Solve the following system of equations:

$$
\begin{aligned}
& x+y^{2}=a \ldots \\
& y+x^{2}=a \ldots
\end{aligned}
$$

For which values of $a$ do we get real solutions?
We subtract the terms of equation (2) from the corresponding terms of equation (1), resulting in

$$
x-y+y^{2}-x^{2}=0 \quad \text { or } \quad(x-y)(1-x-y)=0 \ldots
$$

According to (3), we have

$$
x-y=0 \quad \text { or } \quad 1-x-y=0
$$

Connecting each of these to one of the original system's equations, we arrive at two systems of equations:

$$
\left.\left.\begin{array}{r}
x+y^{2}=a \\
x-y=0
\end{array}\right\} \text { I. } \begin{array}{r}
x+y^{2}=a \\
x+y=1
\end{array}\right\} \text { II. }
$$

The solution for I. is

$$
x=y=\frac{-1 \pm \sqrt{1+4 a}}{2}
$$

The solution for II. is

$$
x=\frac{1 \pm \sqrt{4 a-3}}{2}, \quad y=\frac{1 \mp \sqrt{4 a-3}}{2}
$$

The solutions for I. are real if $1+4 a \geq 0$, i.e., $a \geq-\frac{1}{4}$.

The solutions for II. are real if $4 a-3 \geq 0$, i.e., $a \geq \frac{3}{4}$.

Otto Szakáll (Érseki g. V. o. Bp. II.)[^0]


[^0]:    ${ }^{1}$ According to I., $x^{2}+x-a=0$

    ${ }^{2}$ According to II., $1-y+y^{2}=a$ or $y^{2}-y+(1-a)=0$.