Exercise in algebra with modulo I'm studying Cassels' book Elliptic Curves for a week now, and I'm at the local global principle. I'm trying to prove the first exercise of this chapter, which says 

Let $p > 2$ be prime and let $b, c \in \Bbb Z$ and $ p $ does not divide $b$. Show that $bx^2 + c$ takes
  precisely $\frac {p + 1}{2}$ distinct values $p$ for $x \in \Bbb Z$. 

The problem here is that i don't understand what "values $p$" means. At first i thought that it means $\mod(p)$ values, but if that was the case, then for $x=0\in \Bbb Z$,we have that the $b\cdot 0^2+c=c$ can take $p-1$  values $\mod(p)$, $\{0,1,...,p-1\}$, and then we should have that $p-1=\frac {p + 1}{2}=>p=3$ which restricts the exercise.
So what  does "values $p$" mean?
Thank you!
 A: Since $b$ is not divisible by $p$ it has an inverse under multiplication, call it $b'$ for now, such that $b'b=1$ mod $p$. Then the equation $bx^2=c$ may be multiplied on both sides by $b'$ to get $b'bx^2=b'c,$ which since $b'b=1$ is the same as the equation
$$x^2=b'c \tag{1}$$
Thus since $b,c$ are fixed we may as well define $d=b'c$ and we're looking at the equation $x^2=d$. Now note that $x^2=(-x)^2$ for any $x$, and note that if $d \neq 0$ the values $x,-x$ are distinct, otherwise $2x=0$ and then $x=0$ since $p$ is odd. The solution to $x^2=0$ is only $x=0$, so that counts as one, and the other solutions can be paired up as $x,-x$ as indicated above. The "values taken on" are then the various possible $b$ which can appear on the right side of $x^2=b$, so by the 2-to-1 ness of the solutions for nonzero $b$ there are $(p-1)/2$ with nonzero $b$, and one more solution when $b=0$, for a total of $(p-1)/2+1=(p+1)/2.$
Once you've arrived at $x^2=b$ you are looking for values of $b$ which are either $0$ or else are what is called "quadratic residues". If you google the latter you can likely find sites where this count of the quadratic residues is done more carefully than outlined above.
