## Task 6/85

How many different real solutions does the system of equations

$$
x^{2}+y=a \quad ; \quad x+y^{2}=a
$$

have, depending on the parameter a?
By subtracting one equation from the other and subsequent rearrangement, one obtains

$$
x^{2}+y-x-y^{2}=0 \rightarrow(x-y)(x+y-1)=0
$$

A product is exactly zero if at least one factor is zero: 1. $x=y$, 2. $x=1-y$.

Substituting this into one of the given equations yields $y^{2}+y=a$ or $y^{2}-y+1=0$. The solutions of these quadratic equations in $y$ are

$$
y=-0.5 \pm \sqrt{a+0.25} \quad \text { or } \quad y=0.5 \pm \sqrt{a-0.75}
$$

Real solutions only exist for $a \geq 0.25$ and in the second case for $a \geq 0.75$. Thus, the system has no real solutions for $a<0.75$, where the multiplicity of solutions was not considered.