quadratic congruence

Let $a,b,c$ be known integers and $p$ an odd prime number not dividing $a$.  The number of non-congruent roots of the quadratic congruence

$a{x}^2+bx+c\equiv \mathrm{\hspace{0.33em}0}(modp)$(1)

is

• two, if  $b^2-4ac$  is a quadratic residue modulo $p$;

• one, if  $b^2-4ac\equiv 0(modp)$;

• zero, if  $b^2-4ac$  is a quadratic nonresidue modulo $p$.
  

Proof.  Since  $\gcd (p,\mathrm{\hspace{0.17em}4}a)=1$,  multiplying (1) by $4a$ gives an equivalent (http://planetmath.org/Equivalent3) congruence

$$4a^2{x}^2+4abx+4ac\equiv \mathrm{\hspace{0.33em}0}(modp)$$

which may furthermore be written as

$${(2ax+b)}^2\equiv b^2-4ac(modp).$$

Accordingly, one can obtain the the solution of the given congruence from the solution of the pair of congruences

$\{\begin{array}{cc}{y}^2\equiv b^2-4ac(modp)\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}(2)\hfill & \\ 2ax+b\equiv y(modp).\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.25em}(3)\hfill & \end{array}$

Case 1:  $b^2-4ac$ is a quadratic residue$(modp)$.  Then (2) has a root  $y=y_0\ne 0$,  and therefore also the second root  $y=-y_0$.  The roots  $y=\pm y_0$ are incongruent, because otherwise one had $p\mid 2y_0$  and thus  $p\mid y_0\mid y_0^2\equiv b^2-4ac$  which is not possible in this case.
Case 2:  $b^2-4ac\equiv 0(modp)$.  Now (2) implies that  $y\equiv 0(modp)$,  whence the corresponding root $x_0$ of the linear congruence (3) does not allow other incongruent roots for (1).
Case 3:  $b^2-4ac$ is a quadratic nonresidue$(modp)$.  The congruence (2) cannot have solutions; the same concerns thus also (1).

Example.  Solve the congruence

$$4x^2+6x-3\equiv \mathrm{\hspace{0.33em}0}(mod43).$$

We have  $b^2-4ac=36+4\cdot 4\cdot 3=84\equiv -2(mod43)$  and the Legendre symbol

$$\left(\frac{-2}{43}\right)=\left(\frac{-1}{43}\right)\left(\frac{2}{43}\right)=-1\cdot (-1)=\mathrm{\hspace{0.33em}1}$$

(see values of the Legendre symbol) says that $-2$ is a quadratic residue modulo 43.  The congruence corresponding (2) is

$$y^2\equiv -2(mod43),$$

which is satisfied by  $y\equiv \pm 16(mod43)$ as one finds after a little experimenting.  Then we have the two linear congruences  $2\cdot 4x+6\equiv \pm 16(mod43)$,  i.e.

$$4x\equiv \pm 8-3(mod43)$$

corresponding (3).  The first of them,  $4x\equiv 5(mod43)$,  is satisfied by  $x=12$  and the second,  $4x\equiv -11(mod43)$,  by  $x=8$.  Thus the solution of the given congruence is

$$x\equiv \mathrm{\hspace{0.33em}8}(mod43)\mathit{\hspace{1em}}\text{or}\mathit{\hspace{1em}}x\equiv \mathrm{\hspace{0.33em}12}(mod43).$$