Solve the following equation:

$$
x^{4 n}-4 x^{n}-1=0 \text {. }
$$
We denote $x^{n}$ by $y$, thus reducing our equation to a quartic equation

$$
y^{4}-4 y-1=0
$$

We complete the left side of the equation into two perfect squares.

$$
\begin{gathered}
y^{4}+2 y^{2}+1-2 y^{2}-4 y-2=0 \\
\left(y^{2}+1\right)^{2}-2(y+1)^{2}=0
\end{gathered}
$$

Thus,

$$
\left(y^{2}+1\right)^{2}=2(y+1)^{2}
$$

This reduces the problem to solving two quadratic equations.

$$
\begin{gathered}
y^{2}+1= \pm \sqrt{2}(y+1) \\
y^{2}+\sqrt{2} y+1+\sqrt{2}=0, \quad y^{2}-\sqrt{2} y+1-\sqrt{2}=0
\end{gathered}
$$

The roots of the first quadratic equation are

$$
\begin{aligned}
& y_{1,2}=\frac{-\sqrt{2} \pm \sqrt{2-4-4 \sqrt{2}}}{2} ; \\
& y_{1,2}=\frac{-\sqrt{2} \pm \sqrt{2}(\sqrt{-1-2 \sqrt{2}})}{2}=\frac{\sqrt{2}}{2}[-1 \pm \sqrt{-1-2 \sqrt{2}}]
\end{aligned}
$$

These roots are not real numbers.

The roots of the second equation are

$$
y_{3,4}=\frac{\sqrt{2}}{2}[1 \pm \sqrt{-1+2 \sqrt{2}}]
$$

Given $y$, the value of $x$ can be determined.

$$
x=\sqrt[n]{y}
$$

If we consider the complex solutions as well, the equation has $4 n$ roots.

# Solved by: Dávid P., Seitz K.

Note: It is evident from the graphical solution of the quartic equation that it has only two real roots, which can be seen by plotting $y=x^{4}$ and the line $y=4 x+1$. These two real roots lie in the intervals $(-1,0)$ and $(1,2)$, because the value of the function $y=x^{4}-4 x-1$ at $x=-1$ is positive, and at $x=0$ it is negative, so by the continuity of the function, it must intersect the x-axis somewhere. The same reasoning applies to the interval $(1,2)$.