Show that there are exactly two values in $\{0, 1, ..., N - 1\}$ satisfying $x^{2} \equiv a \pmod{N}$. 
Fix a positive integer $N > 1$. We say that $a$ is a quadratic residue modulo $N$ if there exists $x$ such that $a \equiv x^{2} \pmod{N}$.
Let $N$ be an odd prime and $a$ be a non-zero quadratic residue modulo $N$. Show that there are exactly two values in $\{0, 1, ..., N - 1\}$ satisfying $x^{2} \equiv a \pmod{N}$.

This means to prove that there are exactly two values of $x$ that satisfy $x^{2} \equiv a \pmod{N}$.
I think I should prove that there are at most two values and at least two values that satisfy the above constraints.
I know that because $a$ is a quadratic residue modulo $N$, $a \equiv x^{2} \pmod{N}$, and given in the problem is $x^{2} = a \pmod{N}$.
I can combine these two equations using modular arithmetic to either $x^{2}a = x^{2}a \pmod{N}$ or $x^{2} + a = x^{2} + a \pmod{N}$.
I don't know if this is on the right track or how to continue the proof.
This is a homework question, so I'd be a grateful for a hint of some sort that nudges me in the right direction.

Few remarks just to remind you some theorems about congruences when the modulus is prime:
1- If $p$ is a prime number and $P(x)$ is a polynomial of degree $k$, then $P(x) \equiv 0 \pmod{p}$ has at most $k$ number of solutions.
2- when $p$ is a prime number, $\mathbb{Z}/p\mathbb{Z}$ is an integral domain. That means whenever $a.b \equiv 0 \pmod{p}$ then $a \equiv 0 \pmod{p}$   or   $b \equiv 0 \pmod{p}$. This can be equivalently said as $p \mid ab \implies p \mid a$ or $p \mid b$. 
3- Every finite integral domain is a field (This is more related to abstract algebra than elementary number theory though).
4- Notice that if $n\neq 1>0$ then  $2 \equiv 0 \pmod{n}$ if and only $n=2$. As a corrolary, $-1 \equiv 1 \pmod{n}$ if and only if $n=2$.
The rest is now clear as  André Nicolas has said:
If $a$ is a quadratic residue, then by definition there is at least one $x$ such that $x^2 \equiv a \pmod{N}$. But also $(-x)^2 \equiv a \pmod{N}$. Therefore if $x$ and $-x$ are different then we have found at least two solutions, also because of remark 1 we know that we can have at most 2 solutions because $P(x)=x^2-a$ is of degreee 2 and that proves there are exactly 2 solutions.
To show that $x \not\equiv -x \pmod{N}$ assume the opposite. If $x \equiv -x \pmod{N}$ then $2x \equiv 0 \pmod{N}$, but $2 \not\equiv 0 \pmod{N}$ because $N \neq 2$. That implies $x \equiv 0 \pmod{N}$.But that forces $a$ to be $0$ modulo $N$ which is contradiction.
