Dividing Hypercubes into $n$ smaller Hypercubes

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. – Inactive - Objecting Extremism Dec 13 '18 at 19:33
• How do you divide a square in $6$ squares? – ajotatxe Dec 13 '18 at 19:36
• @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
• $8$ is a bit harder: see here – Robert Israel Dec 13 '18 at 20:05
• @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

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.
• 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