239

Fsociety wrote:

239

Its an open Olympiad held by a school named 239 in Saint Petersburg. The full name is "Open Math Olympiad of 239 Presidential Physics and Mathematics Lyceum in Saint-Petersburg".

Suppose the chips in the corners of the initial arrangement are in set $A$, the chips on the side but not in the corners of the initial arrangement are in set $B$, and the non-bordering chips are in set $C$. Then the chips in set $A$ had two neighbors, the chips in set $B$ had three neighbors, and the chips in set $C$ had four neighbors. In the new arrangement, each type of chip must be in the same region (corners, side but not corners, and non-border) as they were in the initial arrangement. We then know that \begin{align*} |A| &= 4, \\ |B| &= 32, \\ |C| &= 64. \end{align*}In the new arrangement, the set $C$ chips must be arranged so that their perimeter and area stays the same. Their area is $64$, and their perimeter must be $32$, since the set $B$ chips border the set $C$ rectangle. So if $a$ and $b$ are the dimensions of the new set $C$ rectangle, we have \begin{align*} 2a+2b &= 32, \\ ab &= 64. \end{align*}There is only one solution $(a,b)$ to this system, namely $(8,8)$, which forms a square. Because the set $C$ chips form a square, the rest of the arrangement must also be a square, so we are done.