Elementary Number Theory: Solving Quadratic Conguences If possible solve the following congruence, 
$7x^2-4x+1 \equiv 0 \pmod {11}$.
I am using the quadratic equation but I am stuck. The following is where I got stuck:
$x \equiv 4(4\pm \sqrt{-12}\pmod {11}$
I have tried adding $11$ repeatively but I cannot get a perfect square root. Please help!
 A: $$0 \equiv 7x^2-4x+1 \equiv -4x^2-4x-1+2 \equiv 2-(2x+1)^2 \pmod {11}$$
so 
$$(2x+1)^2 \equiv 2\pmod {11}$$
Since for each $a$ we have $$a^2 \equiv 0,1,4,-2,5,3\pmod {11}$$
there is no solution to this congruence.
A: $-12\equiv_{11}-1$ is not a square in $\mathbb{Z}_{11}$, so $\sqrt{-12}$ in your expression has no value, and there are therefore no solutions. 
There are two ways to see why this is the case:


*

*$\left (\frac{-1}{p}\right )=(-1)^{\frac{p-1}{2}}$, the Legendre symbol. So $\left (\frac{-1}{11}\right )=(-1)^5=-1$.

*By brute-force: $0^2\equiv_{11}0$, $(\pm 1)^2\equiv_{11}1$, $(\pm 2)^2\equiv_{11}4$, $(\pm 3)^2\equiv_{11}-2$, $(\pm 4)^2\equiv_{11}5$, $(\pm 5)^2\equiv_{11}4$. So $-1$ is not a square.
A: There is no solution.
Multiply by $8$ and complete the square to get
$$(x+5)^2\equiv 6(\mod 11)$$ and $6$ is a quadratic non residue.
A: Note that since $8$ is the inverse of $7 \pmod {11}$ we have
$$7x^2-4x+1 \equiv 0\iff 8\cdot7x^2-8\cdot4x+8\cdot1 \equiv 8\cdot0\iff x^2+12x+8\equiv 0 \pmod {11}$$
and 
$$x^2+12x+8\equiv0 \iff (x+6)^2 \equiv6 \pmod {11}$$
whic is impossible.
A: So quadratic formula:
$7x^2 -4x + 1\equiv 0 \mod 11\implies$ (using the standard notation abuse)
$x \equiv \frac {4 \pm \sqrt {16 - 28}}{14}\mod 11$
$\frac 12 \equiv 6$ and $\frac 17 \equiv 8$ so $\frac 1{14}\equiv 48 \equiv 4$.
So $x \equiv {16 \pm \sqrt{-12} } \equiv 5\pm \sqrt{-1}$
Now $x^2 \equiv -1 \mod 11$ has no solutions as $11 \not \equiv 1 \mod 4$.
So no solutions.
Or we could try doing this directly:
$7x^2 -4x + 1\equiv 0$
$-4x^2 - 4x \equiv -1$
$4x^2 - 4x \equiv 1$
$4x^2 + 4x \equiv 1$
$4x(x+1)\equiv 1$
$x(x+1)\equiv -x(-x - 1) \equiv 3 \mod 11$.
We could simply test $6$ cases:  $x=0;\pm 1$ are trivial.  $\pm 2;3;4;5$ are easy.
Or we can simply list the quadratic residues mod $11$ are $0, 1,4,9,5,3$.
And completing the square:
$7x^2 - 4x + 1\equiv 0$
$\frac 17(7x^2 - 4x + 1) \equiv 0$
$8(7x^2 - 4x +1) \equiv 0$
$x^2 - 32x + 8 \equiv 0$
$x^2 - 32x \equiv -8$
$x^2 -10x + 25 \equiv 17\equiv 6$
$(x-5)^2 \equiv 6 \mod 11$ and $6$ is not a modulo class.
