On 12/3/06, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
[Blank lines to avoid giving things away]
... but it's easy enough from first principles: expand the product in the LHS to get sum {k,l} of mu(k)/(k^2.l^2); change variable via n=kl to get sum {n} of 1/n^2 . sum {k|n} mu(k); the inner sum is well known (and easily proved) to be 1 when n=1 and 0 otherwise.
Right, that's the bit I (along with Maple) had trouble with. On 12/4/06, Daniel Asimov <dasimov@earthlink.net> wrote:
I may be mistaken, but I have the impression that someone is solving a similar but different question (What is the probability that two random integers are relatively prime?)
Sounds as if Dan has figured out the answer to Gareth's earlier question, if not how the rest of us tackled the problem! Maybe it's time he came clean (after the statutory 40 blank lines, of course). WFL