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

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.
