oVlad wrote: There are $n{}$ stones in a heap. Two players play the game by alternatively taking either 1 stone from the heap or a prime number of stones which divides the current number of stones in the heap. The player who takes the last stone wins. For which $n{}$ does the first player have a strategy so that he wins no matter how the other player plays? Fedor Ivlev https://math.stackexchange.com/questions/4043721/for-which-n-the-first-player-has-a-strategy-so-that-he-wins-no-matter-how-the

We claim that $n=4k$ are the only losing points in this game. We use strong induction on $n$. First, it's easy to check that $n=1,2,3,4$ works, so the base cases works. Now let's assume this is correct for all $n\leq4t$, we will prove that it is also correct for $n=4t+1,4t+2,4t+3,4t+4$. For $4t+1$ just take $1$ candy, for $4t+2$ take two candies, for $4t+3$ there would be a prime $p|4t+3$ satisfying $p\equiv 3\pmod{4}$, take $p$ candies. for $4t+4$ note that there isn't a prime that is a mutiple of $4$, so any move will send your opponent to a winning point. So $4t+1,4t+2,4t+3$ are winning points and $4t+4$ is a losing point, and we're done by strong induction.