# Understanding the set of least residues mod p, p a prime

I would like clarification\intuition about the set of least residues mod p, and the integer $\mu$.

Definition: Consider S= {$-(p-1)/2, -(p-3)/2,\cdots ,-1,1,2,\cdots ,(p-1)2$}.This is called the set of least residues $\mod{p}$

If p does not divide a, let $\mu$ be the number of negative least residues of the integers a,2a,3a,...,(p-1)(1/2)a.

Also, is there an easy way to calculate $\mu$ or is it just used in proofs?

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Is what you write about $(a/p)$ under "I know that..." the definition you have of the Legendre symbol, or a derived property? Because the definition I know of $(a/p)$ would make what you want to prove immediate, and what you describe as "knowing" is a derived property for it... –  Arturo Magidin Feb 9 '12 at 7:07
@ArturoMagidin, I understand, you are right. I will edit my question. –  Edison Feb 9 '12 at 7:13
As you know, $\mu$ is used in one proof of quadratic reciprocity. So far as I know, that is its only use. Here is a project: calculate $\mu$ for some prime like 23 or 29 and all $a$ from 1 to $p-1$, and see whether anything interesting turns up. –  Gerry Myerson Feb 9 '12 at 11:58
@GerryMyerson, Thanks for your comment, I will try this. –  Edison Feb 9 '12 at 15:03

First, we should assume that $p$ is odd.
If $a$ is a quadratic residue mod $p$, then $x^2\equiv a \pmod p$ has exactly two solutions. All you need is to consider this equation in $\mathbb{Z}/p\mathbb{Z}$. Remember that this is a quadratic equation so it has at most two solutions. On the other hand, if a is a root then so is $p-a$. And in this case, $(\frac{a}{p})=1$ by the definition.
This exercise is also true if we replace $p$ by $p^n$. In this case, you need the Hensel's Lemma.
$\LaTeX$ tip: \pmod produces the parenthesis and the correct spacing, and should be used rather than \bmod (which is a binary operator, hence the extra space before the "mod"). Use \pmod{text} to produce $\pmod{\mathrm{text}}$. To remember: b inary mod, \bmod; p arenthetical mod, \pmod. –  Arturo Magidin Feb 9 '12 at 7:19