What is the set $\{x\in\Bbb R\mid \forall q\in\Bbb Q: q^x\in\Bbb Q\}$?

Solution 1:

It is known that if $2^c$, $3^c$, and $5^c$ are all rational then $c$ is a nonnegative integer. It follows from the Six Exponentials Theorem, q.v.

On the 1971 Putnam, it was asked to show that if $2^c,3^c,4^c,\dots$ are all integers then $c$ is a nonnegative integer.

Solution 2:

Expansion of Gerry Myerson/MJD answer.

If $x$ is rational but not integer, then $2^x$ is irrational by unique factorization. Now suppose $x$ is irrational.

Let's use the "Six Exponentials Theorem", which states:
Let $x_1, x_2, x_3, y_1, y_2$ be complex numbers, assume $x_1, x_2, x_3$ are linearly independent over $\mathbb Q$ and $y_1, y_2$ are linearly independent over $\mathbb Q$. Then at least one of $\exp(x_iy_j)$ is transcendental.

Take $x_1 = \log 2, x_2 = \log 3, x_3 = \log 5, y_1 = 1, y_2 = x$. Then $\log 2, \log 3, \log 5$ are linearly independent by unique factorization. And $1,x$ are linearly independent since $x$ is irrational. Conclusion: at least one of $$ 2, 3, 5, 2^x, 3^x, 5^x $$ is transcendental. So, in particular, they are not all rational.

Solution 3:

Note that $-1$ is rational. If $x$ satisfies the condition, in particular $(-1)^x$ is rational. Taking that to mean $e^{i\pi x}$ is rational, and thus real, I conclude that $x$ is an integer.