Prove for each positive integer $n$, there exists $n$ consecutive positive integers none of which is an integral power of a prime number 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.
 A: Hint: Look to numbers near some factorials. 
Example: The numbers near $9!$:


*

*$9!+2 = 2(9!/2+1)$. Since $9!$ has another factor $2$, $9!/2+1$ is
odd and therefore is $9! + 2$ not a prime power. (It is divisible by
$2$, but not a prime power of $2$)

*$9!+3 = 3(9!/3+1)$. Since $9!$ has another factor $3$, $9!/3+1$ is
not divisible by $3$ and therefore is $9! + 3$ not a prime power. (It
is divisible by $3$, but not a prime power of $3$)
In general, you can prove that all numbers form $n!+2$ to $n!+\frac{n}{2}$ cannot be prime powers. 
A: 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. 
A: Consider the system of $n$ congruences 
$x\equiv 0\pmod{(2)(3)}$
$x\equiv -1\pmod{(5)(7)}$
$x\equiv -2 \pmod{(11)(13)}$
and so on, up to
$x\equiv -(n-1)\pmod{(p_{2n-3})(p_{2n-2})}$
where $p_i$ is the $i$-th prime.
By the Chinese Remainder Theorem, this system of congruences has infinitely many positive solutions $x$. 
If $x$ is such a solution, then $x$ is divisible by $(2)(3)$, $x+1$ is divisible by $(5)(7)$, and so on. 
It follows that none of $x$, $x+1$, and so on up to $x+(n-1)$ is a prime power.
