Number of integers $n$ such that $n$ is divisible by every prime belonging to the interval $(1; \sqrt{n})$ -- is it finite? Consider such positive integer $n$ that has the following property: every integer $m$ such that $1 < m < n$ and $\gcd(m, n) = 1$ is a prime number. Is the number of such $n$'s finite?
The above property is equivalent to divisibility of $n$ by any prime in the interval $(1; \sqrt{n})$, because if $p \nmid n$, then $p^2$ is coprime to $n$, so $p^2 > n$ and conversely, if all the primes less than $\sqrt{n}$ are divisors of $n$, then in order for $m$ to be coprime to $n$ it has to have prime divisors greater than $\sqrt{n}$ and as $m<n$, it is the only prime divisor of $m$, and $m$ is prime
I don't know what to do next.
 A: There is only a finite number of such integers $n$. 
Suppose an integer $m$ were divisble by every prime up to $K$. Then it is easy to see that $m \ge \prod_{p \le K} p  \in \omega(K^4)$ [indeed $\theta(K/\log K)$ primes $p \geq 2$ so $m \ge \prod_{p \le K} p \doteq g(K) \in 2^{\Omega(K/\log K)}$] So it follows that  
$m \in 2^{\Omega(K/\log K)} \subset \omega(K^4)$.
However, if there exists such an $n$, then taking $K =\sqrt{n}$, we note that the integer $g(K)$ must satisfy $g(K) \le n = K^2$. But as we observed above $g(K) \in \omega(K^4)$, so there are only a finite number of $K$ s.t $g(K) \le K^2$. Thus as such an $n$ satisfies at least the conditions $n=K^2$ for some $K$ satisfying $g(K) \le K^2$, it follows that there are at most a finite number of $n$.
A: Let $p_1, p_2, p_3...p_k$ be all prime numbers in $(1, \sqrt{n}$). Therefore $$p_1p_2p_3\cdot\ldots\cdot p_k\mid n$$
In virtue of the Fundamental Theorem of Arithmetic, $$n=\prod^j_{i=1}p_i^{\alpha_i},$$ which minimizes when $\alpha_1,\alpha_2,...=1$ and $k=j$, i.e., when $$n=p_1p_2p_3\cdot\ldots\cdot p_k$$
Can you finish now?
