Prove that for any positive integers $m$ and $n$, there exists set on $n$ consecutive positive integers each of which is divisible by a number of the form $a^m$ where $a$ is any integer.
To be honest, I'm not even sure what the question is asking. I mean should I write $a=1$ and move on, or is there an actual non trivial solution.
Any hint/ help is appreciated!
It's a Chinese Remainder Theorem problem. Take the system
$$x\equiv 0 \pmod{2^m}$$
$$x+1 \equiv 0\pmod{3^m}$$
etc.
$$x+n-1 \equiv 0 \pmod{p_k^m}$$
where $p_k$ is the $k$th prime.