## Task 1 - 251211

Determine all pairs $(x ; y)$ of real numbers that satisfy the following system of equations (1), (2).

$$
\begin{aligned}
& x^{2}+y=1 \\
& x+y^{2}=1
\end{aligned}
$$
When a pair $(x ; y)$ of real numbers satisfies the system of equations (1), (2), it follows: From (1) we have

$$
y=1-x^{2}
$$

Substituting this into (2), we get

$$
\begin{array}{r}
x+\left(1-x^{2}\right)=1 \\
x\left(x^{3}-2 x+1\right)=0
\end{array}
$$

Since the equation $x^{3}-2 x+1=0$ clearly has 1 as a solution, the division

$$
\left(x^{3}-2 x+1\right):(x-1)=x^{2}+x-1
$$

leads to the factorization $x^{3}-2 x+1=(x-1)\left(x^{2}+x-1\right)$, so (4) becomes $x(x-1)\left(x^{2}+x-1\right)=0$ or $x^{2}+x-1=0$. Since this quadratic equation has the solutions

$$
-\frac{1}{2} \pm \sqrt{\frac{1}{4}+1}=-\frac{1}{2} \pm \frac{1}{2} \sqrt{5}
$$

it follows that $x$ is one of the numbers

$$
x_{1}=0 \quad, \quad x_{2}=1 \quad, \quad x_{3,4}=\frac{1}{2}(-1 \pm \sqrt{5})
$$

From (3) and because

$$
\frac{1}{4}(-1 \pm \sqrt{5})^{2}=\frac{1}{4}(1 \mp 2 \sqrt{5}+5)=\frac{1}{2}(3 \mp \sqrt{5})
$$

the corresponding values for $y$ are

$$
y_{1}=1 \quad, \quad y_{2}=0 \quad, \quad y_{3,4}=\frac{1}{2}(-1 \pm \sqrt{5})
$$

Thus, the pairs

$$
(0 ; 1) \quad ; \quad(1 ; 0) \quad ; \quad\left(\frac{1}{2}(-1+\sqrt{5}) ; \frac{1}{2}(-1+\sqrt{5})\right) \quad ; \quad\left(\frac{1}{2}(-1-\sqrt{5}) ; \frac{1}{2}(-1-\sqrt{5})\right)
$$

can only satisfy the system of equations (1), (2), as a check confirms. Therefore, (1), (2) is satisfied exactly by the pairs (6).

## Adapted from $[5]$