B2. Find all natural numbers $n$ and prime numbers $p$ for which $\sqrt[3]{n}+\frac{p}{\sqrt[3]{n}}$ is a square of a natural number.
B2. Let $\sqrt[3]{n}+\frac{p}{\sqrt[3]{n}}=k^{2}$, where $k$ is a natural number. We cube the equation to get $n+3 p \sqrt[3]{n}+3 \frac{p^{2}}{\sqrt[3]{n}}+\frac{p^{3}}{n}=k^{6}$, which can be rewritten as $n+3 p k^{2}+\frac{p^{3}}{n}=k^{6}$. From this, it follows that $n$ divides $p^{3}$. Since $p$ is a prime number, $n$ must be $1, n=p, n=p^{2}$, or $n=p^{3}$. If $n=p$ or $n=p^{2}$, we get the equation $p+3 p k^{2}+p^{2}=k^{6}$. From this, it follows that $k$ must be divisible by $p$, so the right side is divisible by $p^{2}$, but the left side is not. The remaining possibilities are $n=1$ and $n=p^{3}$, which we substitute back into the original equation to get $1+p=k^{2}$, thus $p=(k-1)(k+1)$. From this, it follows that $k=2$, so $p=3$. We get the solutions $n=1, p=3$ and $n=27$, $p=3$.
Cubing the initial equation ..... 1 point
Writing the cubed equation as $n+3 p k^{2}+\frac{p^{3}}{n}=k^{6}$ and determining that $n$ divides $p^{3}$.. 1 point
Writing all possible values for $n$. ...................................................................................................
Determining that for $n=p$ or $n=p^{2}$, no solution is obtained $\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point
Writing both solutions when $n=1$ or $n=p^{3}$ ..... 2 points
(If the contestant only writes both solutions, award 1 point.)

2nd method. Let $\sqrt[3]{n}+\frac{p}{\sqrt[3]{n}}=k^{2}$, where $k$ is a natural number. We multiply the equation by $k^{2} \sqrt[3]{n}$ or $\sqrt[3]{n^{2}}$, to get the equations $k^{2} \sqrt[3]{n^{2}}+k^{2} p=k^{4} \sqrt[3]{n}$ and $n+p \sqrt[3]{n}=k^{2} \sqrt[3]{n^{2}}$. Adding these two equations and rearranging slightly, we get $n+k^{2} p=\left(k^{4}-p\right) \sqrt[3]{n}$. From this, it follows that $n$ must be a cube of a natural number. Substituting $n=m^{3}$, where $m$ is a natural number, into the initial equation, we get $m+\frac{p}{m}=k^{2}$. From this, it follows that $m$ divides $p$. Since $p$ is a prime number, $m$ must be $1$ or $m=p$. In both cases, we get the equation $1+p=k^{2}$, thus $p=(k-1)(k+1)$. From this, it follows that $k=2$, so $p=3$. We get the solutions $n=1, p=3$ and $n=27, p=3$.

Rearranging $\mathbf{v} n+k^{2} p=\left(k^{4}-p\right) \sqrt[3]{n} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

Concluding that $n$ is a cube of a natural number ......................................................................

Writing the equation $m+\frac{p}{m}=k^{2}$ and determining that $m$ divides $p \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

Writing both possibilities for $m$ and both solutions ................................ 2 points

(If the contestant only writes both solutions, award 1 point.)