TITLE: Correlation of the Möbius function
QUESTION [11 upvotes]: Assume that $$\sum_{n \leq x}\mu(n) \mu(n+1) \gg x.$$
Question. Does this contradict any well known results, say the prime number theorem or something like that?

REPLY [27 votes]: The relation you state contradicts the recent result of Terence Tao that
$$ \sum_{n\leq x}\frac{\mu(n)\mu(n+1)}{n}=o(\log x).$$
Indeed, if $S(x)$ is the sum on your left hand side, we have
$$ \sum_{n\leq x}\frac{\mu(n)\mu(n+1)}{n}= \frac{S(x)}{x}+\int_1^x\frac{S(t)}{t^2}\,dt.$$