Proof that every quadratic residue has two roots, modulo a prime

Question

Can someone provide a proof that every quadratic residue, when working in $\mathbb Z_p$, where $p$ is a prime, has exactly two roots? Indeed, there cannot be only one root as for any $a^2$, we know $a$ and $-a$ are both roots. So then we need only show that there cannot be more than two- does this follow from some argument for the number of roots a quadratic can have modulo a prime? I'm not well-acquainted with the theorems pertaining to prime moduli.

Answer 1

This nothing special about quadratic residues. Use the theorem about roots of polynomial leading to a linear factor: that is, if $a$ is a root of the polynomial $f(x)$ then $(x-a)$ is a factor of $f(x)$. This theorem is also valid for polynomials with coefficients in the field of $p$ elements. So a polynomial of degree $n$ can't have more than $n$ roots.

Answer 2

Isn't it that the polynomial $X^2-a^2\in \mathbb{K}[X]$ cannot have more than two roots in $\mathbb{K}$. This is the consequence of a reasoning on degrees

Answer 3

The definition of an integer $x$ being a quadratic residue (mod p) is equivalent to the statement that there exists an integer $a$ so that $x$=$a$(mod p). But this in itself is equivalent to saying that $x^2$-$a$=$0$(mod p). So now one can conisder working over the field $\mathbb Z$/$p$$\mathbb Z$[$x$]. But by the fundamental theorem of algebra, such a polynomial must have at most $2$ roots in $\mathbb Z$/$p$$\mathbb Z$[$x$].