# Prove that there exists such a set of $n$ positive integers

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 positive integer not equal to $1$.

I don't know how to approach this question.

• Hint: Chinese remainder theorem. – Wojowu Aug 14 '16 at 17:21
• What if you take $d=1$? – Rick Decker Aug 14 '16 at 17:22
• haven't studied that yet :p. Any other way? – D.K. Aug 14 '16 at 17:22
• @RickDecker that's a loophole. I should have closed that. – D.K. Aug 14 '16 at 17:23
• The first term of such a sequence is precisely what CRT guarantees – Hagen von Eitzen Aug 14 '16 at 17:25

Hint. The trick is to start by choosing the $d_k$s such that $(d_k)^m \mid x+k$. If you make the $d_k$s mutually coprime (say, choose different primes), then the Chinese Remainder Theorem will tell you what $x$ is.
Hint $\$ If $\, S\subset \Bbb Z\,$ contains infinitely many pairwise coprime integers $\,s_i\,$ then for all $\,n> 0\,$ there is a sequence of $\,n\,$ consecutive naturals each of which is a multiple of an element of $\,S.\,$ Indeed, we we can apply CRT = Chinese Remainder Theorem to solve the system
\begin{align} x\equiv -1\pmod{s_1}\\ x\equiv -2\pmod{s_2}\\ \cdots\qquad \quad\cdots\qquad \\ x\equiv -n\pmod{s_n}\end{align}
Therefore we conclude that $\, s_1\mid x\!+\!1,\,\ s_2\mid x\!+\!2,\, \ldots,\ s_n\mid x\!+\!n$
Note $\$ We can force $\,x > 0\,$ by adding to it a large enough multiple of $\, |s_1\cdots s_n|\,$