Equations modulo $143$ Let $x=11$ and $y=13$, and $z=xy=143$. 

(i) Show that $1$, $x+1$, $−1$ and $−(x+1)$ are the $4$ solutions of $n^2 \equiv 1\pmod z$. 

(ii) Find the coset of $U_z(2)$ consisting of solutions to $n^2 \equiv 9\pmod z$.

I do not know at all how to do this problem. I am very iffy about this section.
Hints: For question (i), you should realise that $n^2\equiv1\pmod{11}$ and $n^2\equiv1\pmod{13}$. Can you see why this is? Find $n$ in each of these two settings, and then use (for example) the Chinese Remainder Theorem to prove the result. To find $n$ in these settings, you can (if you want) use the fact that $n^2-1=(n-1)(n+1)$. Note: We could not use this equality upstairs because of zero divisors. For example, $11\times 13\equiv0\pmod{z}$. This does not happen when we are working modulo a prime.

For question (ii), either proceed as above to find all solutions to the given equation, or multiply your solutions to part (i) to find all solutions to the given equation. I phrase this an an either/or because it is not obvious that the second method works (it is not obvious that you get all the solutions). So if you pick the second method (which is quicker) then you should convince yourself that it works (if I was marking your exam and you picked that method I would expect a few lines of working to explain why it worked). Then, suppose the solution set is $\{n_1, n_2, \ldots, n_q\}$. This means that your coset is written $2\{n_1, 2n_2, \ldots, 2n_q\}$, which is the set...what? You tell me. :-)
Hints: For question (i), you should realise that $n^2\equiv1\pmod{11}$ and $n^2\equiv1\pmod{13}$. Can you see why this is? Find $n$ in each of these two settings, and then use (for example) the Chinese Remainder Theorem to prove the result. To find $n$ in these settings, you can (if you want) use the fact that $n^2-1=(n-1)(n+1)$. Note: We could not use this equality upstairs because of zero divisors. For example, $11\times 13\equiv0\pmod{z}$. This does not happen when we are working modulo a prime.

For question (ii), either proceed as above to find all solutions to the given equation, or multiply your solutions to part (i) to find all solutions to the given equation. I phrase this an an either/or because it is not obvious that the second method works (it is not obvious that you get all the solutions). So if you pick the second method (which is quicker) then you should convince yourself that it works (if I was marking your exam and you picked that method I would expect a few lines of working to explain why it worked). Then, suppose the solution set is $\{n_1, n_2, \ldots, n_q\}$. This means that your coset is written $2\{n_1, 2n_2, \ldots, 2n_q\}$, which is the set...what? You tell me. :-)