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.
Best Answer
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.