Given $A,B,C \in \mathbb{N}$ and $A,B,C > 1$ Prove that no solution exists / find existing solutions for:
$$A^A + B^B = C^{A+B}$$
I have 0 clues on where to start.
I think it is known that $A^A + B^B = C^C$ has no solutions as well, does it help? because I did not say that $C = A + B$ ...
Thank you!
It is most likely that there is no solution.
Lemma: $\gcd(A,B)=1$
Proof:
Set $d=\gcd(A,B)$ and $(A,B,C)=(da,db,dc)$ with $a,b$ coprime. Note that this implies that at least one of the numbers $a,b$ is uneven. Given the symmetry in $A,B$, hence $a,b$, we can assume that $a$ is uneven.
Rewriting the Diophantine equation gives
\begin{equation} (d^aa^a)^d+(d^bb^b)^d=(C^{a+b})^d \tag{1} \end{equation}
Case $d>2$: Using Fermat's Last Theorem, we have that $(d^aa^a)(d^bb^b)(C^{a+b})=0$, hence no solution.
Case $d=2$: Then $(1)$ gives a Pythagorean triple, and therefore there exists numbers $k,m,n \in \mathbb{N}$ such that
$$2^aa^a=k\cdot (m^{2}-n^{2}),\ \,2^bb^b=k\cdot (2mn),\ \,2^{a+b}c^{a+b}=k\cdot (m^{2}+n^{2}) \text{ or} \tag{2a}$$ $$2^bb^b=k\cdot (m^{2}-n^{2}),\ \,2^aa^a=k\cdot (2mn),\ \,2^{a+b}c^{a+b}=k\cdot (m^{2}+n^{2}) \tag{2b}$$
with $m>n$ coprime and not both uneven. As $m^2+n^2$ is uneven, we have $k=2^{a+b}h$ for some number $h$. This implies
$$a^a=2^bh\cdot (m^{2}-n^{2}),\ \,b^b=2^ah\cdot (2mn),\ \, \text{ or} \tag{3a}$$ $$b^b=2^ah\cdot (m^{2}-n^{2}),\ \,a^a=2^bh\cdot (2mn),\ \, \tag{3b}$$
As $a$ is uneven, we derive from $(3a)$ that $2^b=1$, hence $b=0$, respectively from $(3b)$ that both relations are not possible. QED
Weak form of the ABC-conjecture: If $a+b=c$ with $a,b$ coprime, then $c < \operatorname{rad}(abc)^2$
The radical of a positive integer $n$, denoted $\operatorname{rad}(n)$, is the product of the distinct prime factors of $n$. For instance $\operatorname{rad}(20)=2.5=10$ and $\operatorname{rad}(A^A)=\operatorname{rad}(A) \le A$.
We have $C<A+B$ as $A^A+B^B<(A+B)^{A+B}$. Assume w.l.o.g. $B>A$. We have
$$B^B < A^A+B^B=C^{A+B} < \operatorname{rad}(ABC)^2 \le (ABC)^2 < (AB(A+B))^2<4B^6$$
Hence $1<A<B\le6$. Verifying all six cases, we conclude that there is most likely no solution.