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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.