Consider the following challenge between two players A and B. They are given the initial terms $a_0= 3^{2014}$ and $b_0= 15^{4028}$ of two sequences, and the scope is to reach $1$ before the other, and this can be done dividing the preceding term by a prime divisor or taking the square root of it if it is a perfect square (e.g. from $81$ one can get to $27$ but also to $9$) . If the two players are optimal and do one of the operations once each, how to prove that if A is the first to play, then A will win out of $a_n$ and B will out of $b_n$?
What confuses me is the optimality of the players. I have guessed that it means both of them leave an even number to the opponent, in view of, e.g., $3^2 \rightarrow 3 \rightarrow 1.$ With this constraint, indeed A won out of $a_n$. However I (hopefully) must be missing something.
In the first case, the numbers will always be $3^k$, starting with $3^{2014}$. $A$ can ensure a victory by always diving the given number by 3. That way $B$ will always get a number of the form $3^{2k+1}$, which is not a square, and will have to return a $3^{2k}$. Repeat until $A$ gets 81. Then square root to pass a 9 to $B$, get a 3 in return, and claim a win.
In the second case, the number is $3^k5^k$, so $A$ does have an option. If he square roots, $B$ should do the same, and then we have $3^{1007}5^{1007}$. Then (or otherwise), $A$ will return a number $3^k 5^l$, where $k, l$ have different parity. $B$ then removes a factor from the even exponent, and $A$ will always have a number of the form $3^{2a+1}5^{2b+1}$, until eventually one of the factors is finished, and $B$ gets to start in a situation of a prime to an even power, which was the first case of the problem.