Given a square table $n\times n$, two players $A$ and $B$ are playing the following game: At the beginning all cells of the table are empty, and the players alternate playing with coins. Player $A$ has the first move, and in each of the moves a player will put a coin on some of the cells that doesn't contain a coin and is not adjacent to any of the cells that already contains a coin. The player who makes the last move is the winner. Which player has a winning strategy and what is the strategy?
Remark. The cells are adjacent if they share an edge.
The way the question is worded I assume that the same player will have the winning strategy regardless of the size of the $n\times n$ board. But this doesn't make sense because if we look at the simplest cases, for a $1 \times 1$ board the first player will always win, for a $2 \times 2$ board the second player will always win, for a $3 \times 3$ board the first player has the winning strategy, the strategy here is that player one will always win unless his in his first move he places a coin in a cell with exactly three adjacent coins.
I wasn't able to find the winning strategy for a $4 \times 4$ board yet but just from trying a few examples it looked like player 2 has the winning strategy so it looks like if $n$ is odd player 1 has the winning strategy and if $n$ is even then player 2 has the winning strategy.
Is this correct and how can I figure out a winning strategy? I tried to look at the number of open cells the player with the winning strategy should leave open at the end of their turn but I could not find a pattern there.