Fix $k∈\Bbb{N}$. An integer $n$ is said to be $k$-th power free if there exists no prime $p$ such that $p^k | n$....

Question

Fix $k∈\Bbb{N}$. An integer $n$ is said to be $k$-th power free if there exists no prime $p$ such that $p^k | n$. Prove that for any $m∈\Bbb{N} $ greater that 1, there exists $m$ consecutive integers which are not $k$-th power free.

My professor said this question involves the Chinese Remainder Theorem and something relating to being square-free. Still not entirely sure what he is looking for, any help as to how to approach this is greatly appreciated.

Answer 1

Hint: Let $p_1,p_2,\dots,p_m$ be distinct primes. Consider the system of congruences

$x\equiv 0\pmod{p_1^k}$

$x+1\equiv 0\pmod{p_2^k}$

$x+2\equiv 0\pmod{p_3^k}$

and so on up to

$x+m-1\equiv 0\pmod{p_m^k}$

By the Chinese Remainder Theorem, this system of congruences has a solution.