On Monday 04 December 2006 07:52, Fred lunnon wrote:
Well, if Dan does have a \pi-free argument, I'll be interested to see it. But I now have a \mu-free one, so here it is [a long way down, to be sure].
Let S(p, q) denote the subset of the lattice S of integer vectors [i,j,k,l], such that hcf(i,j) = p and hcf (k,l) = q; and let c denote the density of S(1, 1).
Then S is the direct union of the S(p,q), so 1 = \sum_{p,q} c / p^2 q^2 = c \zeta(2)^2 = c \pi^2 / 36; the problem demands the density of the direct union of the S(p,p), = \sum_{p} c / p^4 = c \zeta(4) = c \pi^2 / 90, = (c \pi^2 / 36) 2/5 = 2/5 by the above.
Substantially the same as my argument, for what it's worth, but WFL's is more neatly expressed and implicitly includes a proof that P(p,q coprime) = zeta(2), which I took for granted. On the other hand, the more banal way in which mine's expressed *may* make it easier to follow: 1/N^4 # { p,q,r,s in [1,N] : (p,q) = (r,s) } = 1/N^4 sum over d of # { p,q,r,s in [1,N] : (p,q) = (r,s) = d } = 1/N^4 sum over d of (# { p,q in [1,N] : (p,q) = d })^2 = 1/N^4 sum over d of (# { p,q in [1,N/d] : (p,q) = 1 })^2 ~ 1/N^4 sum over d of (6/pi^2 (N/d)^2)^2 = 36/pi^4 sum over d of 1/d^4 = 36/pi^4 pi^4/90 = 2/5 -- g