Here's some answers (in addition you should look at Sol Golomb's Puzzle column in the Dec 2011 issue of the IEEE Information Theory Society Newsletter: http://www.itsoc.org/publications/newsletters/nits_NL_1211-Web.pdf/view which has a bunch of problems like this in even greater generality). Puzzle #1: The event that both n and n-1 are square free arises because mod p^2, for every prime p, you exclude two moduli: 0 and 1. Thus the answer is: prod_p (1-2/p^2) where the product is taken over all primes. Puzzle #2: if we call the above probability, B, and 6/pi2 A. We have a disjoint union: [at least one of n and n-1 are square free] = [n is square free and (n-1) is not square free] union [(n-1) is square free and n is not square free] union [both n and (n-1) are square free] Each of the first two events has probability A-B, and the last has probability B (by part 1), so we get the answer: 2*A - B. Victor ---------- Forwarded message ---------- From: Warren Smith <warren.wds@gmail.com> Date: Thu, Feb 2, 2012 at 10:26 PM Subject: [math-fun] number theory puzzle about squarefree number density To: math-fun@mailman.xmission.com If you pick a random uniform number n with 0<n<=N then the probability n will be squarefree goes to 6/pi^2=60.79% in the limit N-->infinity. You probably already knew that.  Result by Euler.   Kind of a very inefficient way to calculate pi. Now, which are somewhat harder: PUZZLE #1: what is the probability that both n and n-1 are squarefree? PUZZLE #2: what is the probability that at least one of {n,n-1} is squarefree? (Actually, if you solve one puzzle, I think you'll find it very easy to solve the other...) -- Warren D. Smith http://RangeVoting.org  <-- add your endorsement (by clicking "endorse" as 1st step) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun