Positive integer solutions to $a^{a^a}=b^b$

Question

What are all positive integer solutions to $a^{a^a}=b^b$?

$(a,b)=(1,1)$ works. If we take log on both sides, we get $a^a\log a=b\log b$, which is still hard to analyze. (It helps in equations like $a^b=b^a$, where we get $a/\log a=b/\log b$.)

Answer 1

There are no other solutions.

Assume $a$ and $b$ are not both $1$. Clearly, we must then have $b > a > 1$.

Since $a$ and $b$ have powers that are equal, the exponents of the primes appearing in their prime factorizations must be proportional. This shows that $a$ and $b$ are powers of some common integer $c > 1$, say $a = c^r$ and $b = c^s$.

Substituting this into the given relation, we find $c^{rc^r - s} = s/r$. This shows that either $s/r$ or $r/s$ must be an integer, but since $s > r$, it must in fact be $s/r$. Letting $s/r = k$, we have that $b = a^k$, where $k > 1$ is an integer.

Substituting again into the original equation, we find $a^{a-k} = k$. Since $k > 1$, the exponent $a - k$ must be positive. But this has two contradictory consequences. On the one hand, $a > k$. But on the other, $k = a^{a-k} \geq a^1 = a$.

Thus the only possibility is $a = b = 1$.