Problem 1592

1. Identify the problem and the goal:
- We are given two integers $m$ and $n$.
- Ana provides Beto with the values of $x^m + y^m$ and $x^n + y^n$.
- Beto's goal is to determine $xy$ from this information.
- We need to find all pairs $(m, n)$ for which Beto can always determine $xy$.

2. **Case 1: Both $m$ and $n$ are even:**
- Consider $(x, y) = (1, 1)$ and $(x, y) = (1, -1)$.
- For both pairs, $x^m + y^m$ and $x^n + y^n$ will be the same because $m$ and $n$ are even.
- However, $xy$ will be different: $1$ for $(1, 1)$ and $-1$ for $(1, -1)$.
- Therefore, Beto cannot determine $xy$ uniquely if both $m$ and $n$ are even.

3. **Case 2: Both $m$ and $n$ are odd:**
- Consider $(x, y) = (1, -1)$ and $(x, y) = (0, 0)$.
- For both pairs, $x^m + y^m$ and $x^n + y^n$ will be different because $m$ and $n$ are odd.
- Therefore, Beto can determine $xy$ uniquely if both $m$ and $n$ are odd.

4. **Case 3: One of $m$ or $n$ is even:**
- Without loss of generality, let $n = 2p$ be even.
- We need to show that $m$ must divide $p$ for Beto to determine $xy$ uniquely.
- Suppose $m$ does not divide $p$. We will show the existence of pairs $(x_1, y_1)$ and $(x_2, y_2)$ such that $x_1^{2p} + y_1^{2p} = x_2^{2p} + y_2^{2p}$ and $x_1^m + y_1^m = x_2^m + y_2^m$, but $x_1 y_1 \neq x_2 y_2$.

5. **Constructing the function $f(x)$:**
- Consider the function $f(x) = (1-x)^{\frac{2p}{m}} + (1+x)^{\frac{2p}{m}}$.
- If $m$ does not divide $p$, this function has an extremum at $x = 1$.
- Therefore, there exist $c_1 < 1 < c_2$ such that $f(c_1) = f(c_2)$.
- Let $a = f(c_1)$. Then, $(x_1, y_1) = ((1-c_1)^{\frac{1}{q}}, (1+c_1)^{\frac{1}{q}})$ and $(x_2, y_2) = ((1-c_2)^{\frac{1}{q}}, (1+c_2)^{\frac{1}{q}})$ satisfy the conditions.

6. **Conclusion for $m$ and $n$:**
- Therefore, $m = 2k+1$ and $n = 2t(2k+1)$ for $k, t \in \mathbb{Z}$ and $t > 0$.
- If $t \leq 0$, we can use the same argument: extremum at $1$. So, $t > 0$.

7. Verification:
- To prove that such $(m, n)$ work, consider $x^m + y^m = a \implies x = \left(\frac{a}{2} - t \right)^{\frac{1}{2k+1}}, y = \left(\frac{a}{2} + t \right)^{\frac{1}{2k+1}}$ for some $t \in \mathbb{R}$.
- The function $g$ (replace $1$ with $\frac{a}{2}$ in $f$) has the property that $g(t_1) = g(t_2) \implies t_1 = \pm t_2$.
- This implies a fixed value of $xy$. So, we are done.

The final answer is $\boxed{ (m, n) = (2k+1, 2t(2k+1)) }$ for $k, t \in \mathbb{Z}$ and $t > 0$.