## Task 31/78

Determine all prime numbers $p_{1} ; p_{2} ; p_{3}$ that satisfy the equation $p_{1}^{p_{2}}+p_{2}^{p_{1}}=p_{3}$.
Certainly, here is the translation of the provided text into English, preserving the original formatting:

---

Certainly is $p_{3}$ not the smallest of the sought prime numbers and thus not even. It follows that either $p_{1}$ or $p_{2}$ is even and therefore equal to 2. Due to the symmetry in $p_{1}$ and $p_{2}$, we can set $p_{1}=2$ without loss of generality. The given equation then takes the form

$$
2^{p_{2}} + p_{2}^{2} = p_{3}
$$

Since $p^{2} \geq 3$, it follows that $2^{p_{2}} + p_{2}^{2} \geq 2^{3} + 3^{2} = 17$. Clearly, this already provides a solution: $p_{1}=2, p_{2}=3, p_{3}=17$.

The question arises whether there are any other solutions. For $p_{2} > 3$, which is odd, we can set $p_{2} = 2d + 1$ (with $d \in \mathbb{N}, d > 1$). Then,

$$
2^{p_{2}} = 2^{2d+1} = 2 \cdot 2^{2d} = 2 \cdot 4^{d} \equiv 2 \cdot 1^{d} \equiv 2 \pmod{3}
$$

Furthermore, $p_{2} \equiv \pm 1 \pmod{3}$, so $p_{2}^{2} \equiv 1 \pmod{3}$. Therefore,

$$
p_{3} = 2^{p_{2}} + p_{2}^{2} \equiv 2 + 1 \equiv 0 \pmod{3}
$$

which contradicts the prime property of $p_{3}$. Thus, the found solution is the only one.