# can a number be written as another number to its own power

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?

• Clearly $q$ divides $x$ but $q^2$ does not, so $x$ can't be a perfect power.
– lulu
May 9, 2021 at 10:57
• 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. May 9, 2021 at 11:05
• 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 $q^2-1$.
– lulu
May 9, 2021 at 11:06
• I meant is this why it cannot be a perfect power? I do understand that $q^2$ does not divide it. May 9, 2021 at 11:07
• Say $x=y^m$ for some integers $y,m$ with $m>1$. Then $q\,|\,x\implies q\,|\,y\implies q^m\,|\,x$.
– lulu
May 9, 2021 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 $$x$$.