We can easily construct the case when the side of the square is $2+2\sqrt{2}$. We will prove that this is the smallest length possible. Assume there exists a square ABCD with length $x,x<2+2\sqrt{2}$ which can contain five unit circles $(O_1), (O_2), (O_3), (O_4), (O_5)$ of radius 1 with no two of them have a commen interior point. Construct the square EFGH with side of $x-2$ like the figure so that the distances between $EF,FG,GH,HE$ to $AB,BC,CD,DA$ respectively are 1. Then, $O_1,O_2,O_3,O_4,O_5$ must be inside of this square (because for example, if $O_1$ is outside of the square EFGH, then $(O_1)$ wouldn't be completely inside of $ABCD$) Note: We also consider the case of a point lies on the side of a square as inside. We divide $EFGH$ into 4 equal squares like in the figure. By the Pigeonhole principle, one of these squares must contain at least 2 of the 5 points $O_1,O_2,O_3,O_4,O_5$. Without loss of generality, assume that $O_1$ and $O_2$ are inside the square $EIML$. Because $(O_1)$ and $(O_2)$ don't have a common interior point, $O_1O_2\geq 2$. But because $O_1$ and $O_2$ are inside of $EIML$, so $O_1O_2\leq EM=\sqrt{2}.\frac{(x-2)}{2}< \sqrt{2}.\sqrt{2}=2$ (contradiction) Hence, the smallest possible side of a square satisfying the condition is $2+2\sqrt{2}$

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