1. Determine for which real numbers $p$ the system of equations

$$
\begin{aligned}
(x-y)^{2} & =p^{2} \\
x^{3}-y^{3} & =16
\end{aligned}
$$

has exactly one solution in the set of real numbers.
1. The first equation is satisfied if and only if $x=y+p$ or $x=y-p$. Substituting into the second equation of the given system, we obtain, after simplification, in the first case the quadratic equation

$$
3 p y^{2}+3 p^{2} y+p^{3}-16=0,
$$

and in the second case the quadratic equation

$$
3 p y^{2}-3 p^{2} y+p^{3}+16=0
$$

with the unknown $y$. The given system of equations will have exactly one solution in the set of real numbers if and only if one of the two preceding quadratic equations has a single (double) root and the other has no real roots or has the same double root as the first equation (we can assume that $p \neq 0$, because for $p=0$ the given system clearly has no solution). The first quadratic equation has the discriminant $D_{1}=3 p\left(64-p^{3}\right)$, and the second has the discriminant $D_{2}=-3 p\left(64+p^{3}\right)$. We are looking for those $p \neq 0$ for which one of the numbers $D_{1}$, $D_{2}$ is zero and the other is negative (the case $D_{1}=D_{2}=0$ for $p \neq 0$ does not occur).

If $D_{1}=0$, then $p=4$ and $D_{2}y^{3}$, which is equivalent to the inequality $x>y$ (since $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$ and $x^{2}+x y+y^{2}>0$ for any real $x, y$ except the case $x=y=0$). Thus, $x=y+|p|,|p|>0$. Substituting into the second equation of the system (for simplicity, write $q$ instead of $|p|$) we get for $y$ the quadratic equation

$$
3 q y^{2}+3 q^{2} y+q^{3}-16=0
$$

with the discriminant $D(q)=3 q\left(64-q^{3}\right)=3 q(4-q)\left(16+4 q+q^{2}\right)$. For the given system to have a unique solution in the set of real numbers, the discriminant $D(q)$ of the preceding equation must be zero, i.e., $(4-q)\left(16+4 q+q^{2}\right)=0$ (we know that $q=|p|>0$). Since for any real $q$, $16+4 q+q^{2}>0$, it must be that $q=|p|=4$, i.e., $p=4$ or $p=-4$. At the same time, we immediately get that $y=-\frac{1}{2} q=-2$ and $x=y+4=2$.

The given system of equations has exactly one real solution if and only if $p=4$ or $p=-4$, and that solution is $(x, y)=(2,-2)$.

Award 6 points for a complete solution. In the case of a solution analogous to the 1st solution, deduct 2 points if it is not verified that the second quadratic equation has no solutions.