Each pair $(x, y)$ of nonnegative integers is assigned number $f(x, y)$ according the conditions:
$f(0, 0) = 0$;
$f(2x, 2y) = f(2x + 1, 2y + 1) = f(x, y)$,
$f(2x + 1, 2y) = f(2x, 2y + 1) = f(x ,y) + 1$ for $x, y \ge 0$.
Let $n$ be a fixed nonnegative integer and let $a$, $b$ be nonnegative integers such that $f(a, b) = n$. Decide how many numbers satisfy the equation $f(a, x) + f(b, x) = n$.
1. **Understanding the function \( f(x, y) \)**:
   - The function \( f(x, y) \) is defined recursively with the following properties:
     - \( f(0, 0) = 0 \)
     - \( f(2x, 2y) = f(2x + 1, 2y + 1) = f(x, y) \)
     - \( f(2x + 1, 2y) = f(2x, 2y + 1) = f(x, y) + 1 \)
   - This function counts the number of positions in the binary representations of \( x \) and \( y \) where the binary digits are unequal.

2. **Analyzing the given condition \( f(a, b) = n \)**:
   - Let \( a \) and \( b \) be nonnegative integers such that \( f(a, b) = n \).
   - This means there are exactly \( n \) positions in the binary representations of \( a \) and \( b \) where the digits differ.

3. **Condition \( f(a, x) + f(b, x) = n \)**:
   - We need to find how many nonnegative integers \( x \) satisfy \( f(a, x) + f(b, x) = n \).
   - Consider the binary representation of \( a \), \( b \), and \( x \).

4. **Binary representation analysis**:
   - If the binary digits of \( a \) and \( b \) are different at a given position, this position contributes 1 to \( f(a, b) \) and 1 to \( f(a, x) + f(b, x) \) regardless of the binary digit of \( x \) at this position.
   - If the binary digits of \( a \) and \( b \) are equal at a given position, this position contributes 0 to \( f(a, b) \) and either 0 or 2 to \( f(a, x) + f(b, x) \) depending on the value of the binary digit of \( x \) at this position.

5. **Conclusion**:
   - For \( f(a, x) + f(b, x) = n \) to hold, at each position:
     - If the binary digits of \( a \) and \( b \) are equal, then the binary digit of \( x \) must be the same as those of \( a \) and \( b \).
     - If the binary digits of \( a \) and \( b \) are unequal, then the binary digit of \( x \) can be any value (0 or 1).
   - Since there are exactly \( n \) positions where the binary digits of \( a \) and \( b \) are unequal, there are \( 2^n \) possible values for \( x \).

\[
\boxed{2^n}
\]