Solve the following system of equations:

$$
\begin{aligned}
x+y & =a \\
x^{5}+y^{5} & =2 a^{5}
\end{aligned}
$$
Assuming $a \neq 0$, because otherwise for every $x$, $y = -x$ satisfies the system of equations.

The left side of (2) is factored into a product using a known identity:

$$
(x+y)\left(x^{4}-x^{3} y+x^{2} y^{2}-x y^{3}+y^{4}\right)=2 a^{5}
$$

and thus, considering (1),

$$
x^{4}-x^{3} y+x^{2} y^{2}-x y^{3}+y^{4}=2 a^{4}
$$

On the other hand,

$$
(x+y)^{4}=x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4}=a^{4}
$$

and subtracting (3) from this, we get an equation for the product $x y$:

$$
\begin{aligned}
& 5 x y\left(x^{2}+x y+y^{2}\right)=5 x y\left[(x+y)^{2}-x y\right]=5 x y\left(a^{2}-x y\right)=-a^{4}, \\
& 5(x y)^{2}-5 a^{2}(x y)-a^{4}=0
\end{aligned}
$$

From this,

$$
(x y)_{\mathrm{I}}=\frac{a^{2}}{10}(5-3 \sqrt{5}), \quad(x y)_{\mathrm{II}}=\frac{a^{2}}{10}(5+3 \sqrt{5})
$$

Connecting these with (1), we know the sum and product of $x$ and $y$, thus obtaining two independent quadratic equations whose roots are the pairs of $x, y$ values that satisfy (1) and (2). The equation derived from $(x y)_{\mathrm{I}}$ is

$$
z^{2}-a z+\frac{a^{2}}{10}(5-3 \sqrt{5})=0
$$

and thus the resulting $x, y$ pairs are

$$
\frac{a}{2} \pm \frac{a}{10} \sqrt{30 \sqrt{5}-25}
$$

Numerically, $a \cdot 1.1487$ and $-a \cdot 0.1487$.

The equation derived from $(x y)_{\mathrm{II}}$ has no real solutions, so the given system is satisfied by two real pairs of $x, y$ values, which arise by swapping the values.

We performed equivalent transformations, did not lose any roots, and no new roots appeared, so it is not necessary to verify the solutions.

Remark. It is natural to reduce the system to a single-variable equation by introducing a new variable $z$ based on (1):

$$
x=\frac{a}{2}+z, \quad y=\frac{a}{2}-z
$$

Substituting these into (2), the odd powers of $z$ cancel out:

$$
\begin{gathered}
2\left[\left(\frac{a}{2}\right)^{5}+10\left(\frac{a}{2}\right)^{3} z^{2}+5 \frac{a}{2} z^{4}\right]=2 a^{5} \\
z^{4}+\frac{a^{2}}{2} z^{2}-\frac{31 a^{4}}{80}=0, \quad(a \neq 0) \\
z=a \sqrt{-\frac{1}{4} \pm \frac{3}{2 \sqrt{5}}}
\end{gathered}
$$

which, taking the inner root with a positive sign, yields the same result as above.

![](https://cdn.mathpix.com/cropped/2024_05_02_681aab64c5405700e586g-1.jpg?height=648&width=655&top_left_y=2061&top_left_x=716)