## Task $2 / 87$

Determine all solutions of the system of equations

$$
p_{1}+p_{2}=p_{3}^{m} \quad ; \quad p_{1}-p_{2}=p_{3}^{n}
$$

where the $p_{i}(i=1 ; 2 ; 3)$ are prime numbers and $m$ and $n$ are natural numbers. What values can $m$ and $n$ take?
By adding the two given equations, we obtain

$$
2 p_{1}=p_{3}^{n}\left(p_{3}^{m-n}+1\right)
$$

(due to $p_{1}+p_{2}>p_{1}-p_{2}$, it is certain that $m-n>0$). From the uniqueness of the prime factorization and the prime property of $p_{1}$, there are exactly three possibilities:

1. $p_{3}^{n}=p_{1}$; it follows that $p_{2}=0$, which contradicts $p_{2}>0$.
2. $p_{3}^{n}=1$. Since $p_{3} \geq 1$, it follows that $n=0$ and thus $p_{1}-p_{2}=1$, so $p_{1}=3 ; p_{2}=2$ and from the first equation $p_{1}+p_{2}=5=p_{3}^{m}$, hence $p_{3}=5, m=1$.
3. $p_{3}^{n}=2, p_{3}=2, n=1, p_{1}=2^{m-1}+1$. From the second equation, it follows that $p_{2}=2^{m-1}-1$. Among the three consecutive natural numbers $2^{m-1}-1, 2^{m-1}$, and $2^{m-1}+1$, exactly one is divisible by 3; it is certainly not $2^{m-1}$. If $p_{1}=3$, then $p_{2}=1$, which contradicts $p_{2} \geq 2$. Therefore, $p_{2}=3, p_{1}=5, m=3$.

Thus, there are exactly two solutions: $p_{11}=3, p_{21}=2, p_{31}=5$ with $m_{1}=1, n_{1}=0$ and $p_{12}=5, p_{22}=$ $3, p_{32}=2$ with $m=3, n=1$. The verification confirms the correctness, and the solution method excludes further solutions.