Each rational number is painted either white or red. Call such a coloring of the rationals [i]sanferminera[/i] if for any distinct rationals numbers $x$ and $y$ satisfying one of the following three conditions: [list=1][*]$xy=1$,
[*]$x+y=0$,
[*]$x+y=1$,[/list]we have $x$ and $y$ painted different colors. How many sanferminera colorings are there?
To solve the problem, we need to determine the number of valid colorings of the rational numbers such that for any distinct rational numbers \( x \) and \( y \) satisfying one of the conditions \( xy = 1 \), \( x + y = 0 \), or \( x + y = 1 \), \( x \) and \( y \) are painted different colors. Let's proceed step-by-step.

1. **Define the Set \( S \)**:
   Let \( S \) be the set of rational numbers reachable from \( 0 \) by applying the operations \( x \mapsto \frac{1}{x} \), \( x \mapsto -x \), and \( x \mapsto 1 - x \).

2. **Reachability from \( 0 \)**:
   We need to show that all rational numbers can be reached from \( 0 \) using these operations. We start with \( 0 \) and apply the operations iteratively.

3. **Subgroup Analysis**:
   Consider the subgroup \( \langle a \rangle \subset F \) where \( F \) is the free group generated by the operations. If \( q \in S \), then \( q + \mathbb{Z} \in S \). This implies that if \( 0 \in S \), then \( \mathbb{Z} \in S \), and consequently, \( \left\{\frac{1}{n} \mid n \in \mathbb{Z}^{\times}\right\} \in S \).

4. **Euclidean Algorithm and Induction**:
   We use the Euclidean algorithm to show that all rationals can be reached. We prove this by strong induction on \( M(q) = \max(|a|, |b|) \) where \( q = \frac{a}{b} \) and \( (a, b) = 1 \).

   - **Base Case**: When \( q = 0 \), we have reached the base case.
   - **Inductive Step**: For \( q \neq 0 \), one of the sets \( \left(q + \mathbb{Z}, \frac{1}{q} + \mathbb{Z}\right) \) contains an element with a strictly smaller \( M \)-value. Each rational number has a finite \( M \)-value and hence a finite number of steps away from \( 0 \).

5. **Coloring Analysis**:
   We have established that all rationals can be reached from \( 0 \). Now, we need to analyze the parity of the path connecting two arbitrary rationals. 

   - **Parity Preservation**: The operation \( a \) preserves parity when acting on all rationals except for \( 0 \), while \( a^{-1} \) preserves parity when acting on all rationals except for \( 1 \).
   - **Homomorphism to \( C_2 \)**: Consider the homomorphism from \( F \) to \( C_2 \) defined by the generators \( a = 1, b^2 = 1 \). This homomorphism helps us determine the parity of an element of \( F \).

6. **Conclusion**:
   By considering the parity and the homomorphism, we conclude that there are exactly two valid colorings of the rational numbers. These two colorings correspond to the two possible ways to color \( 0 \) and \( 1 \) differently and extend this coloring to all rationals while maintaining the required conditions.

The final answer is \( \boxed{2} \)