Yasnovy E.A.

Investigate how many solutions the system of equations has

$x^{2}+y^{2}+x y=a$, $x^{2}-y^{2}=b$,

where $a$ and $b$ are some given real numbers.
Let's move to new variables: $x+y=u, x-y=v$. Then the system can be written as: $3 u^{2}+v^{2}=4 a^{2}$, $u v=b$.

$(*)$

Obviously, each pair $(x, y)$ corresponds to one pair $(u, v)$, and each pair $(u, v)$ corresponds to a pair $(x, y)$, so we need to determine how many solutions the system (*) has for different values of parameters $a$ and $b$.

Subtract and add to the first equation of the system (*) the second equation multiplied by $2 \sqrt{3}$. We get:

$$
(\sqrt{3} u-v)^{2}=4 a-2 \sqrt{3} b, \quad(\sqrt{3} u+v)^{2}=4 a+2 \sqrt{3} b
$$

It remains to note that the equation $z^{2}=c$ has two solutions if $c>0$, one if $c=0$, and no solutions if $c<0$.

## Answer

Solutions exist if and only if the inequalities $-2 a \leq \sqrt{3} b \leq 2 a$ are satisfied, and if both inequalities are strict ("less than"), there are four solutions; if one of them is an equality, there are two solutions; if both (which is possible only when $a=b=0$), there is one.