Is there $k\in\mathbb{R}-\mathbb{Z}$ such that $2^k\in\mathbb{N}$ and $3^k\in\mathbb{N}?$ Is there $k\in\mathbb{R}-\mathbb{Z}$ such that $2^k\in\mathbb{N}$ and $3^k\in\mathbb{N}?$ My guess is no such $k$ exists but seems hard to prove. Any ideas?
Thanks in advance
 A: I'll show in a pretty elementary way that such a $k$ could not be rational. Using some advanced machinery, it also cannot be any algebraic number. Unfortunately, I don't know how to proceed with the transcendental case.
So if both $2^k$ and $3^k$ are integers, then the quotient $\left(\frac{2}{3}\right)^{k}$ is rational. Let's first suppose that $k = p/q$ (relatively prime) is a rational, but not an integer. Then we're essentially asking when a rational to a rational is itself rational. This is answered in part in this post. Essentially, only the obvious situation can occur - we need $2$ and $3$ to be $q^{th}$ powers in order for $\left(\frac{2}{3}\right)^{p/q}$ to be rational. This comes down to prime factorization of rationals. But since $k \notin \mathbb N$, $q \neq 1$ so $2$ and $3$ are not $q^{th}$ powers.
Now here's the high powered machinery we need: the Gelfond-Schneider theorem. This is very nontrivial - it answers Hilbert's 17th problem! Indeed, it states that if $a, b$ are algebraic numbers, $a \neq 0, 1$, and $b$ irrational, then $a^b$ is transcendental. Taking $a = 2/3$ and $b = k$ says that $k$ cannot be an irrational algebraic number, and we showed above that it cannot be rational, so if such a $k$ existed it would have to be transcendental.
A: Note that $3^k = (2^{\log_2 3})^k = (2^k)^{\log_2 3}$. Therefore, the question is essentially of finding a natural number $a$ such that $a^{\log_2 3}$ is also a natural number : in that case, take $\log_2 a = k$. Note that powers of $2$ are prohibited.

Now, suppose that $a^{\log_2 3} = b$. Then, using the usual log laws, $\log_a b = \log_2 3$, and using $\ln$ we get $\frac{\ln a}{\ln 2} = \frac{\ln b}{\ln 3}$.
This kind of relation between logs can be described as a homogenous quadratic relation. These problems come under the umbrella of the...

Four exponentials conjecture
The four exponentials conjecture(it has not been proved yet) is stated (all notation mine, furthermore I restrict to reals) as:

Let $p,q,r,s$ be positive real numbers. Suppose that $\frac{\ln p}{\ln q}, \frac{\ln r}{\ln s} \notin \mathbb Q$. Then, we have :
$$
\frac{\ln p}{\ln q}\neq \frac{\ln r}{\ln s}
$$

Note how neatly this matches up with our problem.
To prove our result with the conjecture , start with $a,b,2,3$ , which are natural hence algebraic numbers . If $a$ is not a power of $2$, then $\frac{\log a}{\log 2}$ can be shown to be irrational in an elementary manner (Exercise, but for a solution see the other answer which covers this well). Now, if $a$ is not a power of $2$, then $b$ cannot be a power of $3$ (why?) so $\frac{\ln b}{\ln 3}$ is also irrational in a similar manner.
By the conjecture, these are different numbers, so the answer to your question is no.

Note that I may have used a very strong result to prove something that seems very innocuous. However, anything at the surface comes down to the relationship between the logarithms. For example, the Lindemann-Weierstrass theorem and Gelfond-Schneider cover linear relations. The most general result by Baker provides a generalization, along with proofs for a six-exponentials theorem, but I don't think that suffices to work for this problem, unfortunately.
