exponentiation - Is there an irrational number $a$ such that $a^a$ is rational?

It can be proved that there are two irrational numbers $a$ and $b$ such that $a^b$ is rational (see Can an irrational number raised to an irrational power be rational?) and that for each irrational number $c$ there exists another irrational number $d$ such that $c^d$ is rational (see For each irrational number b, does there exist an irrational number a such that a^b is rational?).

My question is: Is there an irrational number $a$ such that $a^a$ is rational (and how could you prove that)?

Answer

Consider the unique (positive) solution $a$ to $x^x = 2$. If $a$ were rational, say, $a = \frac{p}{q}$, $p$ and $q$ are positive integers such that $\gcd(p, q) = 1$, we would have

$$\left(\frac{p}{q}\right)^{p / q} = 2 ,$$ and rearranging gives
$$p^p = q^p 2^q .$$
Since there is no integer $n$ such that $n^n = 2$, we must have $q > 1$ and hence $2 \mid p^p$. Because $2$ is prime, we have $2 \mid p$. So, $2$ occurs an even number of times the prime factorization of $p^p$ and likewise of $q^p$. Since $p^p = q^p 2^q$, we must have $2 \mid q$, but now $2 \mid p, q$, and this contradicts $\gcd(p, q) = 1$. Thus, $a$ is irrational but $a^a$ is rational (in fact, an integer).