Solution of equation $ x^2\equiv 1 \pmod{784}$ How to solve the equation $ x^2\equiv 1 \pmod{784}$ ?
Context
I know the Chinese Remainder theorem, but have no idea how to begin. Could you give me any clue? The only thing that I would like to is learn and understand this subject.
It is very important to me see example solved by person who is experienced.
 A: This breaks down into
$$x^2\equiv 1\pmod{49},$$
that is equivalent to $x\equiv\pm 1\pmod{49}$ since $(\mathbb{Z}/49\mathbb{Z})^*$ is a cyclic group, and into:
$$x^2\equiv 1\pmod{16}$$
whose solutions are $x\equiv \pm 1,\pm 7\pmod{16}$.
A: Hint $\ $ If $\,m,n\,$ are coprime then, by CRT, solving a polynomial $\,f(x)\equiv \,$ mod $\,mn\,$ is equivalent to solving $\,f(x)\equiv 0\,$ mod $\,m\,$ and mod $\,n.\,$ By CRT, each combination of a root $\,r_i\,$ mod $\,n\,$ and a root $\,s_j\,$ mod $\,n\,$ corresponds to a unique root $\,t_{ij}\,$ mod $\,mn\,$ i.e.
$$\begin{eqnarray} f(x)\equiv 0\pmod{mn}&\overset{\rm CRT}\iff& \begin{array}{}f(x)\equiv 0\pmod m\\f(x)\equiv 0\pmod n\end{array} \\ 
&\iff&  \begin{array}{}x\equiv r_1,\ldots,r_k\pmod m\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod n\end{array}\\ 
&\iff& \left\{ \begin{array}{}x\equiv r_i\pmod m\\x\equiv s_j\pmod n\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\
&\overset{\rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\
\end{eqnarray}\qquad\qquad$$
A: 784 = 16*49.  Solve x²=1 in both, and use the remainder therom.
You end up with (1,7,9,15)(1,48), which gives 1; 295, 393; 687., and 784 less these.
