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.



Solving the equation $a^a+b^b=c^c$ in positive integers
The sequence $(n^n)\_{n \ge 1}$ grows too fast for there to exist a solution:
if $b \ge 2$, then $(b+1)^{b+1} > b^{b+1} \ge 2 b^b$ (and also $2^2 > 2 \cdot 1^1$), so if $1 \le a \le b$, then $b^b < a^a + b^b \le 2 b^b < (b+1)^{b+1}$,
and no $c$ can exist such that $a^a + b^b = c^c$.


