Do we know which numbers can be written as another number to its own power? More precisely, if we have a number $x$, when can we write it as $n^n$ for some $n\in \mathbb{N}$?

For example, if $q$ is a prime at least $5$ and $x= (q(q^2-1))^{q^2}$, can we write $x=n^n$ for some $n\in \mathbb{N}$?

Intuition says that writing $n^n$ in its prime factorization, we have each of those primes divides the power by definition. Whereas with $x$, $q$ divides $q^2$ but $q^2-1$ does not divide it, does any of this make sense?

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    $\begingroup$ Clearly $q$ divides $x$ but $q^2$ does not, so $x$ can't be a perfect power. $\endgroup$ – lulu May 9 at 10:57
  • $\begingroup$ Ohhh right, if $q^2$ divides $x$, we could write it as $q^k$ for some $k$ right? Without even considering $q$ and $k$ must be equal here. $\endgroup$ – mandella May 9 at 11:05
  • $\begingroup$ Not sure I follow. I know $q$ divides $x$ because you wrote it as an explicit factor. I know $q^2$ does not divide $x$ because, clearly, $q$ does not divide $q^2-1$. $\endgroup$ – lulu May 9 at 11:06
  • $\begingroup$ I meant is this why it cannot be a perfect power? I do understand that $q^2$ does not divide it. $\endgroup$ – mandella May 9 at 11:07
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    $\begingroup$ Say $x=y^m$ for some integers $y,m$ with $m>1$. Then $q\,|\,x\implies q\,|\,y\implies q^m\,|\,x$. $\endgroup$ – lulu May 9 at 11:09

We want to argue that the given $x$ can not be of the form $n^n$.

For any prime $p$ and a natural number $m$, let $v_p(m)$ denote the order to which $p$ divides $m$. Thus $v_2(24)=3$ since $2^3\,|\,24$ but $2^4\,\nmid \,24$. for example.

Suppose that $q$ is a prime and that $$x=\left(q(q^2-1)\right)^{q^2}=n^n$$ for some natural number $n$. We will derive a contradiction.

Since $q\,\nmid\,(q^2-1)$ we easily see that $v_q(x)=q^2$. It is also easy to see that $v_q(n^n)=v_q(n)\times n$.

Let $v_q(n)=a$. Then $n=q^am$ with $q\,\nmid\,m$. And we see that $$q^2=v_q(x)=v_q(n^n)=a\times q^a\times m$$

But this is impossible. It would imply that $m=1$ since otherwise we'd have $m\,|\,q^2$ which is impossible for $m>1$ (as $m$ is prime to $q$ by construction.). But if $m=1$ then $n=q^a$ which is impossible since $\gcd(q^2-1,n)>1$. And we are done.


By your definition, the list of numbers for which that can be done are $$1^1, 2^2, 3^3, 4^4, 5^5, 6^6...$$

If you broaden it to include non integers, the answer becomes a little more interesting, as the graph of $x^x$ has a minimum at $(1/e, (1/e)^{1/e}) \approx (0.368,0.692)$, so any number $y$ higher than about $0.692$ can be written as $y = x^x$ for some $x$.


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