Prove that for any positive integers $m$ and $n$, there exists a set of $n$ consecutive positive integers each of which is divisible by a number of the form $d^m$ where $d$ is some integer in $\mathbb{N}$.
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1I imagine $d=1$is not intended. Then use the Chinese Remainder Theorem. – André Nicolas May 26 '14 at 02:20
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Hint
Let $p_1,p_2,..,p_n$ be pairwise distinct primes. By the Chinese Remainder Theorem, the system $$x \equiv 0 \pmod{p_1^m} \\ x+1 \equiv 0 \pmod{p_2^m} \\ .... \\ x+n-1 \equiv 0 \pmod{p_n^m}$$ has solutions.
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May I expand on how one can use the Chinese remainder theorem? You say hint is all – Mmm May 26 '14 at 02:22
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Note: If $x$ satisfies all the equations simultaneously, then $p_1^m|x$, $p_2^m|(x+1),\ldots p_n^m|(x+n-1)$, so you have your desired consecutive integers. – vadim123 May 26 '14 at 02:49
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