## Task 17/78

For the equation $x^{n}+p x-q=0, n \geq 2$, it is known that it has a positive rational solution and that $p$ and $q$ are prime numbers. Determine the solution as well as $p$ and $q$!
If $x_{0}$ is the positive rational solution of the given equation, then $x_{0}$ can be expressed in the form $x_{0}=\frac{r}{s}$ with $r ; s \in N$ and $g g T(r, s)=1$, and it holds that

$$
x_{0}^{n}+p x_{0}-y=\frac{r^{n}}{s^{n}}+p \frac{r}{s}-q=0 \quad \text { or } \quad r^{n}+p r s^{n-1}-q s^{n}=0
$$

Since $g g T(r, s)=1$, it follows that $r$ is a multiple of $s$, and it holds that $s=1$, thus $r=x_{0}$, and the equation reduces to

$$
x_{0}^{n}+p x_{0}-q=0 \quad ; \quad x_{0}=r \in N \quad ; \quad x_{0}\left(x_{0}^{n-1}+p\right)=q
$$

Since $x_{0}^{n-1}+p>1$, it follows from the prime property of $q$ that $x_{0}=1$. Thus, we get

$$
x_{0}^{n-1}+p=1+p=q
$$

The only consecutive prime numbers are $p=2$ and $q=3$. The test with $x_{0}=1, p=2, q=3$ confirms the correctness.