Solve the following equation: $[x]=x^{4}-2 x^{2}$.
Solution. Let $[x]=k \in \mathbb{Z}$. Thus, $x^{4}-2 x^{2}-k=0$. The square of the roots of the equation is:

$$
x_{1,2}^{2}=\frac{2 \pm \sqrt{4+4 k}}{2}=\frac{2 \pm 2 \sqrt{1+k}}{2}=1 \pm \sqrt{1+k}
$$

where $1+k \geq 0$ or $k \geq-1$.

1. If $k>0$, then $x>0$, and from the definition of the integer part, $x \geq[x]=k$.

If $x^{2}=1-\sqrt{1+k} \geq 0$ were true, then $1 \geq \sqrt{1+k}$, or $1 \geq 1+k$, which implies $0 \geq k$, but this contradicts the condition, so $x^{2}=1+\sqrt{1+k}$.

Since $x>0$, the condition $x=\sqrt{1+\sqrt{1+k}} \geq k$ must be satisfied. From this,

$$
1+\sqrt{1+k} \geq k^{2}, \quad \text { or } \quad \sqrt{1+k} \geq k^{2}-1
$$

which, with the condition $k^{2} \geq 1$, can be squared: $1+k \geq k^{4}-2 k^{2}+1$. After rearranging and dividing both sides by $k$: $1 \geq k^{3}-2 k$, rearranging and factoring:

$$
0 \geq k^{3}-2 k-1=k\left(k^{2}-1\right)-(k+1)=k(k-1)(k+1)-(k+1)=(k+1)\left(k^{2}-k-1\right)
$$

a) case: $k+1 \geq 0$ and $k^{2}-k-1 \leq 0$, from which

$$
k_{1,2}=\frac{1 \pm \sqrt{1+4}}{2}=\frac{1 \pm \sqrt{5}}{2}, \quad \text { but } 0<\frac{1-\sqrt{5}}{2}<1
$$

b) case: $k+1<0$, but this cannot be true due to the assumption $k>0$.

Therefore, the equation has three solutions: $-1,0$ and $\sqrt{1+\sqrt{2}}$.