<s>[INST] Quadratic Congruence (with Chinese Remainder Thm) How do we solve quadratic congruences such as: 

$x^2 \equiv11 \pmod{39}$

I know I must use the chinese remainder theorem with $p = 13, 3$ but I've only done linear examples and am unsure about how to do quadratic ones. 
 [/INST] $$x^2 \equiv 11 \pmod{39} \implies x^2 \equiv 2 \pmod 3 \implies \text{No solution}$$

EDIT
In general, when you want to solve for $$x^2 \equiv a \pmod n \,\,\, (\spadesuit)$$ and if $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$, the idea is to first solve for $$x^2 \equiv a \pmod {p_l^{a_l}} \,\,\, (\clubsuit)$$ You have a solution for your original problem $(\spadesuit)$ iff you have a solution for each $l \in \{1,2,\ldots,k\}$ in $(\clubsuit)$. Once you find solution for each $l$ in $(\clubsuit)$, put them together using Chinese Remainder theorem.
For instance, if you have $x^2 \equiv 23 \pmod {77}$, then we need to look at $x^2 \equiv 23 \pmod 7$ and $x^2 \equiv 23 \pmod{11}$ i.e. $x^2 \equiv 2 \pmod 7$ and $x^2 \equiv 1 \pmod{11}$.
$$x^2 \equiv 2 \pmod7 \implies x \equiv \pm 3 \pmod 7$$ Similarly, $$x^2 \equiv 1 \pmod{11} \implies x \equiv \pm 1 \pmod{11}$$
Hence,
$$x \equiv 3 \pmod 7 \text{ and } x \equiv 1 \pmod{11} \implies x \equiv 45 \pmod{77}$$
$$x \equiv -3 \pmod 7 \text{ and } x \equiv 1 \pmod{11} \implies x \equiv 67 \pmod{77}$$
$$x \equiv 3 \pmod 7 \text{ and } x \equiv -1 \pmod{11} \implies x \equiv 10 \pmod{77}$$
$$x \equiv -3 \pmod 7 \text{ and } x \equiv -1 \pmod{11} \implies x \equiv 32 \pmod{77}$$
Hence,
$$x \equiv \pm 10, \pm 32 \pmod{77}$$
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