Let $\mu(n)$ be the Mobius function, how to estimate $$\sum_{1\le i<j\le x}\mu(i)\mu(j) $$ as $x$ goes to $\infty$? Are there some references on this?


If $S$ is your sum then

$$ \left\lvert \sum_{1\leq n\leq x}\mu(n)\right\rvert^2 = 2S+ \sum_{1\leq n\leq x}\mu(n)^2.$$

The second sum on the right is $(\frac{6}{\pi^2}+o(1))x$, and hence estimating $S$ is equivalent to estimating $\lvert \sum_{n\leq x}\mu(n)\rvert$, a classical problem of analytic number theory.

In particular, assuming the Riemann Hypothesis, the left-hand side is $O(x^{1+o(1)}$), and hence (assuming RH)

$$ S \ll x^{1+o(1)}.$$

Unconditionally we can show that $S=o(x^2)$, but cannot show $S\ll x^{2-\epsilon}$ for any $\epsilon>0$.

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  • $\begingroup$ "The best we can conclude unconditionally is $S = o(x^2)$". What do you mean by this? I'm no analytic number theorist, but I'd be surprised if $|\sum_{n \le x} \mu(n)| \ll \frac{x}{\log\log x}$ wasn't known unconditionally. $\endgroup$ – mathworker21 Jul 18 at 2:43
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    $\begingroup$ Yes, sorry, I was being imprecise - I meant that in terms of improving the exponent nothing better is known. The standard proof of the prime number theorem can be adapted to show that $S\ll x^2 \exp(-O(\sqrt{\log x}))$, for example, and I believe better quantitative results are known. $\endgroup$ – Thomas Bloom Jul 18 at 9:43

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