Under what all conditions on $(a,b,c)$ where $a,b,c$ are positive integers can we say about the non-existence or existence of any solution to the equation $$a^a+b^b=c^c$$

(Or in other words : Solve $a^a+b^b=c^c$ ) $ $

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About this note that : For a solution we must have $gcd(a,b,c) = 1$ (follows from Fermat's Last Theorem)

and all the elements of the set $\{a,b,c\}$ cannot be prime numbers (proof shown below).

We will prove there are no positive prime numbers $p,q,r$ such that the following equation is satisfied : $$p^p+q^q=r^r$$ This is easy to prove. Note that in fact we have something stronger : The equation $4+m^m=n^n$ has no solutions in positive integers. Assume the contrary, then we get $m>1$ (by checking the case $m=1$ separately) and then note that $n \ge m+2$ and thus $n^n > (m+2)^m > m^m+2^m \ge n^n$ and thus a contradiction.

Prime omega function$\omega(n)$ that counts the number of distinct prime factors in the factorization of $n>1$. My proposal was $a^{\omega(a)}+b^{\omega(b)}=c^{\omega(c)}$ with $a<b<c$ (and now I believe that also I need to require $1<a$ to get a problem more interesting problem). I know also theMSEpost1861380$\endgroup$