How many solutions of $\mod 63$ : $x^2=1 \pmod7 $ and $x^3=1\pmod 9$ How many solutions $\mod 63$ ,  we have for:
$$x^2=1 \pmod 7$$ 
and 
$$x^3=1 \pmod 9$$
Need to find them also. 
 A: Once you find  $\,x^2\equiv 1\iff x\equiv \pm1 \pmod 7\,$ and $\,x^3\equiv 1\iff x\equiv 1,4,7\pmod{9},\,$ you can lift each possible combination of values using the Chinese Remainder Theorem (CRT). It's usually quicker to do it generically first, i.e. solve $\,x\equiv a\pmod 7,\,$ $\,x\equiv b\pmod 9,\,$ then substitute all possible combinations of the above values for $\,a,b,\,$ i.e. for all $\,a\in \{1,-1\}\,$ and all $\,b \in \{1,4,7\}.\ $ So let's do the generic CRT. 
${\rm mod}\ 7\!:\,\ x \equiv a\iff x = a+7j\,$ for some $\,j\in \Bbb Z.\,$
${\rm mod}\ 9\!:\,\ b \equiv x \equiv a+7j\equiv a-2j\iff 2j\equiv a-b\iff j\equiv 4(b-a)$
Thus $\,x = a + 7j = a+7(4(b-a)+9k) \equiv  a+28(b-a)\pmod{\!63}$
For $\,a\ \equiv\ 1\pmod{\! 7}\!:\ x\  \equiv\  1+28(b\!-\!1) \equiv\ \, 1,22,43\pmod{\!63}\, $ for $\, b \equiv 1,4,7 \pmod{\!9}$
For $ a \equiv -1\!\pmod{\! 7}\!:\ x \equiv -1\!+\!28(b\!+\!1) \equiv 55,13,34\pmod{\!63}\, $ for $\, b \equiv 1,4,7 \pmod{\!9}$
Remark $ $ If $\,m,n\,$ are coprime then, by CRT, each combination of a root $\,r_i\,$ mod $\,m\,$ and a root $\,s_j\,$ mod $\,n\,$ corresponds to a unique root $\,t_{ij}\,$ mod $\,mn\,$ i.e.
$$\begin{eqnarray}  \begin{array}{}f(x)\equiv 0\pmod m\\g(x)\equiv 0\pmod n\end{array}  
&\iff&  \begin{array}{}x\equiv r_1,\ldots,r_k\pmod m\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod n\end{array}\\ 
&\iff& \left\{ \begin{array}{}x\equiv r_i\pmod m\\x\equiv s_j\pmod n\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\
&\overset{\large  \rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\
\end{eqnarray}\qquad\qquad$$
A: If you solve $x^2-1=0$ modulo $7$, you get $x=\pm 1$, just by factorization. Since the integers modulo $7$ form a field, you can use all of the algebra manipulations that you're accustomed to, as long as you interpret multiplicative inverses correctly.
Modulo $9$, the situation is a little different, because the integers mod $9$ do not form a field. There aren't many numbers, though, so by inspection, you can see that $1,4,7$ are the only roots of $x^3-1$.
If you want to examine that calculation more closely, note that the units modulo $9$ form a group isomorphic to $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$, so each element is congruent to $(-1)^{t_0}*4^{t_1}$, where $t_0$ is determined modulo $2$ and $t_1$ is determined modulo $3$. Cubing that expression, we obtain $(-1)^{3t_0}(4^3)^{t_1}\equiv (-1)^{t_0}$. Thus, the cubic residues mod $9$ are $1$ and $-1$, and the cube roots of $1$ are precisely the powers of $4$.
Now, you've got six combinations: $x\equiv\pm 1\pmod{7}$ and $x\equiv 1,4,7\pmod{9}$. Putting those together and applying the Chinese Remainder Theorem should get you where you're trying to go.
