Find all solutions mod $19$ to $4x^2+6x+1 \equiv 0$ mod $19$

Question

Find all solutions mod $19$ to $4x^2+6x+1 \equiv 0$ mod $19$

I am not really sure how to start this problem.

Answer 1

Since the modulus $19$ is so small, you could just try all $19$ possibilities and see which of them give the result $0$. That would certainly be faster than trying something more sophisticated. I did that and got the solution $x\equiv 11$ or $x\equiv 16$.

If you want something more algebraic,

$$4x^2+6x+1\equiv 0 \pmod{19}$$ $$5\cdot(4x^2+6x+1)\equiv 0 \pmod{19}$$ $$x^2+11x+5\equiv 0 \pmod{19}$$ $$x^2-8x+5\equiv 0 \pmod{19}$$ $$x^2-8x\equiv -5 \pmod{19}$$ $$x^2-8x+16\equiv 16-5 \pmod{19}$$ $$(x-4)^2 \equiv 11\pmod{19}$$ $$x-4 \equiv \pm 7\pmod{19}$$ $$x\equiv 4\pm 7 \pmod{19}$$ $$x\equiv 11 \text{ or }x\equiv -3 \pmod{19}$$ $$x\equiv 11 \text{ or }x\equiv 16 \pmod{19}$$

Answer 2

Substitute $\,2x = X\,$ to get $\ X^2\! + 3X\,\overbrace{ -\, 18}^{\large \equiv\, 1} = (X\!+\!6)(X\!-\!\color{#c00}3),\, $ and $\,2x \equiv \color{#c00}3 = 22\,\Rightarrow\, x\equiv 11$

Remark $\ $ This can be viewed as a special case of the AC-method.

Answer 3

As usual, but the computations will take place in $\mathbf F_19$: as $4^{-1}=5$, $$4x^2+6x+1=4(x^2-8x+5)=4\bigl((x-4)^2+3+5\bigr).$$ So all you have to do is finding a square root of $-8$ (or $11$, it amounts to the same).You may check by hand that $-8$ has two square roots in $\mathbf F_{19}$ by computing successively: $$(\pm 2)^2, (\pm 3)^2, (\pm 4)^2, (\pm 5)^2, (\pm 6)^2, (\pm7)^2, (\pm 8)^2, (\pm 9)^2. $$ It happens that the square roots are $\pm 7$. Thus: $$x\in \{4\pm 7\}=\{-3,-8\}$$