Solving a quadratic equation in $\Bbb{Z}/19\Bbb{Z}$.

Question

I am working on a problem and I have written down my thoughts but I am having trouble convincing myself that it is right. It very may well be it is also completely wrong, which Is why I thought id post it and my attempt and try to get some feedback.

I want to find all the values c such that the system $xy=c$ and $x+y=1$ has a solution in the field $$\mathbb{Z} / 19 \mathbb{Z}$$

What I know:

I know that for a solution to a quadratic in this sense to exist it must have that its discriminant is a square in the field we are working in.

Rearranging terms I get that $x^2-x+a=0\bmod(19)$

so I concluded that a solution will exist iff and only if $1-4a$ is a square mod $19$.

I then calculated the squares mod 19 they are,

$$0 , 1, 4, 9 ,16 ,6 ,17, 11, 7, 5,$$

and then repeating backwards ie 5, 7, ...

so then I just set $1-4a$ to each of these and solved for $a$.

I also would be interested in knowing if there is an a that gives a unique solution. My initial thoughts are not since all the roots repeat.

Again, I am not sure if this is correct. Does anyone have any insight?

Answer 1

Your approach is entirely correct. There is a unique solution for $a\in\Bbb{Z}/19\Bbb{Z}$ if and only if $$x^2-x+a=(x-\beta)^2,$$ for some $\beta\in\Bbb{Z}/19\Bbb{Z}$. Then in particular $-2\beta=-1$, from which it follows that $\beta=10$ and $a=5$. Indeed this corresponds to the case where the discriminant equals $0$, as $1-4\times5\equiv0\pmod{19}$.

Answer 2

The equation $x^2-x+a=0$ can be rewritten as $$(x+9)^2-5+a=0\iff(x+9)^2=5-a.$$ Hence it has two solutions if $5-a$ is a non-zero square in $\mathbf Z/19\mathbf Z$, has no soultuion if $5-a$ is not a square, has one solution if $a=5$.

Answer 3

It seems to me that you are doing too much work here. You have $y = 1-x$, so $x(1-x) = c$. So now you just have to list all values of $x(1-x)$ mod $19$ for $0 \le x < 19$, and you are done.

This is hardly more difficult than listing all the squares mod $19$, which you did as the first step of your solution.