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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.

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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. –  Arturo Magidin Jul 24 '12 at 16:05
what does miu mean? –  Modded Bear Jul 24 '12 at 16:27
Depends on $n$ in a chaotic way, usually smallish, but can be arbitrarily large. –  André Nicolas Jul 24 '12 at 17:49

1 Answer 1

up vote 3 down vote accepted

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.

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Just beat me - very nice. +1 –  mixedmath Jul 24 '12 at 16:29
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? –  lab bhattacharjee Jul 24 '12 at 16:47
@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). –  joriki Jul 24 '12 at 17:08
Agreed & accepted. –  lab bhattacharjee Jul 25 '12 at 5:41

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