Find all whole numbers $n > 2$ such that $ n^{\frac{n-2}{n}} $ is whole. We are supposed to find all whole numbers n > 2, such that:
$$ \exists x \in \mathbb{N}: n^{n-2} = x^{n} $$
We can modify the expression to:
$$ x = n^{\frac{n-2}{n}} $$
Or perhaps even:
$$ x = \frac{n}{n^{2/n}} $$
Intuitively it makes sense to me that the only number that makes sense as an answer is 4. How would I go about proving that x is not a whole number for all other n?
 A: Note that apart from $x=1,2,4$, we have $x^{\frac{2}{x}}\not\in\mathbb{Q}$. Therefore for $x=3$ and $x\geq 5$, we have $x^{\frac{x-2}{x}}\not\in\mathbb{Q}$. It is easy to see that $x=1,2,4$ satisfies.
A: I have gotten to the answer:
$$ \forall \ n \in \mathbb{N}-\{1,2\} : \exists\ x \in \mathbb{N} : $$
$$ n^{n-2} = x^n $$
$$ n^{\frac{n-2}{n}} = x $$
$$ \frac{n}{n^{2/n}} = x $$
This works for only 4. For 3 it is equal to approximately 1.44.
We will prove by contradiction, that for n > 4 it is irrational:
$$ \exists \ a,b \in \mathbb{N}: gcd(a,b) = 1 \ \land \  n^{2/n} = a/b $$
$$ \exists \ p \in \mathbb{P} : p|a \implies p^n|a^n $$
$$ n^{2/n} = a/b \implies b^nn^2=a^n $$
$$ p^n|a^n \land b^nn^2=a^n \implies p^n|n^2 $$
$$ (\spadesuit)\quad p^n \geq 2^n > n^2 $$
$2^n > n^2$ can be proven by induction ($n\geq5$):
$$ 2^5 > 5^2 \rightarrow 32 > 25 \ \checkmark $$
$$ \exists \ k \in (4;inf) : 2^k > k^2 $$
$$ (\star)\quad 2*2^k > 2k^2 \rightarrow 2^{k+1} > 2k^2 $$
Since k >= 4 , then:
$$ (k-1)^2\geq4^2>2 $$
$$ \begin{align}
k^2-2k+1>&\,2\\
k^2-2k-1>&\,0\\
2k^2-2k-1>&\,k^2\\
2k^2>&\,k^2+2k+1=(k+1)^2,
\end{align} $$
According to $(\star)$ and the preceeding proof:
$$ 2^{k+1} > 2k^2 \implies 2^{k+1} > (k+1)^2 \implies 2^n > n^2$$
This is a contradiction in $(\spadesuit)$ since a larger number can't divide a smaller one, so $n^{2/n}$ is irrational:
$$ (n \in \mathbb{N}-\{1,2,4\}) : n^{2/n} \notin \mathbb{Q} \implies \frac{n}{n^{2/n}} \notin \mathbb{Q} $$
$$ QED $$
