## Task 23/74

Determine all real solutions of the equation

$$
\left[x^{3}\right]=[x]^{4}
$$

(where $[x]$ denotes the greatest integer not greater than $x$).
First, one notices that the equation is satisfied for $0 \leq x \leq 1$; because for $0 \leq x < 1$.

Therefore, only values $x>1$ can be further solutions. To find them, we set $x=[x]+\alpha$ with $0 \leq \alpha<1$. Then we have

$$
[x]^{4}=\left[([x]+\alpha)^{3}\right]<[([x]+1)]^{3}
$$

This inequality has, apart from $[x]=0$ and $[x]=1$, only a solution for $[x]=2$; because if we set $[x]=3+C$ (where $C$ is a non-negative integer constant), it follows that

$$
C^{4}+11 C^{3}+42 C^{2}+60 C+17<0
$$

which is a contradiction.

If we now set $[x]=1$ and $[x]=2$ in sequence, we get

1. $1^{4} \leq(1+\alpha)^{3}<2, \quad$ thus $\quad 1 \leq 1+\alpha \leq \sqrt[3]{2}$
2. $2^{4} \leq(2+\alpha)^{3}<2^{4}+1, \quad$ thus $\quad 2 \sqrt[3]{2} \leq 2+\alpha<\sqrt[3]{17}$

The solutions are therefore

$$
0 \leq x<\sqrt[3]{2} \quad ; \quad 2 \sqrt[3]{2} \leq x<\sqrt[3]{17}
$$