this effect above based on the "dihedral angles"
(angles between faces) of these polyhedra.

regards, mathtalk-ga