# 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?

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.

• Yes but how would you go about proving that $n^{2/n}$ is irrational? Oct 17, 2021 at 10:33
• @PavolKomlos It suffices to prove that they are not integral. This can be done by an easy estimation. Oct 17, 2021 at 11:01
• @Trebor by 'they are not integral' do you mean the function $f(x) = x^{2/n}$ (does not have a closed form integral) or that $n^{2/n}$ for all n other that 1,2,4 is a non integral rational number? Oct 17, 2021 at 11:16
• Integral means "is an integer" in this case. Naturally --- that's the adjective of "integer". Oct 17, 2021 at 13:59
• @Trebor could you please explain how that suffices? I don't quite get that. Proving that $n^{2/n}$ is not an integer is easy, but how does it prove that $\frac{n}{n^{2/n}}$ is not an integer? Oct 17, 2021 at 18:23

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:

$$(n \in \mathbb{N}-\{1,2,4\}) : n^{2/n} \notin \mathbb{Q} \implies \frac{n}{n^{2/n}} \notin \mathbb{Q}$$ $$QED$$