1. Find the real number $m$ such that the equation

$$
\left(x^{2}-2 m x-4\left(m^{2}+1\right)\right)\left(x^{2}-4 x-2 m\left(m^{2}+1\right)\right)=0
$$

has exactly three distinct real roots.
Solution. The given equation is equivalent to the system of equations

$$
\left[\begin{array}{l}
x^{2}-2 m x-4\left(m^{2}+1\right)=0 \\
x^{2}-4 x-2 m\left(m^{2}+1\right)=0
\end{array}\right.
$$

The discriminant of the first equation is

$$
D_{1}=4\left(5 m^{2}+4\right)
$$

and of the second

$$
D_{2}=4\left(4+2 m\left(m^{2}+1\right)\right)
$$

It is clear that $D_{1}>0$. Since the first equation always has two distinct real roots, the following cases are possible:

A) The second equation has exactly one real root, different from the roots of the first.

B) The second equation has two distinct real roots, but one of them is the same as a root of the first equation.

A) In this case, we get the equation

$$
D_{2}=m^{3}+m+2=(m+1)\left(m^{2}-m+2\right)=0
$$

which in the set of real numbers has one solution $m=-1$. Then the solutions of the first equation are 2 and -4, and the solution of the second equation is 2, i.e.,

$$
\left(x^{2}-2 m x-4\left(m^{2}+1\right)\right)\left(x^{2}-4 x-2 m\left(m^{2}+1\right)\right)=0
$$

has two real solutions. Thus, $m=-1$ is not the desired solution.

B) Let $x_{0}$ be the common root of the equations in the system. Then if we subtract

$$
x_{0}^{2}-2 m x_{0}-4\left(m^{2}+1\right)=0
$$

from

$$
x_{0}^{2}-4 x_{0}-2 m\left(m^{2}+1\right)=0
$$

we get

$$
(4-2 m) x_{0}=\left(m^{2}+1\right)(4-2 m)
$$

If $m=2$ then the first and second equations of the system are identical, so

$\left(x^{2}-2 m x-4\left(m^{2}+1\right)\right)\left(x^{2}-4 x-2 m\left(m^{2}+1\right)\right)=0$

has exactly two distinct real roots. Thus, $m \neq 2$, and then $x_{0}=m^{2}+1$. By substituting $x_{0}$ into the first equation, we get the equation $\left(m^{2}+1\right)\left(m^{2}-2 m-3\right)=0$, and its real solutions are $m_{1}=-1$ and $m_{2}=3$. From the discussion under A), $m=-1$ is not the desired solution. If $m=3$ then the roots of the first equation are 10 and -4, and the roots of the second equation are -4 and -6. Thus, the desired solution is $m=3$.