4. For which whole $n>2$ do there exist rational numbers $p$ and $q$ such that $\sqrt[3]{n}=p+q \sqrt[3]{2}$?
Solution. By cubing, we get the equivalent equality

$$
n=\left(p^{3}+2 q^{3}\right)+3 p^{2} q \sqrt[3]{2}+3 p q^{2} \sqrt[3]{4}.
$$

Let's first consider the case $n=4$. If $\sqrt[3]{4}=p+q \sqrt[3]{2}$, then from (1) we get

$$
4=\left(p^{3}+2 q^{3}\right)+3 p^{2} q \sqrt[3]{2}+3 p q^{2}(p+q \sqrt[3]{2})
$$

or $4-p^{3}-2 q^{3}-3 p^{2} q^{2}=\sqrt[3]{2}\left(3 p^{2} q+3 p q^{3}\right)$. Since $\sqrt[3]{2}$ is an irrational number, the last equality is possible only if $4-p^{3}-2 q^{3}-3 p^{2} q^{2}=0$ and $3 p q\left(p+q^{2}\right)=0$. From the second equation, we get $p=0, q=0$ or $p=-q^{2}$, and substituting into the first equation, we get $q^{3}=2, p^{3}=4$ or $q^{6}+q^{3}-2=0$ respectively. Since the numbers $p$ and $q$ are rational, only the third condition of the last triplet is satisfiable, which means that $q^{3}=-2$ or $q^{3}=1$. This gives us the only pair $(p, q)=(-1,1)$, for which (2) holds, but not $\sqrt[3]{4}=p+q \sqrt[3]{2}$. Therefore, the last equality is not satisfied by any rational $p$ and $q$.

From the proven fact, it follows: if (1) holds for some rational $n, p$ and $q$, then the coefficient $3 p q^{2}$ of the term $\sqrt[3]{4}$ must be zero. In the opposite case, (1) could be expressed as

$$
\sqrt[3]{4}=\frac{n-p^{3}-2 q^{3}}{3 p q^{2}}-\frac{p}{q} \cdot \sqrt[3]{2}
$$

which would be a contradiction. Therefore, $3 p q^{2}=0$, i.e., $p=0$ or $q=0$. Then, however, $n=p^{3}$ or $n=2 q^{3}$. If the number $n$ is also an integer, then the numbers $p, q$ must also be integers in the last two equalities. Answer: $n=k^{3}$ or $n=2 k^{3}$, where $k>1$ is an integer.

Guiding and supplementary problems. (N1) Prove that the numbers $\sqrt[3]{2}$ and $\sqrt[3]{4}$ are irrational.

(N2) If $p$ is rational and one of the numbers $p^{3}, 2 p^{3}, 4 p^{3}$ is an integer, then the number $p$ is also an integer, prove it. (N3) Answer the question from the problem first for $n=4$.

(D1) For which integers $n$ is the number $(1+\sqrt[3]{2})^{n}$ rational?

(D2) If the equality $\sqrt[3]{x}=p+q \sqrt[3]{2}+r \sqrt[3]{4}$ holds for rational $x, p, q$ and $r$, then at least two of the numbers $p, q, r$ are zero, prove it. (Hint: $x=A+B \sqrt[3]{2}+C \sqrt[3]{4}$, where the rational coefficients $A, B, C$ depend on $p, q, r$ such that $r B-q C=p\left(4 r^{3}-3 q^{3}\right)$. According to the problem, $B=C=0$ must hold.)