Let $m$ and $n$ two given integers. Ana thinks of a pair of real numbers $x$, $y$ and then she tells Beto the values of $x^m+y^m$ and $x^n+y^n$, in this order. Beto's goal is to determine the value of $xy$ using that information. Find all values of $m$ and $n$ for which it is possible for Beto to fulfill his wish, whatever numbers that Ana had chosen. 
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 \( t > 0 \).