I am looking for some hints please!
Show that if $m = p_1\cdots p_r$ is a product of distinct odd primes, the set of odd $a$ such that
$\left(\dfrac{a}{m}\right) = 1$ are those lying in half of the congruence classes $b$ modulo $m$ such that $\gcd(b,m) = 1$.
Here is what I am thinking:
Want: odd $a_i$ for $i = 1$ to $\frac{m-1}{2}$ (half the congruences of $b \mod m$)
More explicitly I want:
$b \equiv a_1 \mod m$
$b \equiv a_2 \mod m$
$\phantom{b\ \ }\vdots\phantom{\equiv a \mod m}$
$b \equiv a_{\frac{m-1}{2}} \mod m$
Working backwards this could be written with the CRT as
$b \equiv a_i \mod p_i$
for $i = 1$ to $\frac{m-1}{2}$.
However, how could I get to this point and count the number of congruence classes that I have?