[ [ Mixed equations and systems of equations] $[$ Integer and fractional parts. The Archimedean principle ]  Solve the equation $x^{2}-[x]=3$.

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Let $[x]=n$ and $x=n+\alpha$, where $0 \leq \alpha<1$. Then $x^{3}-x+\alpha=3$, and from the constraints on $\alpha$ it follows that $2<x^{3}-x \leq 3$. If $x \geq 2$, then $x\left(x^{2}-1\right) \geq 2 \cdot 3=6$, so the inequality $x^{3}-x \leq 3$ is not satisfied. If $x<-1$, then $x\left(x^{2}-1\right)<0$, Therefore, $-1 \leq x<2$, that is, $[x]=-1,0$ or 1. Accordingly, we get the equations $x^{3}+1=3, x^{3}=3, x^{3}$ $x=\sqrt[3]{2}, x=\sqrt[3]{3}, x=\sqrt[3]{4}$. In this case, $[\sqrt[3]{2}] \neq-1,[\sqrt[3]{3}] \neq 0$, and $[\sqrt[3]{4}]=1$.

## Answer

$x=\sqrt[3]{4}$