TITLE: Prove for each positive integer $n$, there exists $n$ consecutive positive integers none of which is an integral power of a prime number
QUESTION [5 upvotes]: Prove for each positive integer $n$, there exists $n$ consecutive positive integers none of which is an integral power of a prime number.
I'm not getting a single idea of how to approach it. One I found was to show that $n$ consecutive integers should not be divisible by any prime but it didn't worked.
Thanks in advance.

REPLY [3 votes]: I see there is a lot of confusion about the idea described by wythagoras. Let's consider every natural number in the interval $[n!+2,n!+n/2]$. Suppose we want to prove that $n!+k$ from the interval, is not integral power of prime. Let's factor out the greatest power possible of smallest prime factor of $k$ and call it $q^z$. Now $n!+k$ is either integral power of $q$ or it is not integral power of any prime at all.  Now let's consider two cases if there is another prime factor of $q$ in $n!/q^z$, then $q\mid n!/q^z$, but it does not divide $k/q^z$ so it cannot divide $n!/q^z+k/q^z$. The other case, suppose there is no other factor of $q$ in $n!/q^z$, but this cannot be the case as we know that $k\le n/2$, there is $2k$ in the product of $n!$, i.e. $k*2k$ and $k$ ,$q^z\mid k$ we can certainly say that at least $q^{2z}\mid k*2k$, therefore $q^{2z}\mid n!$. I hope this helps.