Integer solutions of $a^{(b^c)}=b^{(a^c)}=c^{(b^a)}$ Let $a,b,c$ be positive integers and let
$$a^{(b^c)}=b^{(a^c)}=c^{(b^a)}.$$
Are there any nontrivial positive integer solutions to this set of equations?
This is a question of my own musings. I know a way to prove there is only one nontrivial integer solution to $x^y=y^x$ by letting $y=kx$ and solve for $x$ and $y$ in terms of $k$. I'm not sure it would be of any help though.
It's not even clear to me there are positive real solutions. If
$$f(x,y,z) = z \log(y) + \log(\log(x)),$$
then we want $f(a,b,c)=f(b,c,a)=f(c,a,b)$. I don't see a division to make to get a function like $g(x)=\log(x)/x$ in the $x^y=y^x$ case.
 A: Presumably "nontrivial" means $a$, $b$, and $c$ are not all equal.  First, fix $d\geq1$ and consider the function $g(x)=\frac{\log x}{x^d}$.  By a bit of calculus, you can show that $g(x)$ is decreasing for $x>e^{1/d}$.  Thus if $e^{1/d}<x<y$, then $\frac{\log x}{x^d}>\frac{\log y}{y^d}$ so $y^d\log x>x^d\log y$ and hence $x^{y^d}>y^{x^d}$.
Now let us return to the original problem; we may suppose WLOG that $a,b\leq c$.  Suppose additionally that $b>2$ and $a>1$.  Then by the result above with $d=1$, $b^c\geq c^b$ so $a^{b^c}\geq a^{c^b}$.  By the result above with $d=b$, $a^{c^b}\geq c^{a^b}$.  Thus $a^{b^c}\geq a^{c^b}\geq c^{a^b}$.  Moreover, the first equality can be equality only if $b=c$ and the second can only be equality if $a=c$.  So if $a$, $b$, and $c$ are not all equal, $a^{b^c}>c^{a^b}$.
Thus any nontrivial solution must have $a=1$ or $b\leq 2$.  It is easy to see that if one variable is $1$, all the others must be $1$.  So the only remaining case is $b=2$.  Since we are assuming $a,b\leq c$, $c=2$ would imply $a\leq2$, and then the equations cannot hold unless $a=2$.  So we may also assume $c>2$.  Moreover, inspection of our equations shows that $c$ is a power of $2$, so this means $c\geq 4$.  This implies $2^c\geq c^2$, with equality only if $c=4$.  Thus $a^{b^c}=a^{2^c}\geq a^{c^2}$, and now the result of the first paragraph with $d=2$ gives $a^{c^2}\geq c^{a^2}=c^{a^b}$.  The only way both inequalities can be equality is if $c=4$ and $a=c$, but it is then easy to check that our equations do not hold.
Thus there are no nontrivial positive integer solutions.
