I have a question on Jacobi symbols, and the problem goes:
Show that is $(a,p)=1$, and $p$ is an odd prime, then prove that:
$$\sum_{n=1}^p \left(\frac{n^2+a}{p}\right)= -1$$
I think I have some idea on how to do it, but not sure if it is entirely correct.
Proof: Well first if we denote $\Lambda(a,p)$ as: $$\Lambda(a,p)=\sum_{n=1}^p \left(\frac{n^2+a}{p}\right)$$ $$\sum_{a=1}^p \Lambda(a,p)=\sum_{a=1}^p\left[\sum_{n=1}^p \left(\frac{n^2+a}{p}\right)\right]=\sum_{n=1}^p\left[\sum_{a=1}^p \left(\frac{n^2+a}{p}\right)\right]=0$$ because it is well established that: $$\sum_{a=1}^p \left(\frac{n^2+a}{p}\right)=0, \space \forall n$$
We can also establish the result: $$\Lambda(p,p) =\sum_{n=1}^p \left(\frac{n^2}{p}\right)=\sum_{n=1}^{p-1}\left(\frac{n^2}{p}\right)+\left(\frac{p^2}{p}\right)=(p-1)+0=p-1$$ $$\sum_{a=1}^{p-1} \Lambda(p,p)=-(p-1)$$
Now if $a$ is a quadratic residue modulo $p$, then $a*1^2, a*2^2,....,a*p^2$ is a permutation of $1^2,2^2,...,p^2 \bmod p$. Then we have the following:
$$\Lambda(a,p)=\sum_{n=1}^p \left(\frac{n^2+a}{p}\right)=\sum_{n=1}^p \left(\frac{an^2+a}{p}\right)=\left(\frac{a}{p}\right)\sum_{n=1}^p \left(\frac{n^2+1}{p}\right)=\sum_{n=1}^p \left(\frac{n^2+1}{p}\right)$$
On the other hand if $a$ is not a quadratic residue of $\bmod p$, then so is $b = a^{'}$ the inverse of $a \bmod p$. Hence we get:
$$\Lambda(a,p)=\sum_{n=1}^p \left(\frac{n^2+a}{p}\right)=\sum_{n=1}^p \left(\frac{abn^2+a}{p}\right)=\left(\frac{a}{p}\right)\sum_{n=1}^p \left(\frac{bn^2+1}{p}\right)=-\sum_{n=1}^p \left(\frac{bn^2+1}{p}\right)$$
Now, if $b$ is an arbitrary quadratic non residue of $\bmod p$, then we have $n^2, 1\le n \le p-1$ takes each of all the quadratic non residues twice (exactly). So together with $p^2 = bp^2 = 0\bmod p$, the numbers $n^2, bn^2, 1 \le n \le p$ take each of the elements in a COMPLETE residue system $\bmod p$ exactly twice, and so are $n^2+1, bn^2+1,$ for $1 \le n \le p$. Therefore we have:
$$\sum_{n=1}^p \left(\frac{bn^2+1}{p}\right) +\sum_{n=1}^p \left(\frac{n^2+1}{p}\right)=2\sum_{n=1}^p \left(\frac{n}{p}\right)=0$$
The above shows that $\Lambda(a,p)$ is the same for each a with $(a,p)=1$.
Hence we have:
$$\Lambda(a,p)=\frac{\sum_{a=1}^{p-1} \Lambda(a,p)}{p-1}=-1$$
I will change your notation a bit:$$\Lambda (a) = \sum\limits_{n = 1}^p {\left( {\frac{{{n^2} - a}}{p}} \right)} $$ The following is useful:
Hint: if $a,b$ are both residue, then $a\equiv br^2$ for some $p\nmid r$.
Let $u$ denote a residue modulo $p$, $v$ a non-residue. You have already shown that $$\sum_{a=1}^p \Lambda (a) = 0$$ This implies $$\frac{p-1}{2} \Lambda (u) + \frac{p-1}{2} \Lambda (v) + p-1 = 0 \implies \Lambda (u)+\Lambda (v) = -2$$
On the other hand, we calculate $$\begin{aligned}\sum_{a=1}^p \Lambda(a^2) &= \sum\limits_{n = 1}^p {\sum\limits_{a = 1}^p {\left( {\frac{{{n^2} - {a^2}}}{p}} \right)} } \\&= \sum\limits_{n = 1}^p {\sum\limits_{a = 1}^p {\left( {\frac{{n - a}}{p}} \right)\left( {\frac{{n + a}}{p}} \right)} } \\&= \sum\limits_{n = 1}^p {\sum\limits_{a = 1}^p {\left( {\frac{n}{p}} \right)\left( {\frac{{n + 2a}}{p}} \right)} } \\&= \sum\limits_{a = 1}^p {\left( {\frac{{n + 2a}}{p}} \right)\left[ {\sum\limits_{n = 1}^p {\left( {\frac{n}{p}} \right)} } \right]} = 0 \end{aligned}$$ this implies $$p-1 + (p-1)\Lambda(u) = 0 \implies \Lambda(u)=-1$$ Can you finish the proof?