It's "How Many", But Not An Inequality

Set $t=\sqrt[12]{x}$ and $u=\sqrt[12]{y}$ and we'll get the following system $\begin{cases} t^4+u^4=1 \\ t^3+u^3=1\end{cases}$ $\Leftrightarrow\begin{cases} t^4+u^4=t^3+u^3 \\ t^3+u^3=1\end{cases}$ Considering the first equation $t^4+u^4=t^3+u^3$ $t^3(t-1)=u^3(1-u) \ (*)$ Since $t,u\geq 0$ and $t^3+u^3=1$ , we get $t,u\in [0;1]$ . So $t^3(t-1)\leq 0$ and $u^3(1-u)\geq 0$ , therefore $(*)$ happens when $t^3(t-1)=u^3(1-u)=0$ $\Rightarrow\begin{cases} t^3(t-1)=u^3(1-u)=0 \\ t^3+u^3=1\end{cases}$ $\Leftrightarrow (t,u)=(1;0) \ and \ (0;1)$ So finally $(x,y)=(1;0),(0;1)$ , $2$ is the correct answer

Well,i discovered that for the equations to be true, x and y must be greater than or equal tozero. So by mere observation, (1,0) and (0,1) seem to be the only valid entry.