Show that there are infinitely many winning positions for the second player

Question

The following solution is not my piece of work. It's a provided solution in 104 Number Theory Problems by Titu Andreescu which I miserably failed to understand. I've written my doubts at the end. Kindly explain me.

QUESTION :

Consider the following two-person game. A number of peb- bles are lying on a table. Two players make their moves alternately. A move consists in taking off the table $x$ pebbles, where $x$ is the square of any positive integer. The player who is unable to make a move loses. Prove that there are infinitely many initial situations in which the player who goes second has a winning strategy.

---

SOLUTION :

Assume to the contrary that there are only finitely many initial situations in which the player who goes second has a winning strategy. Under our assumption, there exists a positive integer $N$ such that if there are $n > N$ pebbles on the table, the player who goes first at the moment has a winning strategy.

Consider the initial situation with $(N + 1)^2 − 1$ pebbles on the table. Let $P_1$ and $P_2$ denote the players who go first and second, respectively. By our assumption, $P_1$ has a winning strategy, which requires $P_1$ to remove $x$ pebbles on his first move. It is clear that $x \neq N^2$, and $P_2$ is left with at least $(N + 1)^2 − 1 − N^2 = 2N > N$ pebbles to make the first move.

By our assumption, at this moment, $P_2$ has a winning strategy. But it is impossible for both players to have a winning strategy for the same initial situation. Hence our original assumption was wrong and there are infinitely many initial situations in which the player who goes second has a winning strategy.

---

MY DOUBTS:—

1. At some point they say that $x\neq N^2$, but why? It may or may not be equal to $N^2$ but it's nothing strict apparently.
2. At the end of the second paragraph, it says that there are $>2N$ pebbles for $\mathcal{P_2}$ to choose and at the third paragraph (beginning), it claims: By our assumption, at this moment, $P_2$ has a winning strategy. And I fall to understand what's the benefit of the $"> 2N"$ fact and where from comes the assumption that $P_2$ wins.

Answer 1

I don't understand why the first move couldn't be $N^2$, but I don't see a problem with that being the first move. The important point is that the first move couldn't be greater than $N^2$.

Suppose the claim was false. That is equivalent to saying that beyond some point only the first player has a winning strategy. Let $N$ be that point, we will derive a contradiction.

Toward that end, consider the arrangement with $(N+1)^2-1$ pebbles. That is greater than $N$ so, by assumption, the first player must have a winning strategy. But what could that strategy be? Clearly the first player can remove at most $N^2$ pebbles. But that would leave player $2$ with at least $(N+1)^2-1-N^2=2N>N$ pebbles. Thus player $2$ would now be the first player in a game with more than $N$ pebbles and thereby would have a winning strategy. A contradiction.

Answer 2

It should be $x\le N^2$ for the first of your comments.

For the second part, $N$ is chosen so that if there are more than $N$ pebbles left, the first player has a winning strategy.

The strategy of the poof is as follows:

(i) If there are finitely many second player wins there is a greatest. Call it $N$

(ii) From any larger position than this, the first player wins

(iii) but we can construct a position where the first player has no move to reduce the position to $N$ or fewer

(iv) the position after the first move is greater than $N$ and is therefore a win for the next player to move (by (i))

(v) But that means the second player can win the position we constructed

(vi) and that is a contradiction with (i), so we can't have a finite number of second player wins