Correlation of a quadratic character with the Mobius function Let $p$ be a prime number and consider the sum $S(x)=\sum_{n\le x}\left(\frac{n}{p}\right)\mu(n)$. For how small an $x$ in terms of $p$ is it known that $S(x)=o(x)$?
I am especially interested in unconditional results.
 A: Wirsing's Theorem tells us that if $f$ is multiplicative and each $f(p)=-1,0 $ or $ 1$ then 
$\sum_{n\leq x} f(n) = o(x)$ as  $\sum_{p\leq x} (1-f(p))/p \to \infty$ (and one cannot do much better). Moreover one can get an explicit upper bound:
$ \sum_{n\leq x} f(n) \ll x \exp( -.32\sum_{p\leq x} (1-f(p))/p)$.
In your case $f(n)=\mu(n)(n/p)$ so that 
$\sum_{p\leq x} (1-f(p))/p = 1/p+ 2\sum_{q\leq x, (q/p)=1} 1/q $.
Therefore to get the bound $o(x)$ you need that a significant number of the $q\leq x$ satisfy $(q/p)=1$. So your question becomes: For what $x$ can we guarantee this?
Or, in other words, is it possible that $(q/p)=-1$ for "most" of the primes $\leq x$ (as may well be the case of one has a Siegel zero)?
If we use quadratic reciprocity, then $(q/p)=-1$ is equivalent to demanding $(p/q)=$ something fixed, and we can find such $p$ for which this holds  for all but one prime $q\ll \log p$, by Dirichlet's Theorem.  But then, by smooth number estimates, one knows that for almost all such $p$ one has $\sum_{n\leq x} \mu(n)(n/p) \gg \rho(A)x$ for $x=(\log p)^A$ (for each $A$).
So we have "proved" that for any fixed $A>0$, the estimate  $\sum_{n\leq x} \mu(n)(n/p) = o(x)$ does not hold uniformly for $x=(\log p)^A$.
The same ideas give, assuming GRH, that $\sum_{n\leq x} \mu(n)(n/p) = o(x)$ does   hold uniformly provided $\log x/\log\log p \to \infty$ as $p\to \infty$.
These ideas can   be found in my paper "Large Character Sums" with Soundararajan, though there we looked at character sums $\sum_{n\leq x} \chi(n)$; it should not take much to modify those ideas for this situation.
A: In general, we can take $x>\exp\{c_\epsilon p^\epsilon\}$ by the Prime Number Theorem for arithmetic progressions. More generally, one can use $L$-functions methods to relate $x$ to zero-free regions. In doing so it is hard to avoiding `losing logarithms'. The following elementary argument though, essentially due to Granville, does the job in an easier way.
Let $f$ be a completely multiplicative such that $|f(n)|\le1$ for all $n$ (so one can think that $f$ is a Dirichlet character), and assume that we know that
$$
\left|\sum_{n\le x} \Lambda(n)f(n) \right| \le Cx\cdot \frac{\log Q}{\log x} \tag{*}
$$
for all $x>Q$ (the size of $Q$ will depend on the available zero-free regions). Then we claim that $(*)$ holds for $\mu(n) f(n)$ too (with a different constant). Note that if suffices to show the result for $g(n)=\prod_{p^e\|n}(-f(p))^e$ (then one can use a convolution argument to pass to $f$). In order to show that $(*)$ holds with $g$ in place of $f\Lambda$, possibly with another constant $C'$ in place of $C$, we use induction, with the induction hypothesis being that 
$$
\left|\sum_{n\le x} g(n) \right| \le C' x\cdot \frac{\log Q}{\log x} \tag{**}
$$
for all $x\le 2^m$. If $2^m\le Q$, this holds trivially (choosing $C'$ appropriately). Next, assume that $2^m>Q$ (and that $Q$ is large). Suppose also that that $(**)$ holds for $x\le 2^m$, and consider $x\in(2^m,2^{m+1}]$.Then
$$
\sum_{n\le x} g(n) \log n 
  = \sum_{n\le x} g(n) \sum_{d|n} \Lambda(d) 
  = \sum_{dm\le x} \Lambda(d) g(d)g(m) .
$$
We apply Dirichlet's hyperbola method:
\begin{align*}
\sum_{n\le x} g(n) \log n 
  &= \sum_{m\le x^{1-\epsilon}} g(m)  \sum_{d\le x/m} g(d)\Lambda(d)
    + \sum_{1<d\le x^{\epsilon}} \Lambda(d) g(d) \sum_{x^{1-\epsilon}<m\le x/d} g(m) \\
&\ll \sum_{m\le x^{1-\epsilon}} \frac{Cx}{m}\cdot \frac{\log Q}{\epsilon \log x}
    + \sum_{d\le x^{\epsilon}} \Lambda(d) \frac{C'x}{d} \cdot \frac{\log Q}{\log x} \\
   &\ll \left( \frac{C}{\epsilon} + \epsilon C'\right) x\log Q
\end{align*}
Then applying partial summation and choosing $\epsilon$ and $C'$ appropriately completes the inductive step and thus the proof of $(**)$. 
In the special case that $f(n)=(n/p)$, we know that
$$
\sum_{n\le x}\Lambda(n) \left(\frac{n}{p}\right) 
= -\sum_{\substack{\rho=\beta+i\gamma\\L(\rho,(\cdot/p))=0,\,|\gamma|\le p}} \frac{x^{\rho}}{\rho} + O\left(xe^{-c\sqrt{\log x}}\right),
$$
(see e.g. eq (13), p. 120 in Davenport's book "Multiplicative Number Theory"). There is a $c>0$ such that the first sum has at most one summand with $\beta\ge1-c/\log p$, for which one then necessarily has that $\gamma=0$ (i.e. $\rho=\beta$ is a Siegel zero). The sum over the zeroes with $\beta\le 1-c/\log p$ can be shown to be $\ll x^{1-c'/\log q}$ for some absolute constant $c'>0$, using zero-density estimates (see e.g. equation (18.9) in p. 428 of the book "Analytic Number Theory" by Iwaniec and Kowalski). We conclude that
$$
\sum_{n\le x}\Lambda(n) \left(\frac{n}{p}\right) 
= -\frac{x^{\beta}}{\beta} + O\left( x^{1-c'/\log p} + xe^{-c\sqrt{\log x}}\right) .
$$
So $(*)$ holds with $Q=1/(1-\beta)$ if $\beta$ exists and with $Q=p$ otherwise.
