1. Find all triples of real numbers $a, b, c$ with the property: Each of the equations

$$
\begin{aligned}
& x^{3}+(a+1) x^{2}+(b+3) x+(c+2)=0, \\
& x^{3}+(a+2) x^{2}+(b+1) x+(c+3)=0, \\
& x^{3}+(a+3) x^{2}+(b+2) x+(c+1)=0
\end{aligned}
$$

has three distinct real roots, but altogether there are only five different numbers.
SOLUTION. Let's assume that the numbers $a, b, c$ have the required property. First, we notice that any two of the given equations must have a common root, otherwise they would have six different roots in total.

The common roots of any two of the given three cubic equations are the roots of the quadratic equations obtained by subtracting them. Let's write down all three "difference" equations, which do not depend on the parameters $a, b, c$ (this is a positive finding in the solution), and factorize their left sides into root factors:

$$
\begin{aligned}
x^{2}-2 x+1=(x-1)^{2} & =0, \\
2 x^{2}-x-1=(2 x+1)(x-1) & =0, \\
x^{2}+x-2=(x-1)(x+2) & =0 .
\end{aligned}
$$

We see that equations (1) and (2) have only one common root $x=1$, so they have exactly five different roots in total. Therefore, each root of equation (3) must be a root of at least one of the equations (1) or (2). From the given factorizations, it follows that the number $x=1$ is also a root of equation (3).

Let's explain why the other two roots of equation (3) cannot be roots of the same equation (1) or (2). In the opposite case, one of the equations (1), (2) would have the same triple of roots as equation (3), and therefore they would have the same coefficients not only for the cubic term. However, this is not true, since for any value of the parameter $c$ the numbers $c+1, c+2, c+3$ (i.e., the constant terms of the equations) are all different.

Therefore, equation (3) has, in addition to the root $x=1$, one common root with equation (1) and one common root with equation (2); according to the factorizations (3-1) and (3-2), these are the numbers $x=-\frac{1}{2}$ and $x=-2$. The left side of equation (3) therefore has the factorization

$$
(x-1)(x+2)\left(x+\frac{1}{2}\right)=x^{3}+\frac{3}{2} x^{2}-\frac{3}{2} x-1 .
$$

From this, by comparing with the coefficients written in (3), we obtain $a=-\frac{3}{2}, b=-\frac{7}{2}, c=-2$.

From our procedure, it follows that for the found values of $a, b, c$, equation (3) has the triple of roots $1, -\frac{1}{2}$, and -2, that the numbers $1, -\frac{1}{2}$ are roots of equation (1), and that the numbers $1, -2$ are roots of equation (2). We still need to verify that the third roots of equations (1) and (2) are two other (different) numbers.

These third roots can be conveniently found using Viète's formulas. Since the product of the three roots of equation (1) is the number opposite to the constant term $c+2$ equal to zero, the number zero is the third root of equation (1). Similarly, the product of the three roots of equation (2) is equal to -1, so the third root is the number $x=\frac{1}{2}$.

Conclusion. The only solution to the problem is the numbers $a=-\frac{3}{2}, b=-\frac{7}{2}, c=-2$.

SUPPLEMENTARY AND ADDITIONAL PROBLEMS:

N1. Find (if they exist) all common roots of the cubic equations

$$
\begin{array}{r}
x^{3}+(2 \sqrt{2}-1) x^{2}-(9 \sqrt{2}+10) x+10 \sqrt{2}+16=0 \\
x^{3}+(2 \sqrt{2}-3) x^{2}+(\sqrt{2}-6) x-10 \sqrt{2}+16=0
\end{array}
$$

[The common roots must satisfy the quadratic equation $x^{2}-(5 \sqrt{2}+2) x+10 \sqrt{2}=0$, which we obtain by subtracting the given equations and dividing the result by two. This equation has a discriminant $54-20 \sqrt{2}$ equal to $(5 \sqrt{2}-2)^{2}$, so its roots are the numbers 2 and $5 \sqrt{2}$. By substitution, we can verify that the common root is only the number 2. For practice, it is possible to calculate the remaining roots of the given cubic equations by reducing their degree using the usual method of dividing by the root factor (in our case equal to $x-2$). For the first one, these are the numbers $\sqrt{2}+1$ and $-2-3 \sqrt{2}$, for the second the numbers $\sqrt{2}-1$ and $2-3 \sqrt{2}$. The more challenging part of the calculations is the square root of discriminants of the form $a+b \sqrt{2}$ by solving the equation $a+b \sqrt{2}=(u+v \sqrt{2})^{2}$ with unknown integers $u, v$.]

N2. Determine for which real number $p$ the equations

$$
\begin{aligned}
x^{3}+x^{2}-36 x-p & =0 \\
x^{3}-2 x^{2}-p x+2 p & =0
\end{aligned}
$$

have a common root in the set of real numbers. [52-A-S-3]

D1. Determine for which real numbers $p$ the system

$$
\begin{aligned}
x^{2} y-2 x & =p, \\
y^{2} x-2 y & =2 p-p^{2}
\end{aligned}
$$

has exactly three solutions in the set of real numbers. [B-51-I-5]