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Prove that for every integer m there exist n consecutive positive integers such that each of them is divisible by some number of the form a^m. a,m are natural number.

Attempt- i tried to prove that there are solutions to (a^mb+n)/(c^m) has is natural number for manh values of a and c but in vain.

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closed as off-topic by Namaste, Isaac Browne, Shailesh, Xander Henderson, José Carlos Santos Jul 27 '18 at 16:40

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    $\begingroup$ You should clarify your quantifiers. I think you mean "for every $m\in \mathbb N$ and every $n\in \mathbb N$ there exist $n$ consecutive natural numbers each of which is divisible by some $m^{th}$ power." That is, your "$a$" changes from number to number in the consecutive string. Thus $8,9$ is a good sequence for $m=n=2$, for example. If I am correct, the Chinese Remainder Theorem quickly solves your problem. $\endgroup$ – lulu Jul 16 '18 at 12:31
  • $\begingroup$ @lulu You are correct! $\endgroup$ – Peter Jul 16 '18 at 12:36
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Denote the $k$-th prime number with $p_k$

Given $m$ and $n$ , use the chinese remainder theorem to construct a number $N$ , which is congruent to $0$ modulo $2^m$ , congruent to $-1$ modulo $3^m$ , congruent to $-2$ modulo $5^m$ and so on until congruent to $1-n$ modulo $p_n\ ^m$.

Then, the consecutive $n$ numbers $N,N+1,N+2,\cdots N+n-1$ are divisible by a $m$-th prime power.

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