Find a value to satisfy an equation

Plugging in $x_1, x_2, x_3, x_4$ into the original equation gives $\begin{cases} x_1^4 + 2mx_1 ^2 + 4 = 0 \\ x_2^4 + 2mx_2 ^2 + 4 = 0 \\ x_3^4 + 2mx_3 ^2 + 4 = 0 \\ x_4^4 + 2mx_4 ^2 + 4 = 0 \\ \end{cases} \quad \implies \quad \underbrace{(x_1 ^4 + x_2 ^4 + x_3^4 + x_4^4)}_{=\, 32} + 2m(x_1^2 + x_2^2 + x_3^2 + x_4^2) + 16 = 0 \quad \implies \quad x_1^2 + x_2^2 + x_3^2 + x_4^2 = - \frac{24}m$ And by Vieta's formula , $x_1^2 + x_2 ^2 + x_3^2 + x_4^2 = (\underbrace{x_1 + x_2 + x_3 + x_4}_{=\, 0})^2 - 2(\underbrace{x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4}_{=\, 2m}) = -4m$ Hence, $-\dfrac{24}m = 4m \implies m^2 = \boxed{6} .$

But is it true that $m = \pm \sqrt 6$ satisfy condition? Turns out $m = \sqrt6$ is not satisfied. Here's why:

Because $f(x):= x^4 + 2mx^2 + 4$ has 4 distinct roots, then $f'(x) = 4x^3 + 4mx$ has 3 distinct roots, namely $0, \pm \sqrt{-m}$ , hence $m < 0$ , so $m = -\sqrt6$ only.

The 4 unordered values of $(x_1, x_2, x_3, x_4)$ are $\pm \sqrt{ \sqrt6 \pm \sqrt2 } , \mp \sqrt{ \sqrt6 \pm \sqrt2 } .$

By Newton's sum or identities , we have

$\begin{aligned} \sum_{k=1}^4 x_k^4 & = 0 \cdot \sum_{k=1}^4 x_k^3 - 2m \sum_{k=1}^4 x_k^2 + 0 \cdot \sum_{k=1}^4 x_k - 4 \cdot 4 \\ 32 & = -2m \left(x_1^2+x_2^2+x_3^2+x_4^2 \right) - 16 \\ 24 & = - m \left(\left(x_1+x_2+x_3+x_4 \right)^2 - 2(2m)\right) \\ 24 & = - m \left(0^2 - 4m\right) \\ 24 & = 4m^2 \\ \implies m^2 & = \boxed 6 \end{aligned}$

Now we have to prove if $m=\pm \sqrt 6$ are there four real roots. There are no real roots for $m=\sqrt 6$ . For $m=-\sqrt 6$ , we have:

$\begin{aligned} x^4 - 2\sqrt 6 x^2 + 4 & = 0 \\ x^4 - 2\sqrt 6 x^2 + 6 & = 2 \\ (x^2 - \sqrt 6)^2 & = 2 \\ x^2 & = \sqrt 6 \pm \sqrt 2 \\ \implies x & = \pm \sqrt{\sqrt 6 \pm \sqrt 2} \end{aligned}$

There are four real roots.

Let $t=x^2$ ( $t \geq 0$ )

Therefore, we can rewrite the equation as $t^2 + 2mt + 4 = 0$ $(1)$

In order for the original equation has 4 distinct real solutions, $(1)$ must have two distinct real solutions.

Therefore, according to Vieta's formula for quadratic equation we have:

$\left\{\begin{matrix} \Delta ' >0 \Leftrightarrow m^2 - 4 > 0 \Leftrightarrow m < -2 \text{ OR } m>2 \\ t_1 + t_2 = -2m > 0 \Leftrightarrow m<0 \\ t_1 \times t_2 = 4 \end{matrix}\right.$

$\Rightarrow m<-2$ .

With $m<-2$ , let $t_1, t_2$ be the solutions of $(1)$ .

Therefore, the original equation has 4 distinct solutions: $(x_1, x_2, x_3, x_4) = (\sqrt{t_1},-\sqrt{t_1},\sqrt{t_2},-\sqrt{t_2})$

Substituting these results in $S$

$S= 2(t_1^2+t_2^2)= 32 \Leftrightarrow t_1^2 + t_2^2 = 16$

$\Leftrightarrow (t_1+t_2)^2 - 2t_1t_2 = 16 \Leftrightarrow (-2m)^2 -2\times4 = 32$

$4m^2 = 24 \Leftrightarrow m^2 = \boxed{6}$

As $m<-2$ , there is only one solution, which is $m=-\sqrt{6}$