Problem 1658

To solve the problem, we need to find all triples $(p, x, y)$ where $p$ is a prime number and $x$ and $y$ are positive integers, such that both $x^{p-1} + y$ and $x + y^{p-1}$ are powers of $p$.

Let's denote:

$$x^{p-1} + y = p^a \quad \text{and} \quad x + y^{p-1} = p^b$$

for some positive integers $a$ and $b$.

### Case 1: $p = 2$

In this scenario, the equations become:

$$x^1 + y = 2^a \quad \text{and} \quad x + y^1 = 2^b$$

This simplifies to:

$$x + y = 2^a \quad \text{and} \quad x + y = 2^b$$

For these equations to hold simultaneously, it must be the case that $a = b$, which means $x + y = 2^k$ for some $k$.

Thus, for $p = 2$, the solutions are given by:

$$(p, x, y) = (2, n, 2^k - n) \quad \text{where} \quad 0 < n < 2^k$$

### Case 2: $p = 3$

For $p = 3$, we have:

$$x^2 + y = 3^a \quad \text{and} \quad x + y^2 = 3^b$$

We shall try small values of $x$ and check for integer $y$.

**Subcase $x = 2$:**

- $x^2 = 4$, so $4 + y = 3^a$.
- This implies $y = 3^a - 4$.

$x + y^2 = 3^b$:

- $2 + (3^a - 4)^2 = 3^b$.

Checking possible small values for $a$:

- $a = 1$ gives $y = 3 - 4 = -1$, which is not valid.
- $a = 2$ gives $y = 9 - 4 = 5$, and thus:
- $x + y^2 = 2 + 5^2 = 27 = 3^3$. This works, leading to a solution $(3, 2, 5)$.

**Subcase $x = 5$:**

- $x^2 = 25$, so $25 + y = 3^a$.
- This implies $y = 3^a - 25$.

$x + y^2 = 3^b$:

- $5 + (3^a - 25)^2 = 3^b$.

Checking for valid values for $a$:

- $a = 3$ gives $y = 27 - 25 = 2$, and thus:
- $x + y^2 = 5 + 2^2 = 9 = 3^2$. This works, leading to another solution $(3, 5, 2)$.

There are no other small values of $x$ yielding valid $y$ as a power of 3 sum.

### Conclusion

Considering all cases, the complete set of solutions is:

$$\boxed{(p, x, y) \in \{(3, 2, 5), (3, 5, 2)\} \cup \{(2, n, 2^k - n) \mid 0 < n < 2^k\}}$$