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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?

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  • $\begingroup$ 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. $\endgroup$ – Inactive - Objecting Extremism Dec 13 '18 at 19:33
  • $\begingroup$ How do you divide a square in $6$ squares? $\endgroup$ – ajotatxe Dec 13 '18 at 19:36
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    $\begingroup$ @ajotatxe: Divide a $3 \times 3$ square into one $2 \times 2$ and five $1 \times 1$. $\endgroup$ – Robert Israel Dec 13 '18 at 19:40
  • $\begingroup$ $8$ is a bit harder: see here $\endgroup$ – Robert Israel Dec 13 '18 at 20:05
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    $\begingroup$ @RobertIsrael Wouldn't a $4\times4$ divided into one $3\times3$ and seven $1\times1$ squares also be a dissection into $8$ squares? $\endgroup$ – David K Dec 13 '18 at 20:17
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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.

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  • $\begingroup$ 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 $a,b$. $\endgroup$ – ajotatxe Dec 13 '18 at 19:43

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