###### Asked in Math and ArithmeticAlgebraGeometry

# What is the total number of cubes of any size with integral side lengths in the 5x5x5 cube?

## Answer

###### Wiki User

###### November 23, 2007 7:51PM

This is a cracking question. [I'm assuming that 'integral
**side lengths**' is intended also to mean integral
**positions**. That is to say, the position vectors of the
cubes' vertices should consist only of integer components...such
that we are effectively drawing our cube in a 3D grid. If this was
not a condition, the answer would be infinity.] So then: first,
there are 53 = 125 *1x1x1* cubes. Then 43 = 64 *2x2x2*
cubes. 33 = 27 *3x3x3* cubes. 23 = 8 *4x4x4* cubes. And,
stunningly, 13 = 1 *5x5x5* cube. So far we have 225 cubes.
That would not be interesting. But we have only counted the cubes
with edges parallel to the edges of the main cube. Suppose there
are some more cubes formed by diagonal lines? Since edge lengths
and vertex positions both have to be integral, and we are working
in 3 dimensions, we are actually looking for Pythagorean quadruples
- integer solutions to **a2 + b2 + c2 = d2** Up to now we have
only used trivial solutions like this: **52** + 02 + 02 =
**52** Our edges have only moved in one dimension. We might
consider edges that move in two dimensions, using the smallest
Pythagorean triple: **32 + 42** + 02 = **52** But diagonal
edges of length five are clearly not going to fit. So introduce the
third component, and we find this: **12 + 22 + 22 = 32** By
using diagonal edges it is possible to constuct a few more valid
*3x3x3* cubes. It turns out, I think, that four of these can
be formed. So we have 229 cubes in total.