2
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ Hint: Chinese remainder theorem. $\endgroup$ – Wojowu Aug 14 '16 at 17:21
  • $\begingroup$ What if you take $d=1$? $\endgroup$ – Rick Decker Aug 14 '16 at 17:22
  • $\begingroup$ haven't studied that yet :p. Any other way? $\endgroup$ – D.K. Aug 14 '16 at 17:22
  • $\begingroup$ @RickDecker that's a loophole. I should have closed that. $\endgroup$ – D.K. Aug 14 '16 at 17:23
  • 1
    $\begingroup$ The first term of such a sequence is precisely what CRT guarantees $\endgroup$ – Hagen von Eitzen Aug 14 '16 at 17:25
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\, |s_1\cdots s_n|\,$

$\endgroup$

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.