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I was solving some number theory problems and couldn't solve this one. I've already proved that if $a^n|b^n$, then $a|b$ (since positive integer power of a rational number is an integer implies that the rational number is an integer). But in this case, the exponents are different so that result is not applicable. Also, since $x^x$ is an increasing function, so $n<m$ in this case, so the exponent of the denominator is lesser (if it would be higher, we could've reduced the exponent of the denominator and then the previous result would be applicable.)

Please help.

Bill Dubuque
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Martund
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    No, there is a popular counterexample $4^4 | 10^{10}$. – Sarvesh Ravichandran Iyer Jan 28 '21 at 05:10
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    Thank you, @TeresaLisbon, please post it as an answer, I'll accept. – Martund Jan 28 '21 at 05:12
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    I usually don't jump to quickly vote close a question as a duplicate when I have answered the older incarnation when I have the required dupehammer. Basically because then I want to give a window of opportunity to eventual objections. This time the duplication is so exact that I did it anyway. Also, objections are still possible, for closure is reversible! – Jyrki Lahtonen Jan 28 '21 at 05:34
  • A pretty counterexample: $8^8 \mid 12^{12}$ – Joffan Jan 28 '21 at 17:59

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Note that $$ 4^4 | 10^{10} $$

but $4 \nmid 10$. The basic reason why this doesn't work is that if we consider a prime $p$ and look at $\nu_p(k)$, the maximum power of $p$ which divides $k$, then $n^n | m^m$ implies that $n\nu_p(n) \leq m \nu_p(m)$ i.e. $\nu_p(n) \leq \frac{m}{n} \nu_p(m)$ for every prime $p$, while $n|m$ implies $\nu_p(n) \leq \nu_p(m)$ for every prime $p$. You can see that going from the weaker to the stronger statement is a jump in an inequality that could be violated for small primes.

In fact, as pointed out in a deleted answer, $n=4$ and $m=2k$ for any odd $k>4$ works, and you can try to verify this yourself. I've picked out the smallest counterexample.

EDIT : Suppose that $n= p^2$ for a prime $p$ then $n^n | m^m$ for $m>p^2$ a multiple of $p$ but not $p^2$. However, $n \nmid m$. I thank Jyrki in the comments for this example. This can be extended to $n$ being a multiple of $p^2$ and $m$ being a multiple of $\frac np$ without being a multiple of $p^2$ for any prime $p$.

Essentially, one creates contradictions in the given statement by realizing that for every prime $p$, $\nu_p(n) \leq \frac{m}{n} \nu_p(m)$ can be ensured by raising $m$ so that $\frac mn$ becomes large enough. This can be done, even if $\nu_p(m)$ is just conditioned to stay non-zero and below $\nu_p(n)$, in which case $n \nmid m$.

Sarvesh Ravichandran Iyer
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    This is all it needs. If $n$ is divisible by the square of some prime $p$, that creates the opportunity for $m$ only needs to be large enough and simply divisible by $p$. The mechanism was not immediately clear, but not too difficult to spot either, so +1. – Jyrki Lahtonen Jan 28 '21 at 05:38
  • @JyrkiLahtonen Oh yes, thank you for pointing it out. – Sarvesh Ravichandran Iyer Jan 28 '21 at 05:39