The original problem of IMO $1985$ was as follows-
Given any $1985$ positive integers all of which contains no prime factor $>23,$ prove that you can find four of them whose multiple is a perfect fourth power.
I solved this problem earlier.There are total $9$ primes $≤23$.Then every given integer is of the form $2^a3^b5^c7^d11^e13^f17^g19^h23^k$. Now consider the ordered $9$-tuple of the powers of the $9$ primes corresponding to every given numbers whose coordinates are either $0$ or $1$ accordingly as the power is even or odd. Then there are total $2^9=512$ such $9$-vectors.So by Pigeon Hole Principle there are two vectors exactly with the same configuration $(1985>512)$. The multiplication of these two numbers is a perfect square.We can have more than $512$ such pairs and among these pairs,e.i.among these perfect squares we can have similarly two numbers with exactly the same pre-defined vector configuration. Then their multiplication is a perfect fourth power.It is easy to check that the number $1985$ can be replaced by $1537$.
Now, this problem can be generalized as below:
Let $N(m,n,p)=$ the smallest positive integer $k$ such that for any given set of $k$ positive integers with no prime factor $>m$ such that there exist $n$ out of them whose multiple is a perfect $p$th power.
In the above problem we just proved that $N(23,4,4)=1537$.We can even prove by similar argument that $N(m,4,4)=(2^p+1)2+2^p-1$ where $p=π(m)$.
Now my questions are what is the explicit/most general formula,can it be calculated? Are there any asymptotic approximations?