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.