Let $x$ be a positive real number such that $n^x\in\mathbb{N}$ for each $n\in\mathbb{N}.$ Show that $x\in\mathbb{N}.$
I saw this here: https://mathoverflow.net/questions/17560/if-2x-and-3x-are-integers-must-x-be-as-well. OP says that it is a "fun little puzzle", so I thought I'd try it, but haven't gotten far.
Firstly, $n^x\in\mathbb{N}\text{ }\forall n\in\mathbb{N}\iff p^x\in\mathbb{N}\text{ }\forall p\in\mathbb{P},$ where $\mathbb{P}$ is the set of all primes.
So, we might as well consider just the primes. But, now what? I think I've handled the case where $x$ is rational.
Suppose it is given that $x\in\mathbb{Q}.$ Put $x=\frac{p}{q}$ for co-prime $p,q\in\mathbb{N}.$ We wish to show $q=1.$
Now, $2^{\frac{p}{q}}=a,$ for some $a\in\mathbb{N}.$ This gives $2^p=a^q.$ This implies that $a=2^k,$ for some $k\in\mathbb{N}.$ This gives $2^p=2^{kq}.$ Hence, $p=kq.$ Now, $\gcd(p,q)=1=\gcd(kq,q).$ Since $q≤\gcd(kq,q)=1,$ we get $q=1.$ Well, as soon as we get $p=kq,$ we get $\frac{p}{q}=k,$ and we are done. However, I show $q=1$ for completeness. Anyway, $x$ is a natural number. Is this right?
What do we do when $x$ is a positive irrational? Obviously in this case, we need to use all primes (or atleast $2$ primes), not just $2.$ This is because of existence of things like $2^{\log_2(3)}=3.$
PS: I don't know what tag to put for this, but number theory seems appropriate.
