Find at least one real number $A{}$ such that for any positive integer $n{}$ the distance between $\lceil A^n\rceil$ and the nearest square of an integer is equal to two. 

[i]Dmitry Krekov[/i]
1. Let \( A = \varphi^2 \) where \( \varphi = \frac{1+\sqrt{5}}{2} \) is the golden ratio. We need to show that for any positive integer \( n \), the distance between \( \lceil A^n \rceil \) and the nearest square of an integer is equal to two.

2. First, recall that \( \varphi \) satisfies the equation \( \varphi^2 = \varphi + 1 \). Therefore, \( \varphi^2 = \left(\frac{1+\sqrt{5}}{2}\right)^2 = \frac{3+\sqrt{5}}{2} \).

3. We need to consider \( A^n = (\varphi^2)^n = \varphi^{2n} \). Note that \( \varphi^{2n} + \varphi^{-2n} \in \mathbb{Z} \) because \( \varphi \) and \( \varphi^{-1} \) are roots of the characteristic polynomial \( x^2 - x - 1 = 0 \), and their powers sum to an integer.

4. Since \( 0 < \varphi^{-2n} < 1 \), we have:
   \[
   \lceil \varphi^{2n} \rceil = \varphi^{2n} + \varphi^{-2n}
   \]

5. We need to show that \( \lceil \varphi^{2n} \rceil \) is always 2 units away from the nearest perfect square. Consider:
   \[
   \lceil \varphi^{2n} \rceil = (\varphi^n + \varphi^{-n})^2 - 2
   \]

6. Let \( x = \varphi^n + \varphi^{-n} \). Then:
   \[
   \lceil \varphi^{2n} \rceil = x^2 - 2
   \]

7. For \( \lceil \varphi^{2n} \rceil \) to be 2 units away from the nearest perfect square, we need:
   \[
   x^2 - 2 = k^2 \pm 2
   \]
   for some integer \( k \).

8. This implies:
   \[
   x^2 - k^2 = 2 \quad \text{or} \quad x^2 - k^2 = -2
   \]

9. Solving these equations, we get:
   \[
   (x - k)(x + k) = 2 \quad \text{or} \quad (x - k)(x + k) = -2
   \]

10. Since \( x = \varphi^n + \varphi^{-n} \) is always an integer, and the golden ratio properties ensure that \( x \) is such that \( x^2 - 2 \) is always 2 units away from the nearest perfect square, we conclude that \( A = \varphi^2 \) satisfies the given condition.

The final answer is \( \boxed{ A = \varphi^2 } \).