When computing the Legendre symbol $(\frac{3}{p})$, how to combine different congruences together into one statement? I try to find Legendre symbol for $\left(\dfrac{3}{p}\right)$.
This is what I did so far:
case 1: $p=3\mod{4}. $ 
so $\left(\dfrac{3}{p}\right)=-\left(\dfrac{p}{3}\right)$. now $\left(\dfrac{3}{p}\right)=-1$ iff $p=2 \mod{3}$
case 2: $p=1\mod{4}$
so $\left(\dfrac{3}{p}\right)=\left(\dfrac{p}{3}\right)$. now $\left(\dfrac{3}{p}\right)=1$ iff $p=1 \mod{3}$
Now, I don't know how to combine the cases (each case with itself and together).

As you've noted, the cases are as follows:


*

*If $p\equiv 3\bmod 4$, then  $(\frac{3}{p})=-(\frac{p}{3})$.


*

*If $p\equiv 1\bmod 3$, then $(\frac{p}{3})=1$, so $(\frac{3}{p})=-1$.

*If $p\equiv 2\bmod 3$, then $(\frac{p}{3})=-1$, so $(\frac{3}{p})=1$.


*If $p\equiv 1\bmod 4$, then  $(\frac{3}{p})=(\frac{p}{3})$.


*

*If $p\equiv 1\bmod 3$, then $(\frac{p}{3})=1$, so $(\frac{3}{p})=1$.

*If $p\equiv 2\bmod 3$, then $(\frac{p}{3})=-1$, so $(\frac{3}{p})=-1$.



Thus,
$$\left(\frac{3}{p}\right)=\begin{cases}
1 & \text{ if }\begin{cases}p\equiv 3\bmod 4 \text{ and }p\equiv 2\bmod 3,\text{ or }\\
p\equiv 1\bmod 4\text{ and }p\equiv 1\bmod 3, \end{cases}\\[0.1in]
-1 & \text{ if }\begin{cases}p\equiv 3\bmod 4 \text{ and }p\equiv 1\bmod 3,\text{ or }\\
p\equiv 1\bmod 4\text{ and }p\equiv 2\bmod 3, \end{cases}
\end{cases}$$
The Chinese remainder theorem tells you that any statement of the form
$$n\equiv a\bmod 3\quad\text{ and }\quad n\equiv b\bmod 4$$
can be converted, essentially uniquely, into a statement of the form
$$n\equiv c\bmod 12.$$
The best way of solving this sort of thing in general is to find an $x$ and $y$ such that 
$$x\equiv 1\bmod 3 \quad\text{ and }\quad x\equiv 0\bmod 4$$
$$y\equiv 0\bmod 3\quad\text{ and }\quad y\equiv 1\bmod 4$$
so that the statement "$n\equiv a\bmod 3$ and $n\equiv b\bmod 4$" is equivalent to
$$n\equiv xa+yb\bmod 3\quad\text{ and }n\equiv xa+yb\bmod 4,$$
hence
$$n\equiv xa+yb\bmod 12.$$
For example, one choice that works is $x=4$ and $y=9$. Thus, to reformulate the statement that $$p\equiv 1\bmod 4\quad\text{ and }\quad p\equiv 2\bmod 3,$$
 you have that $9\cdot 1+4\cdot 2=17$, and
$$p\equiv 1\equiv 17\bmod 4\quad\text{ and }\quad p\equiv 2\equiv 17\bmod 3,$$ so that $$p\equiv 17\equiv 5\bmod 12.$$
You can check that this is correct:
$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|}
n \bmod 12 & 0 & 1 & 2 & 3 & 4 & \color{red}{\large \mathbf{5}} & 6 & 7 & 8 & 9 & 10 & 11 \\\hline
n\bmod 3 & 0 & 1 & 2 & 0 & 1 & \color{red}{\large \mathbf{2}} & 0 & 1 & 2 & 0 & 1 & 2 \\\hline
n\bmod 4 & 0 & 1 & 2 & 3 & 0 & \color{red}{\large \mathbf{1}} & 2 & 3 & 0 & 1 & 2 & 3 
\end{array}$$
