$A, B$ are reals. Find a necessary and sufficient condition for $Ax + B[x] = Ay + B[y]$ to have no solutions except $x = y$.
To find a necessary and sufficient condition for \( Ax + B[x] = Ay + B[y] \) to have no solutions except \( x = y \), we will analyze the given equation step by step.

1. **Rewrite the given equation:**
   \[
   Ax + B[x] = Ay + B[y]
   \]
   Subtract \( Ay + B[y] \) from both sides:
   \[
   Ax + B[x] - Ay - B[y] = 0
   \]
   Factor out common terms:
   \[
   A(x - y) + B([x] - [y]) = 0
   \]

2. **Consider the case when \( A = 0 \):**
   \[
   B([x] - [y]) = 0
   \]
   Since \( B \neq 0 \) (otherwise the equation is trivial), we have:
   \[
   [x] = [y]
   \]
   This implies \( x \) and \( y \) are in the same integer interval, but not necessarily equal. For example, \( x = 0 \) and \( y = \frac{1}{2} \) are solutions with \( x \neq y \). Therefore, \( A = 0 \) is not a sufficient condition.

3. **Consider the case when \( A \neq 0 \):**
   \[
   A(x - y) = -B([x] - [y])
   \]
   Divide both sides by \( A \):
   \[
   x - y = -\frac{B}{A}([x] - [y])
   \]
   Let \( n = [y] - [x] \), where \( n \) is an integer:
   \[
   x - y = -\frac{B}{A}n
   \]
   Rearrange to find \( y \):
   \[
   y = x + \frac{B}{A}n
   \]

4. **Analyze the condition for \( y = x \):**
   For \( y = x \) to be the only solution, \( \frac{B}{A}n \) must be zero for all integers \( n \) except \( n = 0 \). This implies:
   \[
   \left| \frac{B}{A} \right| \ge 1
   \]
   Therefore:
   \[
   |A + B| \ge |A|
   \]
   Since \( A \neq 0 \), we also have:
   \[
   |A| > 0
   \]

5. **Verify the sufficiency:**
   If \( |A + B| \ge |A| \) and \( |A| > 0 \), then:
   \[
   \left| \frac{A + B}{A} \right| \ge 1
   \]
   This implies:
   \[
   \left| 1 + \frac{B}{A} \right| \ge 1
   \]
   Therefore, the only possible solution for \( n \) is \( n = 0 \), which means \( x = y \).

Thus, the necessary and sufficient condition for \( Ax + B[x] = Ay + B[y] \) to have no solutions except \( x = y \) is \( |A + B| \ge |A| > 0 \).

The final answer is \( \boxed{|A + B| \ge |A| > 0} \)