Are there any pairs of numbers (a,b) such that $a^a=b$ and $b^b=a$?

Question

There is the trivial solution if we allow a=b=1. Are there any nontrivial solutions where $a\ne b$? The question is a "challenge" question from Tanton's Arithmetic and I haven't been able to solve it.

Answer 1

We have $$\left(a^a\right)^{a^a}=a,$$ which for $a>0$ and $a\neq1$ gives $$a^{a^{a+1}}=a$$ or $$a^{a+1}=1$$ or $$a+1=0,$$ which is impossible.

Answer 2

$$a^{ab}=(a^a)^b=b^b=a.$$ So either $a=1$ or $ab=1$. The second condition gives us $a^a=\frac{1}{a}$ implying $a=1$ or $a=-1$. Thus $(1,1)$ or $(-1,-1)$ are the only solutions.

Answer 3

$$ a ln(a) = ln(b)$$ $$ b ln(b) = ln(a)$$ substituting $ln(a) = \frac{ln(b)}{a}:

$$ b ln(b) = ln(a) = \frac{ln(b)}{a}$$ $$b = \frac{1}{a}$$

WLOG, we can assume $a \ge 1$, and $b \le 1$. However, $a^a = b$ doesn't make sense in the case where $a>1$. The reason why this $b = \frac{1}{a}$ despite none of the other values working is because we are multiplying by $ln(a) = 0$, which allows there to be extra solutions than the initial problem had.

Answer 4

The second equation leads to $b = a^{1 \over b}$, so since $b = a^a$ we have $$a^{1 \over b} = a^a$$ Thus $a = {1 \over b}$. Then $b^b = a$ leads to $b^b = b^{-1}$, which in turn implies $b^{b+1} = 1$. Thus either $b = 1$ or $b = -1$. Thus we have either $(a,b) = (1,1)$ or $(-1,-1)$, both of which can be directly plugged in and seen to solve the equations. (I am assuming $x^{-1}$ is given the appropriate definition for $x < 0$).