[list=a]
[*]Determine all real numbers $x{}$ satisfying $\lfloor x\rfloor^2-x=-0.99$.
[*]Prove that if $a\leqslant -1$, the equation $\lfloor x\rfloor^2-x=a$ does not have real solutions.
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### Part (a)
Determine all real numbers \( x \) satisfying \( \lfloor x \rfloor^2 - x = -0.99 \).

1. Let \( \lfloor x \rfloor = n \), where \( n \) is an integer. Then we have:
   \[
   n^2 - x = -0.99 \implies x = n^2 + 0.99
   \]

2. Since \( \lfloor x \rfloor = n \), it follows that:
   \[
   n \leq x < n+1
   \]

3. Substituting \( x = n^2 + 0.99 \) into the inequality:
   \[
   n \leq n^2 + 0.99 < n+1
   \]

4. We need to solve the inequalities separately:
   - For the left inequality:
     \[
     n \leq n^2 + 0.99 \implies 0 \leq n^2 - n + 0.99
     \]
     This is always true since \( n^2 - n + 0.99 \) is a quadratic expression that is always non-negative for all integers \( n \).

   - For the right inequality:
     \[
     n^2 + 0.99 < n + 1 \implies n^2 - n + 0.99 < 1 \implies n^2 - n - 0.01 < 0
     \]

5. Solve the quadratic inequality \( n^2 - n - 0.01 < 0 \):
   \[
   n = \frac{1 \pm \sqrt{1 + 4 \cdot 0.01}}{2} = \frac{1 \pm \sqrt{1.04}}{2}
   \]
   \[
   n = \frac{1 \pm 1.02}{2}
   \]
   \[
   n = 1.01 \quad \text{or} \quad n = -0.01
   \]

6. Since \( n \) must be an integer, the possible values for \( n \) are \( n = 0 \) or \( n = 1 \).

7. Check each case:
   - For \( n = 0 \):
     \[
     x = 0^2 + 0.99 = 0.99
     \]
     \[
     \lfloor 0.99 \rfloor = 0 \quad \text{and} \quad 0^2 - 0.99 = -0.99
     \]
     This is a valid solution.

   - For \( n = 1 \):
     \[
     x = 1^2 + 0.99 = 1.99
     \]
     \[
     \lfloor 1.99 \rfloor = 1 \quad \text{and} \quad 1^2 - 1.99 = -0.99
     \]
     This is a valid solution.

Therefore, the solutions to the equation are \( x = 0.99 \) and \( x = 1.99 \).

### Part (b)
Prove that if \( a \leq -1 \), the equation \( \lfloor x \rfloor^2 - x = a \) does not have real solutions.

1. Assume \( \lfloor x \rfloor = n \), where \( n \) is an integer. Then we have:
   \[
   n^2 - x = a \implies x = n^2 - a
   \]

2. Since \( \lfloor x \rfloor = n \), it follows that:
   \[
   n \leq x < n+1
   \]

3. Substituting \( x = n^2 - a \) into the inequality:
   \[
   n \leq n^2 - a < n+1
   \]

4. We need to solve the inequalities separately:
   - For the left inequality:
     \[
     n \leq n^2 - a \implies a \leq n^2 - n
     \]

   - For the right inequality:
     \[
     n^2 - a < n + 1 \implies a > n^2 - n - 1
     \]

5. Combine the inequalities:
   \[
   n^2 - n - 1 < a \leq n^2 - n
   \]

6. Since \( a \leq -1 \), we have:
   \[
   n^2 - n - 1 < -1 \leq a \leq n^2 - n
   \]
   \[
   n^2 - n - 1 < -1 \implies n^2 - n < 0
   \]

7. The quadratic inequality \( n^2 - n < 0 \) holds for \( 0 < n < 1 \). However, \( n \) must be an integer, so there are no integer solutions for \( n \).

Therefore, if \( a \leq -1 \), the equation \( \lfloor x \rfloor^2 - x = a \) does not have real solutions.

\(\blacksquare\)

The final answer is \( \boxed{ x = 0.99 } \) and \( x = 1.99 \) for part (a), and no solutions for part (b).