How to solve $2x^2+2x+3=0,\:x\in \mathbb{F}_5$

Question

How to solve $2x^2+2x+3=0$ when $x\in \mathbb{F}_5$

the answers for $2x^2+2x+3=0$ When $x\in \mathbb{C}$ is:

$$x=-\frac{1}{2}+i\frac{\sqrt{5}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{5}}{2}$$

but how to solve it when $x\in \mathbb{F}_5$

Answer 1

Let's see how you arrived at the formula, and how can we apply that to $\mathbb F _5$:

$$ 2x^2 + 2x + 3 = 0 $$ $$ \iff 4x^2 + 4x + 6 = 0 \text { (2 is invertible, in both } \mathbb C \text{ and } \mathbb F_5 \text{)}$$ $$ \iff (4x^2 + 4x + 1) = -5 \text { (completing the square)} $$ $$ \iff (2x+1)^2 = -5 $$

Now, in $\mathbb C$ we have $ 2x+1 = \pm \sqrt {-5} $, but here in $\mathbb F_5$ we can say that

$$ (2x+1)^2 = -5 \iff (2x+1)^2 = 0 $$

From that we have

$$ \iff 2x+1 = \sqrt 0 $$ $$ \iff 2x+1 = 0 $$ $$ \iff 2x = 4 $$ $$ \iff x = 4 \times \frac 1 2 $$ $$ \iff x = 4 \times 3 $$ $$ \iff x = 2 $$

Be careful while calculating $2^{-1}$ in $\mathbb F_5$.

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In this particular case, as noticed in the comments, test every possible values of $x$ is pretty easy, but this will be useful in larger (finite or not) fields.

Answer 2

Same methode, except you have to find the square roots.

However, in $\mathbf F_5$, it is simpler: the equation can be rewritten as $$2x^2+2x-2=0\iff x^2+x-1=0.$$ Its discriminant is $\Delta=1+4=0$, so it has a double root which is (with the standard notation) $$x=-\frac b{2a}=-2^{-1}=\color{red}2.$$ Indded in $\mathbf F_5$, one has $x^2+x-1=(x-2)^2.$

Answer 3

The roots are, in fact, $x=-\frac{1}{2}+i\frac{\sqrt{5}}{2}$ and $x=-\frac{1}{2}-i\frac{\sqrt{5}}{2}$!

More precisely, if you choose $\alpha \in \mathbb{F}_5$ satisfying $\alpha^2 = -1$ and you choose $\beta \in \mathbb{F}_5$ such that $\beta^2 = 5$, then the roots are $-\frac{1}{2} \pm \frac{1}{2} \alpha \beta$.

Or to make things more simple, you might instead choose $\gamma$ satisfying $\gamma^2 = -5$; the two roots are then $-\frac{1}{2} \pm \frac{1}{2} \gamma$.

In case it's not clear, by $\frac{1}{2}$, I mean the result of dividing $1$ by $2$ in the field $\mathbb{F}_5$, which you can do because $2$ is nonzero, and thus invertible.

You can verify these are roots simply by plugging them into the equation. Unfortunately, both of these solutions turn out to be the same number, so before you can say you've solved the problem, you need to find a way to verify that it is actually a double root. For example, factoring the polynomial.

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I mainly went through this to show that intuition derived from real and complex numbers can often suggest what is true for finite fields and other rings as well, so it's often worth just trying to figure out an analogy, and see if the result is what you're looking for.

That said, the "right" way to approach this is to realize that the method you used to find the roots in $\mathbb{C}$ applies perfectly well to finding the roots in $\mathbb{F}_5$.

For example, maybe you completed the square. You can do that over $\mathbb{F}_5$ too; in fact, it works over any field whose characteristic is not $2$.

(it works in characteristic 2 if you can complete the square; the problem is that you can't always do so)

The same goes for the quadratic formula; it's true over $\mathbb{F}_5$, and in fact over any field whose characteristic is not $2$.

And note, like for real polynomials, that sometimes you need to take the square root of something that isn't a square; this will produce roots in a quadratic extension. In this case, those would be in $\mathbb{F}_{25}$.