
Problem 10a.1. Find all pairs $(a ; b)$ of integers such that the system

$$
\left\lvert\, \begin{gathered}
x^{2}+2 a x-3 a-1=0 \\
y^{2}-2 b y+x=0
\end{gathered}\right.
$$

has exactly three real solutions.


Solution: Let $(a ; b)$ satisfy the condition of the problem. Since there are exactly three real solutions the first equation has two distinct roots $x_{1}0$. The discriminant of the second equation in respect to $y$ equals $D_{2}=4\left(b^{2}-x\right)$ and therefore there are exactly three real solutions iff $x_{1}0$ and $b^{2}=-a+\sqrt{a^{2}+3 a+1}$. It follows from the latter equality that $a^{2}+3 a+1=c^{2}$, where $c$ is a positive integer. Therefore the discriminant of $a^{2}+3 a+1-c^{2}=0$ is a perfect square, i.e. $9-4\left(1-c^{2}\right)=d^{2}$, where $d$ is nonnegative integer. The last equality can be written in the form $(d-2 c)(d+2 c)=5$ which implies that $d-2 c=1, d+2 c=5$, i.e. $d=3, c=1$. Hence, $a^{2}+3 a+1=1$ with roots $a=0$ and $a=-3$. Respectively $b= \pm 1$ or $b= \pm 2$. Direct verification shows that all pairs $(a, b)=(0,1),(0,-1),(-3,2)$ and $(-3,-2)$ satisfy the condition of the problem.
