
1. Find all triples $a, b, c$ of real numbers with the following 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 these roots form just five distinct numbers.


Solution. Assume that the numbers $a, b, c$ have the required property. Observe first of all that any two of the given equations must have a common root, otherwise they would have altogether at least six distinct roots.

The common roots of the given three cubic equations are roots of the quadratic equations obtained by subtracting them. These three quadratic equations turn out to be independent of the parameters $a, b, c$ :

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

We see that the equations (1) and (2) have only one common root $x=1$, thus together they have exactly five distinct roots. Hence each of the roots of the equation (3) must also be a root of at least one of the equations (1) or (2). From the subtracted equations it follows that number $x=1$ is also a root of the equation (3).

We claim that the other two roots of the equation (3) cannot be at the same time also roots of (1) or (2). Otherwise one of the equations (1), (2) would have the same three roots as the equation (3), and thus would have the same coefficient not only at the cubic term. This is however not the case, since for any value of the parameter $c$ the constant terms $c+1, c+2, c+3$ of the three equations are always mutually distinct.

The equation (3) has thus, in addition to $x=1$, one more common root with the equation (1) and one common root with the equation (2); from (3-1) and (3-2) we see that these common roots are $x=-\frac{1}{2}$ and $x=-2$. The left hand side of the equation (3) thus decomposes as

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

Comparing this with the coefficients of (3) we already obtain $a=-\frac{3}{2}, b=-\frac{7}{2}, c=-2$.

From our argument it follows that for these values of $a, b, c$ the equation (3) has the roots $1,-\frac{1}{2}$ and -2 , that the numbers $1,-\frac{1}{2}$ are roots of the equation (1) and that the numbers $1,-2$ are roots of the equation (2). What remains to be checked is that the third roots of the equations (1) and (2) are another two (distinct) numbers. These third roots can be conveniently found using Vièta's relations. Because the product of the three roots of (1) is the number opposite to the constant term $c+2=0$, the third root of (1) must be the number 0 . Similarly the product of the roots of (2) equals -1 , thus the third root of the equation (2) is the number $x=\frac{1}{2}$.

Conclusion. The problem has the unique solution $a=-\frac{3}{2}, b=-\frac{7}{2}, c=-2$.
