# For integers $a\ge b\ge 2$, is $f(a,b) = a^b + b^a$ injective?

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.

• Clearly $f(a,b)=f(b,a)$, so this function is clearly not injective. EDIT: Sorry, I missed $a\ge b$. – Martin Sleziak Aug 5 '14 at 11:12
• This question was asked here within the last month, but I can't find the older question. – Thomas Andrews Aug 5 '14 at 11:18
• Best to link to that original question in your question, so people know you know it "the same but different." @Mathmo123 – Thomas Andrews Aug 5 '14 at 11:21
• Have you made any more progress? It's a tricky question... (P.s. always nice to see a fellow mathmo :D) – Shakespeare Aug 9 '14 at 14:33
• 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

## 1 Answer

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.