I am trying to follow a proof of Gauss' lemma in Number Theory by George Andrews. I have a few problems with a couple assumptions made. Let g.c.d.$(m,p)=1$ where $p$ is an odd prime, and let $\mu$ be the number of integers in $$\lbrace m,2m,..., \frac12(p-1)m \rbrace$$ whose least residues modulo $p$ are negative. The notation for the least residue of $m$ modulo $p$ is $LR_p(m)$. The first problem I have is when it is stated that for any $n$, since $0 \lt |LR_p(nm)| \lt p/2$ as $n$ takes all integral values in $(0,p/2)$, so does $|LR_p(nm)|$. Why is this so? I don't see why the least residue would behave like this.
Besides that, I can follow the proof to where it is stated $$ \left(\frac mp\right) \equiv(-1)^\mu \pmod p$$ where $\left(\frac mp\right)$ is the Legendre symbol. Following that, it is stated that the congruence implies $$\left(\frac mp\right)=(-1)^\mu$$ which is where I fail to see the connection. I know each side of the congruence will equal either $1$ or $-1$, but how is it that $\mu$ is necessarily even if $m$ is a quadratic residue modulo $p$ ($\left(\frac mp\right)=1$) and necessarily odd otherwise? Apparently these are supposed to be obvious, so I would appreciate any help in understanding them.
Edit: Of the two questions I have figured out the former, but not the latter.
Although you may know the answer to your first question already, I give it here for other people. The list of positive integers
$|LR_p(m)|$, $|LR_p(2m)|$, ... $|LR_p(\frac{p-1}{2}m)|$
does not contain any repeated value, because when $|LR_p(n_1m)|$ = $|LR_p(n_2m)|$, then either $n_1m \equiv n_2m$, hence ( remember that gcd(m,p)=1) $n_1 \equiv n_2$, or $n_1m \equiv -n_2m$, hence $n_1 \equiv -n_2$, both of which cannot be the case for two different values in the range of $n_j$ values used in the list. So $n_1 = n_2$.
Then, summarizing the proof in Andrews' book, when the product of $m$, $2m$, ..., $\frac{p-1}{2}m$ is taken, this clearly equals $(\frac{p-1}{2})!\cdot m^{\frac{p-1}{2}}$
And when the product of $LR_p(nm)$ is taken for n ranging from 1 to $\frac{p-1}{2}$, this is clearly equal to the product of the signs of these least residue values, times the product of their absolute values - and we just saw that taking the product of their absolute values is congruent (mod p) to taking the product of the numbers 1 to $\frac{p-1}{2}$, which is again equal to $(\frac{p-1}{2})!$.
Striking out left and right the factor $(\frac{p-1}{2})!$ (allowed as this value is not a multiple of p), leaves us with the result that
$m^{\frac{p-1}{2}} \equiv (-1)^\mu$.
Now combine this with the fact that $(\frac{m}{p}) \equiv m^{\frac{p-1}{2}}$, and we get
$(\frac{m}{p}) \equiv (-1)^\mu$
In your second question you overlooked a very simple thing: the numbers left and right in the above congruence equation can only take values 1 or -1, so (p is odd) they must be equal in value. So we get the desired result:
$(\frac{m}{p}) = (-1)^\mu$