Problem 1058

[Proof] Let $p, q, r$ be distinct prime numbers, and $\sqrt[3]{p}, \sqrt[3]{q}, \sqrt[3]{r}$ be three terms in an arithmetic sequence.
Assume
$$\begin{aligned} \sqrt[3]{p} & =a, \\ \sqrt[3]{q} & =a+m d, \\ \sqrt[3]{r} & =a+n d, \end{aligned}$$

where $m, n$ are positive integers.
Eliminating $a, d$ yields
$$\begin{array}{c} \frac{\sqrt[3]{q}-\sqrt[3]{p}}{\sqrt[3]{r}-\sqrt[3]{p}}=\frac{m}{n} \\ m \sqrt[3]{r}-n \sqrt[3]{q}=(m-n) \sqrt[3]{p} \end{array}$$

Cubing and expanding gives
$$\begin{array}{c} \frac{\sqrt[3]{q}-\sqrt[3]{p}}{\sqrt[3]{r}-\sqrt[3]{p}}=\frac{m}{n} \\ m \sqrt[3]{r}-n \sqrt[3]{q}=(m-n) \sqrt[3]{p} \end{array}$$
$$m^{3} r-n^{3} q+3 m n \sqrt[3]{r q}(m \sqrt[3]{r}-n \sqrt[3]{q})=(m-n)^{3} p .$$

Substituting (1) into the above equation gives
$$m^{3} r-n^{3} q-(m-n)^{3} p=-3 m n(m-n) \sqrt[3]{p q r} \text {. }$$

Since $p, q, r$ are all prime numbers, $\sqrt[3]{p q r}$ is an irrational number. Therefore, the left side of (2) is a rational number, while the right side is an irrational number, making (2) impossible.
Thus, $\sqrt[3]{p}, \sqrt[3]{q}, \sqrt[3]{r}$ cannot be three terms in an arithmetic sequence.