For which values of $m \in \mathbb{R}$ does the equation:

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

have exactly three distinct roots?
The first quadratic $x^{2}-2 m x-4\left(m^{2}+1\right)$ always admits two distinct real roots. Therefore, either the second quadratic must admit a double root, or it must share a root with the first.

It admits a double root when $-2 m\left(m^{2}+1\right)=4$, which simplifies to $\mathfrak{m}^{3}+m+2=0$. This polynomial has an obvious root $\mathfrak{m}=-1$, so it can be factored as: $(\mathfrak{m}+1)\left(\mathfrak{m}^{2}-\mathfrak{m}+2\right)$, and $\mathfrak{m}^{2}-\mathfrak{m}+2$ has no real roots. For $m=-1$, the double root 2 of $x^{2}-4 x-2 m\left(m^{2}+1\right)$ is also a root of $x^{2}-2 m x-4\left(m^{2}+1\right)$, so the equation in the statement has only two distinct roots (one of which is a triple root).

If $x^{2}-2 m x-4\left(m^{2}+1\right)$ and $x^{2}-4 x-2 m\left(m^{2}+1\right)$ share a common root, this root is a root of their difference: $(4-2 m)\left(x-\left(m^{2}+1\right)\right)$. Either $m=2$, in which case the two quadratics are identical, equal to $x^{2}-4 x-20$, and the equation has only two distinct roots (two double roots), or $x=m^{2}+1$, in which case each quadratic is equal to $\left(m^{2}+1\right)\left(m^{2}-2 m-3\right)$ and $s'$ vanishes if $m=-1$ or $m=3$. $m=-1$ is excluded (it is the triple root already seen), leaving $m=3$ for which the equation $\left(x^{2}-6 x-40\right)\left(x^{2}-4 x-60\right)$ indeed has three roots: $-4, -6$, and 10 (a double root). This is therefore the unique solution to the exercise.