Find all solutions to $4x^2+6x+1 \equiv 0 \pmod {13}$ Find all solutions to $4x^2+6x+1 \equiv 0 \pmod {13}$
I think it has no solutions but I am not sure how to show this.
 A: Here is an algebraic solution.
$$4x^2+6x+1\equiv 0 \pmod{13}$$
$$10\cdot(4x^2+6x+1)\equiv 0 \pmod{13}$$
$$x^2+8x+10\equiv 0 \pmod{13}$$
$$x^2+8x+16\equiv 6 \pmod{13}$$
$$(x+4)^2\equiv 6 \pmod{13}$$
Then we look for a number from $0$ to $12$ whose square modulo $13$ is $6$. However, there is none. This could be confirmed with quadratic reciprocity, which would be need for numbers much larger than $13$.
Therefore, your equation has no solution.
A: We have $4x^2+6x+1 = (2x+3/2)^2-5/4$. Hence,
\begin{align}
4x^2+6x+1 \equiv 0 \pmod{13} \implies (2x+3/2)^2-5/4 \equiv 0 \pmod{13}
\end{align}
This simplifies into
\begin{align}
(4x+3)^2 \equiv 5 \pmod{13}
\end{align}
This has no solutions, since
$$5^{(13-1)/2}\pmod{13} \equiv 5^6\pmod{13} \equiv (25)^3 \pmod{13} \equiv-1\pmod{13}$$
A: Complete the square: $4x^2+6x+1=(2x+3)^2-8$
$2x+3$ is an integer so we can let $a=2x+3$.
Then: 
$$
a^2\equiv 8 \pmod {13} 
$$
You can check if this has a solution with a simple remainder table:

A: ${\rm mod}\ 13\!:\,\ 0\equiv (2x)^2\overbrace{+\,3}^{\large -10}\,(2x)+1\equiv X^2-10X+1\equiv (X\!-\!5)^2+2$
But $\ (X\!-\!5)^2\equiv -2\,\underset{(\ \ )^{\Large 6}}\Rightarrow\,(X\!-\!5)^{12}\equiv (-2)^6\equiv -1\not\equiv 1,\,$ contra Fermat. So it has no roots.
