$\left(\frac{a}{n}\right)=1$ does not necessarily imply that, $a$ is a quadratic residue $\mod n$ 
$\left(\frac{a}{n}\right)=1$ does not necessarily imply that, $a$ is a quadratic residue $\mod n$, where $\left(\frac{.}{.}\right)$ is the Jacobi symbol.

For example I know that $\left(\frac{2}{15}\right)=\left(\frac{2}{3}\right)\left(\frac{2}{5}\right)=(-1)^{\frac{3^2-1}{8}}(-1)^{\frac{5^2-1}{8}}=(-1)^1(-1)^3=1$
but $x^2\equiv2\mod 15$ has no integer solutions http://www.wolframalpha.com/input/?i=quadratic+residues+mod+15
Is there a generalization ?
$\textbf{EDIT}$: How can I find all quadratic residues $\mod 15 $ without wolframalpha ?

For the second part of your question which remains unanswered, one can find all the quadratic residues $\pmod n$ by considering all numbers up to $n$ modulo $n$, squaring all of them and considering the residues $\pmod n$. Notice that any other numbers, for example $n+2$, would have the same quadratic residue as the number reduced $\pmod n$, in this example $2$, would, hence this list would cover all of the residues (In the taken example, $(n+2)^2 \equiv 2^2 \equiv 4 \pmod n$, provided $4<n$ or it would be reduced further). Notice that we can write the last $\lfloor\frac{n}{2}\rfloor$ numbers in our set as negative counterparts of previous numbers which would have the same quadratic residue $\pmod n$. This means that we only need to check upto $\lceil\frac{n}{2}\rceil$ and would cover all residues.
To make this clearer I will do this procedure for $7$.
$\begin{array}{| c | c | c |}\hline \text{n} & n^2 \pmod  7 \\ \hline 1 & 1\\ \hline 2 & 4 \\ \hline 3 & 2 \\ \hline 4 & 2 \\ \hline
\end{array}$
The above table implies that the only quadratic residues$\pmod 7$ are $1,2,4$. One can repeat this process for $15$ to get that the quadratic residues are $1,4,6,9,10$. Notice that this does not include $2$, hence $2$ is a quadratic nonresidue.
