
Problem 8.1. Find all values of the real parameter $m$ 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 roots.


Solution: Suppose $m$ satisfies the conditions of the problem and the equations

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

share the root $x_{0}$. After subtracting we get $(2 m-4) x_{0}=(2 m-$ 4) $\left(m^{2}+1\right)$ and so $x_{0}=m^{2}+1$ (note that if $m=2$, the two equations coincide). Substituting $x_{0}$ in any of the equations gives the equation $\left(m^{2}+1\right)\left(m^{2}-2 m-3\right)=0$ with roots $m=-1$ and $m=3$. Direct verification shows that the condition is satisfied only for $m=3$.

Let now (1) and (2) share no roots. Since $D_{1}=4+5 m^{2}>0$ (1) always has two distinct roots and therefore (2) should have equal
roots. Thus $D_{2}=4+2 m\left(m^{2}+1\right)=0$ and so $m=-1$. But this case has already been considered. Thus we determine that $m=3$.
