Dividing Hypercubes into $n$ smaller Hypercubes

Question

Name a positive integer $n$ nice if a square can be divided into $n$ smaller squares. The smaller squares do not need to be of the same size. Since you can always divide a square into $4$ smaller squares it immediately follows, that if $n$ is nice $n+3$ has to be aswell. Since $6, 7$ and $8$ are nice all natural numbers greater than $8$ have to be nice.

This got me thinking about the same problem in higher Dimensions. Let $n_d$ be nice if it divides a Hypercube in $d$ Dimensions into $n_d$ smaller Hypercubes.

Does for all Dimensions $d$ exist a $n_d$ such that all numbers greater than $n_d$ are nice? Is there a simple way to determine wether a number is nice in $d$ Dimensions or not?

By the same argument you get that in dimension $d$, if $n$ is nice also $n+2^d-1$ is nice. Also, the only number smaller than $2^d$ that is nice is $1$. But this doesn't help a lot, I guess. — Servaes, Dec 13 '18 at 19:33
@ajotatxe: Divide a $3 \times 3$ square into one $2 \times 2$ and five $1 \times 1$. — Robert Israel, Dec 13 '18 at 19:40
@RobertIsrael Wouldn't a $4\times4$ divided into one $3\times3$ and seven $1\times1$ squares also be a dissection into $8$ squares? — David K, Dec 13 '18 at 20:17
It is not hard to see that there exists $n_d$ such that all $n\ge n_d$ are nice. The hard problem is, finding all nice $n\lt n_d$: For what natural n does there exist a cube composed of n cubes and more. — Vepir, Jan 12 '20 at 15:56

score 5 · Accepted Answer · answered Dec 13 '18 at 19:38

5

For any $k$, you can divide a hypercube into $k^d$ equal hypercubes. Thus if $n$ is nice, so is $n + k^d-1$. Now $2^d-1$ and $(2^d-1)^d-1$ are coprime, so any sufficiently large integer can be expressed as $1 + m (2^d-1) + n ((2^d-1)^d-1)$ for some $m$ and $n$, and thus is nice.

answered Dec 13 '18 at 19:38

Robert Israel

448,999

If $p$ and $q$ are coprime, any positive integer $N\ge (p-1)(q-1)$ can be written as $N=ap+bq$ for non-negative integers $a,b$. – ajotatxe Dec 13 '18 at 19:43

Dividing Hypercubes into $n$ smaller Hypercubes

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