Given two integers $a \ge b \ge 2$, can we encode them as a unique integer $a^b + b^a$?
This question was asked a few weeks ago, but did not rule out the trivial cases. For example, if we allow one of the integers to be $1$, then since $$a^b + b^a =(a^b +b^a-1)^1 + 1^{a^b +b^a-1}$$we get a trivial solution. However, for $a\ge b\ge2$, I believe this expression will be unique, but I haven't been able to prove it.
So far, supposing $a^b + b^a = c^d + d^c$, I've tried working modulo $b$ to no avail. If $ b$ is prime, then using Fermat's Little Theorem, we have $a \equiv c^d + d^c \pmod b$, but I can't see how that is going to help too much.
Update: After asking this on Mathoverflow, it seems like an unconditional solution is probably out of the reach of modern mathematics. However, assuming a generalisation of the abc conjecture, it is possible to prove that there are at most finitely many repeats.
Integers of the form $a^b + b^a$, $a,b > 1$ were named Leyland numbers by Crandall & Pomerance in honor of Paul Leyland. See https://oeis.org/A076980. See also http://en.wikipedia.org/wiki/Leyland_number for more about these numbers, including a project to factor them.