Solving quadratic equations in the field $F_5$

Question

Let $y = x^2 + 2x + 2 = 0$. Solve the equation in the field $F_5$.

So I used the common $b^2 - 4ac$ formula and got that $x$ is either $-1/2$ or $-3/2$ but I'm not sure if this is in the field...

Answer 1

We have $a=1$, $b=2$, $c=2$, so $b^2-4ac=4-8=-4=1$. Moreover $2^{-1}=3$, so $$ x=2^{-1}(-2\pm1)=3(-2\pm1) $$ which means $x=3(-3}=-9=1$ or $x=3(-1)=-3=2$. The formula $$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ indeed means $$ x=(2a)^{-1}(-b\pm\sqrt{b^2-4ac}) $$

Check: $$ 1^2+2\cdot1+1=0,\qquad 2^2+2\cdot2+2=0 $$

Answer 2

Notice that $$(x+ \overline{1})^2 + \overline {1} = (x + \overline{1})^2 -\overline{1}^2 + 2 = x^2+\overline{2}x+\overline{2}$$

Then you want to find $x \in F_5=\mathbb{Z}_5$ such that $(x+\overline{1})^2 = \overline{-1}$. That is, $\overline{1}$ and $\overline{2}$.