I need to solve $3x^2 + 6x +1 \equiv 0 \pmod {19}$
I saw the same problem here –
Solving the congruence $3x^2 + 6x + 1 \equiv 0 \pmod {19}$
but didn't understand how he got to the conclusion
$x(x+2) \equiv 6 \pmod {19}$
and anyway im trying to solve it how we learned in class –
multiply both sides and the modulo by $4a$ then solve the two equations
$y^2 \equiv b^2 -4ac \pmod {4an}$
$2ax + b \equiv y \pmod {4an}$
so I tried multiplying the hole thing by $4a$ , that is $4 \times 3$
and got to $(2 \times 3x + 6)^2 \equiv 36 -4 \times 3 \pmod {19 \times 4 \times 3}$
now I am stuck . any help will be appreciated
Best Answer
That is equivalent to: $$ 3x^2+6x+39\equiv 0\pmod{19} $$ or to: $$ x^2+2x+13\equiv 0\pmod{19} $$ or to: $$ (x+1)^2 \equiv 7\pmod{19}. $$ Since $\left(\frac{7}{19}\right)=-\left(\frac{5}{7}\right)=-\left(\frac{2}{5}\right)=+1$ and $19$ is a prime of the form $4k-1$, a square root of $7$ is given by $$ 7^{\frac{19+1}{4}}\equiv 7^{5}\equiv 11\pmod{19} $$ and the solutions are $x+1\equiv \pm 11\pmod{19}$, or $x\in\{7,10\}\pmod{19}$.