Find all reals $a, b$ for which $a^b$ is also real The title is pretty much clear, but here is a more precise formulation:
Find all pairs $(a,b)\in\mathbb{R^2}$ for which $a^b$ is also real.
I used a CAS to solve the problem and it says that the solution is
$$(a=0\land b>0)\lor
   \left(c_1\in
   \mathbb{Z}\land a<0\land
   b=c_1\right)\lor a>0$$
But i think the correct answer is
$$(a=0\land b>0)\lor
   \left(c_1\in
   \mathbb{Q}\land \nu_2(c_1)\ge0\land a<0\land
   b=c_1\right)\lor a>0$$
where $\nu_p$ is the $p$-adic valuation because the computer says, for example, that $(-1)^{\frac{1}{3}}=\cos(60)+\sin(60)i$.
 A: There is only a commonly agreed definition of $a^b$ in the following cases:


*

*When $a>0$, then $a^b$ is defined to be $e^{b \ln a}$ for all $b\in\mathbb{R}$.

*When $a=0$, we typically define $a^b$ to be zero for all $b>0$.

*When $a<0$, then $a^b$ is only well-defined when $b$ is an integer.
If $a<0$ and $b$ is not an integer, there are sometimes several competing definitions for $a^b$.  For example, some computer algebra systems define $(-1)^{1/3}$ to be $-1$, while others define it to be the principal cube root $e^{\pi i/3} = \cos(\pi/3)+i\sin(\pi/3)$.
If $n$ is an odd integer, it is reasonable to define $a^{1/n} = - |a|^{1/n}$ when $a<0$, and this can be extended to defining $a^{m/n} = (a^{1/n})^m$ for any integer $m$.  However, this definition does not obey the law $(a^b)^c = a^{bc}$.  For example, if we define $(-1)^{1/3} = -1$, then $\left((-1)^2\right)^{1/6}\ne (-1)^{1/3}$.  Indeed, I have seen computer programs that return different graphs for $y = x^{2/6}$ and $y = x^{1/3}$.
