How to read a proof? As I go deeper and deeper into upper division math courses, I find some proofs to be very challenging to understand. Right now I am trying to understand Gauss's lemma in number theory and I can't understand most of the arguments. Can you suggest some tips to read and understand mathematical literature?.
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A: Since you mentioned Gauss' Lemma, I'll try to give a proof with some annotation:

$\textbf{Gauss' Lemma}$. Let $p$ be an odd prime such that $p \nmid a$ for $a \in \mathbb{Z}$.  If 
$A=\{ka \mod p \mid 1 \leq k \leq \frac{p-1}{2}\}$, 
$B=\{-k \mod p \mid 1 \leq k \leq \frac{p-1}{2}\}$, and 
$n=|A \cap B|$, 
then $\left(\frac{a}{p}\right)=(-1)^n$

$\textbf{Proof}$: For simplicity, let all equivalences be congruence modulo $p$.  The general strategy will be to simplify the product $(a)(2a)(3a)\cdots \left(\frac{p-1}{2}a\right)$ in two ways.  
$1)$ $(a)(2a)(3a)\cdots \left(\frac{p-1}{2}a\right)=\left(\frac{p-1}{2}\right)! \cdot a^{\frac{p-1}{2}}=\left(\frac{p-1}{2}\right)! \left(\frac{a}{p}\right)$ (Here we have used Euler's Criterion for that last equality)
$2)$ Before we simplify the product, we define an "absolute value" on the set $\{0, 1, 2, \ldots , p-1\}$.  When taken modulo $p$, this set is the same as
$\{-\frac{p-1}{2}, \ldots,-2, -1, 0, 1, 2, \ldots \frac{p-1}{2} \}$, which we will call $S$.  
Thus, for $x \in S$, we can define 
$|x|=x$ when $0 \leq x \leq \frac{p-1}{2}$, and 
$|x|=-x$ when $-\frac{p-1}{2} \leq x \leq -1$
This function has three important properties.  First, $0 \leq |x| \leq \frac{p-1}{2}$ for all $x \in S$, and second, $|x|=-x$ iff $x \in B$ (recall that $B$ is just the "negatives" of $S$).  
The third property is that this function maps $A \rightarrow \{1, 2, 3, \ldots , \frac{p-1}{2}\}$ as a bijection.  To see this, we will show the function is one-to-one, and since they have the same number of elements, it must be a bijection.  Note that $1 \leq |ka| \leq \frac{p-1}{2}$ for all $k$.  
Recall that to show a function $f$ is one-to-one, assume $f(x)=f(y)$ and show $x=y$.  So assume $|ka|=|la|$ for $1 \leq k, l \leq \frac{p-1}{2}$ (remember, think of this as $k$ and $l$ being "positive").  Then $ka=\pm la \implies k=\pm l$.  This implies that $k=l$ since $k$ and $l$ have the same "sign".  Therefore, $ka=la$, so $||$ is one-to-one.  
This means each $ka$ can be replaced with either $|ka|$ or $-|ka|$, the latter being only when $ka \in B$.  
Finally, the original product is just $$(a)(2a)(3a)\cdots \left(\frac{p-1}{2}a\right)=$$ $$(-1)^{A \cap B}|a||2a||3a|\dots \left|\frac{p-1}{2}a\right|=$$ $$(-1)^n(1)(2)(3)\dots \left(\frac{p-1}{2}\right)=$$ $$\left(\frac{p-1}{2}\right)!(-1)^n$$
Equating this with the first simplification gives us $\left(\frac{p-1}{2}\right)! \left(\frac{a}{p}\right)=\left(\frac{p-1}{2}\right)! (-1)^n \implies$ $$\left(\frac{a}{p}\right)=(-1)^n$$

This is one of my favorite theorems, so if any part still doesn't make sense, I would be happy to edit my answer.  
