Observation on $a^x+b^y=c^z$ Let $D(u,v)$ be the function defined as, sum of digits of $v$ in base $u$.
Example 
To find $D(5,39)$
$39= (124)_5$
So $D(5,39)=1+2+4=7$
Problem


If $a,b,c,x,y,z\in \mathbb{N}$ and $x,y,z\ge 2$
and $$a^x+b^y=c^z$$
Then show that $$gcd(D(c,a^x),D(c,b^y))=1$$


Generalization above problem
If $$\sum_{i=1}^{k}t_{i}^{r_i}=t^r$$  where $t,K,r,t_i,r_i\in \mathbb{N}$ and $r,r_i\ge2$ for $k\ge i\ge1$ then
$$gcd(D(t,t_{1}^{r_1}),D(t,t_{2}^{r_2}),...,D(t,t_{k}^{r_k}))=1$$
Python programming for calculate $D$ function.
    n1=6
    n2=5**3
    rem_array = []
    while n2 != 0:
        mod = n2%n1
        if mod != 0:
          rem = mod
          n2 = n2 - rem
          rem_array.append(round(rem))
          n2=n2/n1
        else:
            n2 = n2/n1
            rem_array.append(0)
    print(rem_array[::-1])
    print("D(n1,n2)=",sum(rem_array))

 A: Consider $a=3$, $b=18$, $c=81$, $x=6$, $y=3$, and $z=2$.  Clearly, $a^x+b^y=c^z$.  We have
$$a^x=9\cdot c\text{ and }b^x=72\cdot c\,.$$
Thus, $$D(c,a^x)=9\text{ and }D(c,b^y)=72\,,$$
which means
$$\gcd\big(D(c,a^x),D(c,b^y)\big)=9\neq 1\,.$$
Here is a counterexample for a general $k\geq 2$.  Fix a $(k+1)$-tuple $(r_1,r_2,\ldots,r_k,r)\in\mathbb{Z}_{\geq 2}^{k+1}$.  Let $(\tau_1,\tau_2,\ldots,\tau_k,\tau)\in\mathbb{Z}_{>0}^{k+1}$ be such that $\left(t_1,t_2,\ldots,t_k,t\right)=(\tau_1,\tau_2,\ldots,\tau_k,\tau)$ is a solution to
$$t_1^{r_1}+t_2^{r_2}+\ldots+t_k^{r_k}=t^r\,.\tag{*}$$
Let $L:=\text{lcm}(r_1,r_2,\ldots,r_k,r)$, and define $l_i:=\frac{L}{r_i}$ for $i=1,2,\ldots,k$ as well as $l:=\frac{L}{r}$.  
Observe that $$(T_1,T_2,\ldots,T_k,T):=\left(\tau_1\tau^{rl_1},\tau_2\tau^{rl_2},\ldots,\tau_k\tau^{rl_k},\tau^{1+rl}\right)$$
is also a solution to (*).  We can show that the base-$T$ representation
$$T_i^{r_i}=\sum_{j=0}^{r-1}\,A_{i,j}\,T^j$$
for each $i=1,2,\ldots,k$ with $A_{i,j}\in\{0,1,2,\ldots,T-1\}$ for all $j=0,1,2,\ldots,r-1$ satisfies $$A_{i,j}=\left\{\begin{array}{ll}0&\text{if }j\in\{0,1,2,\ldots,r-2\}\,,\\
\tau_i^{r_i}\tau^{L-r+1}&\text{if }j=r-1\,.\end{array}\right.$$ for all $i\in\{1,2,\ldots,k\}$.  That is,
$$D\left(T,T_i^{r_i}\right)=\tau_i^{r_i}\tau^{L-r+1}\text{ for every }i=1,2,\ldots,k\,,$$
whence
$$\tau^{L-r+1}\mid \gcd\Big(D\left(T,T_1^{r_1}\right),D\left(T,T_2^{r_2}\right),\ldots,D\left(T,T_n^{r_n}\right)\Big)\,.$$
Because $\tau>1$ and $L\geq r$, this gives us a counterexample.
In particular, we also have a counterexample when $x=y=z=2$.  Take $a=75$, $b=100$, and $c=125$.  We have
$$a^x=45\cdot c\text{ and }b^y=80\cdot c\,,$$
implying that
$$\gcd\big(D(c,a^x),D(c,b^y)\big)=\gcd(45,80)=5\neq 1\,.$$
