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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.

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Mathmo123
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  • Clearly $f(a,b)=f(b,a)$, so this function is clearly not injective. EDIT: Sorry, I missed $a\ge b$. – Martin Sleziak Aug 05 '14 at 11:12
  • This question was asked here within the last month, but I can't find the older question. – Thomas Andrews Aug 05 '14 at 11:18
  • @ThomasAndrews yes it was. Please see this meta discussion http://meta.math.stackexchange.com/questions/16502/how-best-to-re-ask-a-question-that-wasnt-fully-answered – Mathmo123 Aug 05 '14 at 11:19
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    Best to link to that original question in your question, so people know you know it "the same but different." @Mathmo123 – Thomas Andrews Aug 05 '14 at 11:21
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    Have you made any more progress? It's a tricky question... (P.s. always nice to see a fellow mathmo :D) – ShakesBeer Aug 09 '14 at 14:33
  • I have a feeling it's very hard... Perhaps the folks over at mathoverflow would appreciate it if it doesn't get answered here? More elementary stuff like this does get posted there and you can mention nobody solved it here even after a bounty. Do mention you've come up with it though (or at least you have no idea as to its solvability). – ShakesBeer Aug 09 '14 at 20:35
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    This reminds me of Catalan's conjecture which took quite some time to prove. I may be wrong though. – Dejan Govc Aug 10 '14 at 15:13

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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.

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