Complete representative set of squares modulo $15$. 
$2.\,\,$Do the following computations.
$\text{(a)}$ Solve the equation $x^2\equiv 1\mod15$
Solution: We only need to choose a complete representative set modulo $15$ and verify the equation over such a set. In the following table, we choose the representative set 
  $$\{0,\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm7\}$$ 
  and verify the equation as follows: $$
\begin{array}{c|c}
x   & 0 & \pm 1 & \pm 2 & \pm 3 & \pm 4 & \pm 5 & \pm 6 & \pm 7 \\
\hline
x^2 & 0 &    1  &   4   &  -6   &   1   &   -5  &   6   &    4   \\
\end{array}
$$
We see that the equation has four solutions: $\pm1$ and $\pm 4$.
(Note that $15$ is not a  prime, so we do not just have two square roots!)

I understand that if you do $8^2 \mod 15 $, $9^2  \mod 15$, $10^2 \mod 15$, and so forth, you get repetitions of the above representative set.  What I don't understand is how you could know this before doing those calculations manually.   

If you really wanted to, you could choose the set $\{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14\}$, but notice that $8 \equiv -7 \mod 15$, $9 \equiv -6 \mod 15$, etc. So then it is more convenient to choose the representative set $\{0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6, \pm 7\}$. This cuts down the amount of work from $15$ computations to $8$, as clearly $a^2 \equiv (-a)^2$.
