0
$\begingroup$

Find the largest positive integer $k$, such that $\mu(n+r)=0$ for all $1\leq r\leq k$ where $r,n$ are positive integers.

As far as I could make out, we need to find out the maximum range(if nay) of numbers where each has a square divisor.

I have gone through the theory of square-free numbers here and there, but could not proceed much.

$\endgroup$
  • 5
    $\begingroup$ By now, over a month on the website, you should know that (i) you will get better answers if you present context, and you explain what you have done or where you are stuck; and (ii) that many people find it at least mildly annoying to have requests for help phrased as orders. $\endgroup$ – Arturo Magidin Jul 24 '12 at 16:05
  • $\begingroup$ what does miu mean? $\endgroup$ – Jorge Fernández Hidalgo Jul 24 '12 at 16:27
  • 1
    $\begingroup$ en.wikipedia.org/wiki/M%C3%B6bius_function $\endgroup$ – lab bhattacharjee Jul 24 '12 at 16:48
  • 1
    $\begingroup$ Depends on $n$ in a chaotic way, usually smallish, but can be arbitrarily large. $\endgroup$ – André Nicolas Jul 24 '12 at 17:49
3
$\begingroup$

There is no such largest positive integer. Given $k$ primes, by the Chinese remainder theorem we can find a number $m$ that has remainders $1$ through $k$ with respect to their squares. Then $m-k$ through $m-1$ all have square divisors.

$\endgroup$
  • 1
    $\begingroup$ Just beat me - very nice. +1 $\endgroup$ – davidlowryduda Jul 24 '12 at 16:29
  • $\begingroup$ Are you saying we can always find integral solutions of $a_{r+1}p^2_{r+1}-a_rp^2_r=1$ for any range 1≤r≤n for different primes $p_r$ for any positive integral value of n? $\endgroup$ – lab bhattacharjee Jul 24 '12 at 16:47
  • 1
    $\begingroup$ @lab: Do you know the Chinese remainder theorem? It's not just about two primes at a time; we can find numbers with arbitrary remainders for arbitrary coprime integers (e.g. prime powers). $\endgroup$ – joriki Jul 24 '12 at 17:08
  • $\begingroup$ Agreed & accepted. $\endgroup$ – lab bhattacharjee Jul 25 '12 at 5:41

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.