This is a proof from Burton's Number Theory.
Theorem: If $p$ is an odd prime, then $(2/p)= \begin{cases} 1, & \text{if $p = 1 (\mod 8)$ or $7 (\mod 8)$} \\ -1, & \text{if $p = 3 (\mod 8)$ or $5 (\mod 8)$}\end{cases}$\
According to Gauss' lemma, $(2/p)=(-1)^n$, where $n$ is the number of integers in the set \begin{align*} S=\{1\cdot 2, 2\cdot 2,3\cdot 2,...,(\frac{p-1}{2})\cdot 2\} \end{align*} which upon division by $p$, have remainders greater than $p/2$. The members of $S$ are all less than $p$, so that it suffices to count the number that exceed $p/2$. For $1\leq k\leq (p-1)/2$, we have $2k<p/2$ if and only if $k<p/4$. If $[ ]$ denotes the greatest integer function, then there are $[p/4]$ integers in $S$ less than $p/2$; hence, \begin{align*} n=\frac{p-1}{2}-[\frac{p}{4}] \end{align*} is the number of integers that are greater than $p/2$.\ Now, we have four possibilities; for any odd prime has one of the form $8k+1, 8k+3, 8k+5,$ or $8k+7$.
This is where I got confused:\ (1) For $1\leq k\leq (p-1)/2$, we have $2k<p/2$ if and only if $k<p/4$.\ (2) If $[ ]$ denotes the greatest integer function, then there are $[p/4]$ integers in $S$ less than $p/2$
Thank you for your help!
See my answer on a similar question: Using gauss's lemma to find $(\frac{n}{p})$ (Legendre Symbol)
This hopefully clarifies to you how Gauss' lemma can best be used to calculate $(\frac{n}{p})$.