Problem 913

$$\begin{aligned} & \sqrt[5]{x}+\sqrt[5]{y}=7 \\ & x+y=3157 \end{aligned}$$

Let's substitute $x$ and $y$ with $u^{5}$ and $v^{5}$. The equations then become:

$$\begin{gathered} u+v=7 \\ u^{5}+v^{5}=3157 \end{gathered}$$

If we raise the third equation to the fifth power and subtract the fourth equation from it, we get the following equations:

$$\begin{gathered} 5 u^{4} v+10 u^{3} v^{2}+10 u^{2} v^{3}+5 u v^{4}=13650 \\ u v\left(u^{3}+2 u^{2} v+2 u v^{2}+v^{3}\right)=2730 \\ u v(u+v)\left(u^{2}+u v+v^{2}\right)=2730 \\ u v\left(u^{2}+u v+v^{2}\right)=390 \\ u v\left[(u+v)^{2}-u v\right]=390 \\ u v(49-u v)=390 \\ (u v)^{2}-49(u v)+390=0 \end{gathered}$$

From this,

$$\begin{gathered} u v=\frac{49 \pm \sqrt{49^{2}-4 \cdot 390}}{2} \\ u v=\frac{49 \pm 29}{2} \\ u_{1} v_{1}=39 \\ u_{2} v_{2}=10 \end{gathered}$$

Thus, $u$ and $v$ are the roots of the following quadratic equations:

$$\begin{aligned} & z^{2}-7 z+39=0 \\ & z^{2}-7 z+10=0 \end{aligned}$$

That is,

$$\begin{aligned} & z=\frac{7 \pm \sqrt{49-4 \cdot 39}}{2} \\ & z^{\prime}=\frac{7 \pm \sqrt{49-4 \cdot 10}}{2} \end{aligned}$$

Thus,

$$\begin{gathered} u_{1}=\frac{1}{2}(7+\sqrt{-107}) \\ v_{1}=\frac{1}{2}(7-\sqrt{-107}) \\ u_{2}=5 \\ v_{2}=2 \end{gathered}$$

Therefore,

$$\begin{gathered} x_{1}=\frac{1}{32}(7+\sqrt{-107})^{5} \\ y_{1}=\frac{1}{32}(7-\sqrt{-107})^{5} \\ x_{2}=3125 \\ y_{2}=32 \end{gathered}$$

Finally, they can also be written as:

$$\begin{aligned} & x_{1}=1578.5-905.5 \sqrt{-107} \\ & y_{1}=1578.5+905.5 \sqrt{-107} \end{aligned}$$

Schönner Odilo and Seidner Mihály Losoncz; Sztrapkovits István, S.-A.-Ujhely.