From: kramsay@aol.commangled (Keith Ramsay)
Subject: Re: question on squarefree numbers
Date: 28 Dec 1999 06:01:21 GMT
Newsgroups: sci.math.research
Keywords: arithmetic progressions of squarefree integers
In article <848878$909$1@news.si.fct.unl.pt>, "manuel"
writes:
| Can we find arbitrarily long arithmetic progressions( or even an
|arithmetic progression) formed of different integers such that all elements
|are squarefree integers?
The squarefree numbers have a density of 6/pi^2 by a theorem of
Gegenbauer. (If f(n) is the number of squarefree numbers in 1,...,n,
then the limit as n goes to infinity of f(n)/n is 6/pi^2.) Any set of
natural numbers having a positive density has arbitrarily long
arithmetic sequences in it by a theorem of Szemeredi.
On the other hand, for any integers A, B>0, there exists a prime p
not dividing A and not dividing B. The congruence A+Bt=0 (mod p^2)
is soluble, hence the sequence A, A+B, A+2B,... does not consist
entirely of squarefree numbers.
Keith Ramsay
==============================================================================
From: Fred Galvin
Subject: Re: question on squarefree numbers
Date: Mon, 27 Dec 1999 15:21:59 -0600
Newsgroups: sci.math.research
On Mon, 27 Dec 1999, manuel wrote:
> Can we find arbitrarily long arithmetic progressions( or even
> an arithmetic progression) formed of different integers such that
> all elements are squarefree integers?
Arbitrarily long (finite) arithmetic progressions, yes, because the
set of squarefree numbers has positive density.
No infinite arithmetic progressions of squarefree numbers. Consider
the arithmetic progression a, a+d, a+2d, ..., where a and d are
positive integers. Let p be a prime that is not a divisor of d; find
positive integers x and y such that (p^2)x-dy = 1; let n = ay; then
a+nd = (p^2)ax is not squarefree.