## Task 22/88

Determine all pairs $(p ; n)$, where $p$ is a prime number and $n$ is a natural number, for which the solutions of the equation

$$
x^{2}+2\left(p^{n}+2\right) x+p^{2 n}=0
$$

are integers!
If the equation has integer solutions $x_{1}, x_{2}$, then by Vieta's root theorem, we have

$$
-\left(x_{1}+x_{2}\right)=2 p^{n}+4 ; \quad x_{1} x_{2}=p^{2 n}
$$

From the second equation, it follows that $x_{1}$ and $x_{2}$ have the same sign, and from the first equation, that $x_{1} ; x_{2}<0$. Therefore, by the uniqueness of the prime factorization, $x_{1}=-p^{k} ; x_{2}=-p^{l}$ with $k ; l \in \mathbb{N}, k+l=2 n$.

Without loss of generality, let $l \leq k$. Then we have (since $p^{l} \neq 0$)

$$
p^{k}+p^{l}=2 p^{n}+4 \quad(*) ; \quad p^{k-l}+1=2 p^{n-l}+\frac{4}{p^{l}}
$$

From the integrality of the powers of $p$, it follows that $p^{l} \mid 4$. Therefore, there are three cases to consider:

1. $p^{l}=1$, so $l=0, k=2 n$. The equation $(*)$ takes the form $p^{2 n}+1=2 p^{n}+4$. The solution is $p^{n}=3$, thus $p=3, n=1$. The verification confirms the correctness of this solution.
2. $p^{l}=2$, so $l=1, p=2, k=2 n-1$. From the equation $(*)$ we get $p^{2 n-1}+2=2 p^{n}+4$. This equation has no integer solutions for $p^{n}$.
3. $p^{l}=4$, so $p=2, l=2, k=2 n-2$. We obtain analogously $2^{2 n-2}+4=2 \cdot 2^{n}+4$, thus $n=3$. This solution is also confirmed by the verification.

Since the case distinction is complete, the solutions are exactly the pairs $(3 ; 1)$ and $(2 ; 3)$.