Solve $x^2\equiv -3\pmod {\!91}$ by CRT lifting roots $\!\bmod 13\ \&\ 7$ Question 1) Solve $$x^2\equiv -3\pmod {13}$$
I see that $x^2+3=13n$. I don't really know what to do? Any hints?
The solution should be $$x\equiv \pm 6 \pmod {13}$$
Question 2) $\ $ [note $\bmod 7\!:\ x^2\equiv -3\equiv 4\iff x\equiv \pm 2.\,$ Here we lift to $\!\!\pmod{\!91}\ $ -Bill] 
Given $x\equiv \pm 6 \pmod {13}$ and $x\equiv \pm 2 \pmod {7}$ find solutions $\pmod {91}$. I see that $91=13 \times 7$, does it mean I have to use chinese remainder theorem on 4 equations? If,so $x=6\times 13\times 7 \times 7\times (13\times 7 \times 7)^{-1}...$

For Question 2), you can use the Chinese Remainder Theorem on the one equation
$$
13x+7y=1\tag{1}
$$
Using the Extended Euclidean Algorithm, an implementation of which is in this answer, compute
$$
\begin{array}{r}
&&1&1&6\\\hline
1&0&1&-1&7\\
0&1&-1&2&-13\\
13&7&6&1&0
\end{array}\tag{2}
$$
that is,
$$
13(-1)+7(2)=1\tag{3}
$$
From $(3)$, we can get
$$
\begin{align}
14&\equiv1\pmod{13}\\
14&\equiv0\pmod{7}
\end{align}\tag{4}
$$
and
$$
\begin{align}
-13&\equiv0\pmod{13}\\
-13&\equiv1\pmod{7}
\end{align}\tag{5}
$$
Now adding $6$ times $(4)$ to $2$ times $(5)$, we get
$$
\begin{align}
58&\equiv6\pmod{13}\\
58&\equiv2\pmod{7}
\end{align}\tag{6}
$$
Adding other multiples of $(4)$ and $(5)$, we can get the other answers you are looking for.
