Find all solutions for $\begin{cases} x^{\frac{1}{4}}+y^{\frac{1}{5}}=5\\x^{\frac{1}{2}}+y^{\frac{2}{5}}=13\end{cases}$ I've been struggling with a quite difficult problem the past few days. This is my class's 'Exponentials and Logarithms' unit and my peers and I are stumped. Here it is:

Find all integer solutions to this system for x and y: $\begin{cases} x^{\frac{1}{4}}+y^{\frac{1}{5}}=5\\x^{\frac{1}{2}}+y^{\frac{2}{5}}=13\end{cases}$.

So far, I've tried to get y isolated, giving me $y=(x^{\frac{1}{4}}-5)^5$ and $y=\pm \sqrt{(x^{\frac{1}{2}}-13)^5}$, but I don't know what to do next. Could someone please give me guidance? Thanks!

Given
$$\begin{cases} x^{\frac{1}{4}}+y^{\frac{1}{5}}=5\\x^{\frac{1}{2}}+y^{\frac{2}{5}}=13\end{cases}$$
then let $t = x^{1/4}$ and $u = y^{1/5}$ to obtain
$$ \begin{cases} t + u = 5 \\ t^2 + u^2 = 13 \end{cases}. $$
Square the first equation, and use the second, to obtain
$$25 = t^2 + u^2 + 2 u t = 13 + 2 u t $$
gives $u t = 6$. Now use this in the first equation to obtain $ t^2 - 5 t + 6 = (t-3)(t-2) = 0$. Choosing $t = 3$ leads to $u = 2$, and finally $ x = 3^4$ and $ y = 2^5$. A second set can be obtained by setting $t = 2$ and $u = 3$ which gives $x = 2^4$ and $y = 3^5$.
