Balti Tee 1995 · Ülesanne 13

Solution:

Suppose that there is an $n$ such that the first player always wins if there are initially more than $n$ pebbles. Consider the initial situation with $n^{2}+n+1$ pebbles. Since $(n+1)^{2}>n^{2}+n+1$, the first player can take at most $n^{2}$ pebbles, leaving at least $n+1$ pebbles on the table. By the assumption, the second player now wins. This contradiction proves that there are infinitely many situations in which the second player wins no matter how the first player plays.