I asked: << I found this problem someone had posed, and after having fun solving it, I thought it might amuse some folks on math-fun: ------------------------------------------------------------------ Let f(N) be the probability that 4 random integers i,j,k,m in the range 1 <= i,j,k,m <= N satisfy gcd(i,j) = gcd(k,m) . Find the limit of f(N) as N -> oo.

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It looks like a number of folks have solved this just fine, but here is my solution: ------------------------------------------------------------------------------------ Suppose given integer N >= 1, then for each prime p, and k = 0,1,2,.... the probability that two random integers i,j in the range 1 <= i,j <= N have p^k as the power-of-p factor of gcd(i,j) is Prob(p^k | i && p^k | j && !(p^(k+1) | i && p^(k+1) | j) ) = (1/p^k) * (1/p^k) *(1 - 1/p^2), (where ! denotes falsity), at least in the limit as N -> oo. (All probabilities are calculated in the limit as N -> oo.) Thus Prob(gcd(i,j) and gcd(k,m) have the same highest power-of-p factor = p^k) = (1/p^4k) * (1 - 1/p^2)^2, and summing over k we get Prob(gcd(i,j) and gcd(k,m) have the same highest power-of-p factor) = (1 - 1/p^2)^2 * Sum{k = 0,...,oo} of 1/p^4k) = (1 - 1/p^2)^2 * 1/(1 - 1/p^4). Taking the product over all p gives via the Euler product Prob( gcd(i,j) = gcd(k,m) ) = zeta(4) / zeta(2)^2 = 36/90 = 2/5. ----------------------------------------------------------------------------------- The problem, by the way, comes from Noam Elkies's interesting Mathematical Miscellany website, at < http://www.math.harvard.edu/~elkies/Misc/index.html >, where it is mentioned (as someone asked about) that it's a long-standing unsolved problem to show Prod{p prime} of (p^2 - 1)/(p^2 + 1) = 2/5 by elementary means. --Dan