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.



Correlation of a quadratic character with the Mobius function
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.


