Does equation $a^m+b^n =c^{m+n}$ have integral solutions?
Equation $a^m+b^n=c^t $, for a and b as a power of 2 have solutions, for example $4^7+4^7=8^5$ , or $4^{13}+4^{13}=8^9$ . Also may have solution like $7^1+1^n=2^3$. But for:
$\begin{cases} a, b \neq 2^k\\a, b \neq 1\\(a, b, c)=1\\(m, n)=1\\m\neq n\end {cases}$
equation $a^m+b^n=c^t$ has no solution. Equation $a^m+b^n=c^{m+n}$ has solutions of the form $(2^k-1)^1+1^2=2^k$, only with $k=3$ without theses restrictions. With these restrictions I could not find any by brute force. If it is true is there a way or article showing this analytically?
Update: We can consider particular cases to modify the equation to linear form so that the solution become easier. For example suppose:
$a=ma_1+1 \Rightarrow a^m=km+1$
$b=nb_1-1; n \text{ is odd} ; \Rightarrow b^n=l n -1$
$c=(m+n)s+1 \Rightarrow c^{m+n}=t(m+n)+1$
then we get:
$km+ln=tm+tn+1$
We must optimize the solution of this equation subjected to our initial equation, that is following system of Diophantine equation mus be consistent:
$\begin{cases}(k-t)m+(l-t)n=1\\a^m+b^n=c^{m+n}\end{cases}$
For an arbitrary Pythagorean tripple $(a, b, c)$ with $a^2 + b^2 = c^2$, we have $(ca)^2 + (bc)^2 = c^4$, so $(ca, cb, c)$ forms a solution with $m = n = 2$.
Another solution:
If $a^2+b^3=c$, then multiplying by $c^{24}$ we get:
$$(ac^{12})^2+(bc^8)^3=(c^5)^5$$