50 Followers Problem

We have $x^5=y^9+z^{11}$

Let us choose $x=2^{r_1},y=2^{11r_2},z=2^{9r_2}$ $r_1,r_2,r_3>0$

Substituting we get ,

$2^{5r_1}=2^{99r_2}+2^{99r_2}$

$\implies 2^{5r_1}=2^{99r_2+1}$

$\implies 5r_1=99r_2+1$ , it is sufficient to prove that there are infinitely many integers $(r_1,r_2)$ which satisfy this relation. Suppose $r_2=10k+1$ where $k\in[1,\infty)$ is any integer.

$\implies 5r_1=990k+100\implies r_1=198k+20$

So there are infinitely many solutions of the form $(x,y,z)=(2^{198k+20},2^{11(10k+1)},2^{9(10k+1)})$ where $k\in(0,\infty)$ is an integer.