Find the smallest possible side of a square in which five circles of radius $1$ can be placed, so that no two of them have a common interior point.
1. **Construct the Case**: We start by constructing a square with side length \(2 + 2\sqrt{2}\). We will show that this is the smallest possible side length that can contain five unit circles without any two circles having a common interior point.

2. **Assume a Smaller Square**: Assume there exists a square \(ABCD\) with side length \(x\), where \(x < 2 + 2\sqrt{2}\), that can contain five unit circles \((O_1), (O_2), (O_3), (O_4), (O_5)\) of radius 1, such that no two circles have a common interior point.

3. **Construct Inner Square**: Construct a smaller square \(EFGH\) inside \(ABCD\) with side length \(x-2\). The distances between the sides of \(EFGH\) and \(ABCD\) are 1 unit each. This means that the circles must be completely inside \(EFGH\) because if any circle were outside \(EFGH\), it would not be completely inside \(ABCD\).

4. **Divide Inner Square**: Divide the square \(EFGH\) into four equal smaller squares. Each smaller square has a side length of \(\frac{x-2}{2}\).

5. **Apply Pigeonhole Principle**: By the Pigeonhole Principle, at least one of these four smaller squares must contain at least two of the five circle centers \(O_1, O_2, O_3, O_4, O_5\). Without loss of generality, assume that \(O_1\) and \(O_2\) are inside the smaller square \(EIML\).

6. **Distance Between Centers**: Since the circles \((O_1)\) and \((O_2)\) do not have a common interior point, the distance between their centers must be at least 2 units, i.e., \(O_1O_2 \geq 2\).

7. **Maximum Distance in Smaller Square**: The maximum distance between any two points inside the smaller square \(EIML\) is the length of its diagonal, which is \(\sqrt{2} \cdot \frac{x-2}{2}\). Therefore, \(O_1O_2 \leq \sqrt{2} \cdot \frac{x-2}{2}\).

8. **Derive Contradiction**: Since \(x < 2 + 2\sqrt{2}\), we have:
   \[
   \sqrt{2} \cdot \frac{x-2}{2} < \sqrt{2} \cdot \sqrt{2} = 2
   \]
   This implies \(O_1O_2 < 2\), which contradicts the requirement that \(O_1O_2 \geq 2\).

9. **Conclusion**: Therefore, the assumption that \(x < 2 + 2\sqrt{2}\) is false. Hence, the smallest possible side length of the square that can contain five unit circles without any two circles having a common interior point is \(2 + 2\sqrt{2}\).

\(\blacksquare\)

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