Conjecture similar to Fermat's Theorem.

I was wondering about a problem which i could reduce to asking the following

Does there exist a set $a,b,c$ of prime numbers such that $$a^a+b^b=c^c$$

Is it really a tough problem or do you think it can be solved with some amount of work . Thank you for your ideas and help .

Solution 1:

Assume $a, b, c$ are primes with $a<b<c$. Then $b$ and $c$ are both odd, so that $a^a$ must be even, so that $a=2$.

Then we have $4+b^b=c^c$, but with $c$ being at least 2 more than $b$, this is impossible, so there are no solutions with all primes.

Edit:

A slightly different argument shows that there are no solutions in natural numbers even if we do not insist that $a, b, c$ be prime:

$$(n+1)^{n+1} > n^{n+1} = n\times n^n$$

so that for $n > 2$ we have $$(n+1)^{n+1} > 2n^n \geq a^a + b^b$$ for any $a, b \leq n$, so it is impossible to add 2 terms of the sequence $1^1, 2^2, 3^3, \ldots$ to get another term of the sequence.