For what prime numbers $p$ and $q$ and natural number $n$ does the equation

$$
\sqrt[n]{p+q}=p-q
$$

hold?
From the equation, on one hand $p>q$, and on the other hand

$$
p+q=(p-q)^{n}
$$

On the left side, $(p-q)+2q$ stands, therefore

$$
(p-q) \mid 2q
$$

Since $q$ is prime, the divisors of $2q$ are only $1, 2, q$, and $2q$. If $p-q=1$, then in (1) $p+q=1$, which is impossible. If $p-q=q$ or $p-q=2q$, then $p=2q$ or $p=3q$, so $p$ is not a prime number. Thus, only

$$
p-q=2
$$

is possible, meaning $p$ and $q$ are twin primes.

Solving the system of equations (1) and (2)

$$
\begin{aligned}
& p=2^{n-1}+1 \\
& q=2^{n-1}-1
\end{aligned}
$$

Since among the numbers $2^{n-1}-1, 2^{n-1}, 2^{n-1}+1$, exactly one is divisible by $3$ and this is not the middle $2^{n-1}$, one of $p$ and $q$ must be divisible by $3$, and thus can only be prime if it is $3$. If $p=3$, then $q=1$, which is not prime; if $q=3$, then $p=5$.

From equation (1) in this case,

$$
8=p+q=(p-q)^{n}=2^{n}
$$

from which $n=3$.

Thus, the only solution to the problem is

$$
p=5, q=3, n=3
$$

Dombi Gergely (Pécs, JPTE I. sz. Gyak. Ált. Isk., 6. o. t.)