How do I use the Chinese remainder theorem to find all the square roots of 11 in $\mathbb Z_{35}$? I understand what the Chinese remainder theorem is. However I am not sure how to apply it to my question. Can someone explain please?
 A: We apply CRT as explained in the Remark below: first we compute the square roots $\!\bmod 5\ \&\ 7$ then we take all possible combinations of roots and lift them to roots $\bmod 35\,$ using CRT.
$$\begin{align}&\overbrace{{\rm mod}\ 5\!:\,\ x^2\equiv 11\equiv 1}^{5\mid (x-1)(x+1)\iff \color{#0a0}{5\mid x-1}\ {\rm or}\ \color{#0a0}{5\mid x+1}\!\!\!\!\!\!\!\!\!\!}\!\iff \color{#0a0}{x\equiv \pm1}\equiv: \color{#0af}{a}\\[.2em] 
&{\rm mod}\ 7\!:\,\ x^2\equiv 11\equiv 4\iff x\equiv \pm 2\equiv: \color{#0af}{b}\end{align}\qquad\qquad$$

Next we solve the system $\, x\equiv \color{#0af}a\pmod 5\,$ and $\,x\equiv \color{#0af}b\pmod 7$ for  $\rm\color{#0af}{generic}$ (any)  $\rm\color{#0af}{a,b}$
${\rm mod}\ 5\!:\ a \equiv x\equiv b+7\color{#c00}k\equiv b+2k\iff 2k\equiv a\!-\!b\!\overset{\times\,(-2)}\iff \color{#c00}{k\equiv 2(b\!-\!a)}$
therefore $\ x = b+7\color{#c00}k = b+7(\color{#c00}{2(b\!-\!a)+5n}) = b+14(b\!-\!a)+35n$

For example the case $\, a,b \equiv \color{#0a0}1, 2\,$ yields that $\, x = 2 + 14(2\!-\!1) = 16+35n$
Its negation  $\,-16\equiv 19\pmod{\!35}\,$ is also a square root - the case $\,a,b \equiv -1,-2$.
Do similarly for  $\,a,b \equiv -1,2\,$ and its negative $\,1,-2\,$ to get all $(=4)$ square roots, by below.
Remark $\ $ If $\,m,n\,$ are coprime then, by CRT, solving a polynomial $\,f(x)\equiv 0\pmod{\!mn}\,$ is equivalent to solving $\,f(x)\equiv 0\,$ both $\!\bmod m\,$ and $\!\bmod n.\,$ By CRT, each combination of a root $\,r_i\bmod m\,$ and a root $\,s_j\bmod n\,$ corresponds to a unique root $\,t_{ij}\bmod mn\,$ i.e.
$$\begin{eqnarray} f(x)\equiv 0\!\!\!\pmod{\!mn}\!\!&\overset{\small\rm CRT\!}\iff& \begin{array}{}f(x)\equiv 0\pmod{\! m}\\f(x)\equiv 0\pmod{\! n}\end{array} \\ 
&\iff&  \begin{array}{}x\equiv r_1,\ldots,r_k\pmod{\!m}\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod{\!n}\end{array}\\ 
&\iff& \left\{ \begin{array}{}x\equiv r_i\pmod{\! m}\\x\equiv s_j\pmod{\! n}\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\
&\overset{\small\rm CRT\!\!}\iff&\, \left\{\, x\equiv t_{i j}\!\!\!\!\pmod{\!mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\
\end{eqnarray}\qquad\qquad\qquad$$
