Problem for the 241st birthday of Gauss

At first we consider only real solutions of the equation $f (x) = 0$ . Obviously $x = 0$ is a zero. Now consider only the remainder of the polynomial and its derivatives: $\begin{aligned} g(x) = \frac{f(x)}{x} &= x^4 + 4 x^3 + 8 x^2 + 8 x + 4 \\ g'(x) &= 4 x^3 + 12 x^2 + 16 x + 8 \\ g''(x) &= 12 x^2 + 24 x + 16 \\ &= 12 \left((x + 1)^2 + \frac{1}{3} \right) \end{aligned}$ Obviously, $g '' (x)> 0$ for $x \in \mathbb {R}$ , so that $g '(x)$ represents a strictly monotone increasing function that has exactly one zero. By trying, one finds the zero $x = -1$ for the derivative $g '(x)$ . This zero corresponds to a minimum of the function $g (x)$ with the function value $g(-1) = 1 - 4 + 8 - 8 + 4 = 1$ . Thus, we have proven, that $g (x) \geq 1> 0$ for $x \in \mathbb{R}$ . Thus, the function $g (x)$ has no real zeros.

The remaining zeros of $f (x)$ are therefore complex and each two of the zeros are complex conjugate to each other: $\begin{aligned} f(x) &= x \left(x - z_1 \right) \left(x - z_1^\ast \right) \left(x - z_2 \right) \left(x - z_2^\ast \right) \\ &= x \left(x - a_1 - i b_1 \right) \left(x - a_1 + i b_1 \right) \left(x - a_2 - i b_2 \right) \left(x - a_2 + i b_2 \right) \end{aligned}$ where $z_i = a_i + i b_i \in \mathbb{C}$ , $a_i,b_i \in \mathbb{R}$ , $b_i > 0$ .

Let's suppose we have a double zero with $z_1 = z_2 = a + ib$ . (If this approach fails, we know that all zeros of the polynomial must be different.) $\begin{aligned} g(x) &= \left[\left(x - a - i b \right) \left(x - a + i b \right)\right]^2 \\ &= \left[(x - a)^2 + b^2 \right]^2 \\ &= (x - a)^4 + 2 b^2 (x - a)^2 + b^4 \\ &= x^4 - 4 a x^3 + 2(3 a^2 + b^2) x^2 - 4a (a^2 + b^2) x + (a^2 + b^2)^2 \\ &\stackrel{!}{=} x^4 + 4 x^3 + 8 x^2 + 8 x + 4 \end{aligned}$ By coefficient comparison, the solution $a = -1, b = 1$ can be found. Therefore, our approach was valid and we have three different zeros $\begin{aligned} z_0 &= 0 \\ z_1 &= -1 + i\\ z_2 &= -1 - i \end{aligned}$ The function can be written as $f(x) = x (x + 1 - i)^2 (x + 1 + i)^2$