For which values of the parameter $a$ are the following systems of equations equivalent:

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

$(\{z\}=z-[z]$, where $[z]$ is the greatest integer not greater than $z$. $)$
I. solution. If $a<0$, then neither of our systems of equations has a solution, so the two solution sets are identical - the empty set - in this case, therefore I. and II. are equivalent.

For the rest, let $a \geq 0$. The second equation of II. states that $x+y$ is an integer. Therefore, the two systems of equations are equivalent for those values of $a$ for which this integer is 0. Thus, we need to exclude those values of $a$ for which, under the condition $x^{2}+y^{2}=a$, $x+y$ can take an integer value different from 0.

Let us introduce the notation $x+y=m$ (where $m$ is an integer), and examine how the solutions of II. depend on $a$ and $m$. Substituting $x=m-y$ into the first equation, after rearrangement, we get

$$
2 y^{2}-2 m y+m^{2}-a=0
$$

I. and II. are exactly equivalent if (2) does not hold for any $y$ when $m$ is an integer different from 0, that is, if the discriminant of (2), $4\left(2 a-m^{2}\right)$, is negative for such values of $m$.

If $m \neq 0$ and is an integer, then $m^{2}$ is at least 1. The discriminant will therefore be negative for all integers $m$ different from 0 precisely when $2 a<1$, that is, $a<\frac{1}{2}$.

László Dringó (Budapest XI., Petőfi S. Secondary School, 8th grade)

II. solution. Again, we assume that $a \geq 0$ and represent the solutions of the individual systems of equations in a Cartesian coordinate system. The points whose coordinates satisfy $x^{2}+y^{2}=a$ lie on a circle centered at the origin with radius $\sqrt{a}$. The points that satisfy the condition $x+y=m$ lie on parallel lines.

![](https://cdn.mathpix.com/cropped/2024_05_02_d3f0f08c888d66fcf177g-1.jpg?height=876&width=979&top_left_y=1015&top_left_x=562)

Since $a \geq 0$, the two systems of equations are equivalent for those values of $a$ for which the circle with radius $\sqrt{a}$ centered at the origin has common points only with the line corresponding to $m=0$, that is, the circle lies within the region bounded by the lines $x+y=1$ and $x+y=-1$ corresponding to $m=1$ and $m=-1$.

The origin exactly bisects the distance between these two lines, which is $\sqrt{2}$, so the obtained condition is precisely satisfied if the radius of the circle is less than $\sqrt{2} / 2$, or $a<1 / 2$.