solving $3x^2 +4x-2= 0\pmod{31}$

Question

I tried multiplying both sided by 4a which leads to $(6x+4)^2=40 \pmod{372}$ now I'm stuck with how to find the square root of a modulo.

Answer 1

Might be easier to factor or use the quadratic formula.

$3x^2 + 4x - 2\equiv 0\pmod {31}$ so abusing notation where $\sqrt {k}$ will mean the congruence $a$ where $a^2 \equiv k \pmod {31}$ and $\frac 1{m}=m^{-1}$ is the congruence where $m(m^{-1})\equiv 1 \pmod {31}$ then

$x \equiv \frac {-4\pm \sqrt{16 +24}}{6}\equiv$

$(-4 \pm \sqrt{40})\cdot 6^{-1}\equiv$

$(27\pm \sqrt{9})\cdot 6^{-1}\equiv $

$(27\pm 3)\cdot 6^{-1}\equiv $

$\begin{cases}30\cdot 6^{-1}\equiv 5\cdot 6\cdot 6^{-1}\equiv 5\\ 24\cdot 6^{-1}\equiv4\cdot 6\cdot 6^{-1}\equiv 4\end{cases}\pmod{31}$

So $x \equiv 4\pmod {31}$ or $x\equiv 5\pmod{31}$

Answer 2

First multiply by $3^{-1} \equiv 21 \mod 31$ to get $x^2 + 22 x + 20 \equiv 0 \mod 31$. Then complete the square to get $(x+11)^2 \equiv 101 \equiv 8 \mod 31$. Now the square roots of $8$ mod $31$ are $15$ and $16$, so $x+11 \equiv 15$ or $16$ and $x \equiv 4$ or $5$ mod $31$.

Why, you ask, are those the square roots of $8$ mod $31$? Well, $2^5 = 32 \equiv 1$, so $8 = 2^3 \equiv 2^{-2}$, and one of the square roots of that is $2^{-1} \equiv 16$. The other is $-16 \equiv 15$.

Of course it might have been simpler to just compute $3x^2 + 4 x - 2 \mod 31$ for each $x$ from $0$ to $31$ until you find the roots.

Answer 3

Completing the square, we have: $$5x^2\equiv 2(x-1)^2$$ Since $36\equiv 31+5\equiv 5$, we have: $$36x^2\equiv2(x-1)^2$$ $$18x^2\equiv(x-1)^2$$ Adding $31$ again: $$49x^2\equiv(x-1)^2$$ Then $7x\equiv x-1$ or $-7x\equiv x-1$. Which is the same as $6x\equiv-1\equiv30$ or $8x\equiv1\equiv 32$.

Giving us the final answer of $x\equiv4,5$.

Answer 4

$$3x^2+4x-2\equiv0\operatorname{mod}31$$it's$$ 21(3x^2+4x-2)\equiv0\operatorname{mod}31$$ or$$x^2-9x+20\equiv0\operatorname{mod}31$$ or$$(x-4)(x-5)\equiv0\operatorname{mod}31,$$ which gives $$x\equiv4\operatorname{mod}31$$ or $$x\equiv5\operatorname{mod}31.$$