I have to show that $(E):x^2+12x-37 \equiv 0\;mod\;p$ where $p=421 \in \mathbb{P}$ have solutions, or not.
I would like to double check if my reasoning is correct. There is something that I feel uneasy about, although I am not sure why exactly.
Here's what I have done:
We find the discriminant and roots of $(E)$:
$\delta = 12^2 - 4 \cdot (1) \cdot (-37) = 292 > 0$
Therefore, $(E)$ has two real solutions. We are going to show that these two real solutions are not in $\mathbb{Z}_{p}$ and that therefore, the equation has no solutions.
We have $x_i = \frac{-12 + (-1)^i \sqrt{\delta}}{2}$ where $i = 0, 1$
$x_0^2 = 109 - 12 \cdot \sqrt{73}$
$x_1^2 = 109 + 12 \cdot \sqrt{73}$
Observe that, $x_1^2, x_2^2 \notin \mathbb{Z}_{p}$
$\therefore (E)$ has no solutions modulo $p$
I suspect that this could be extended to say that $\delta$ has to be a quadratic residue in $\mathbb{Z}_{p}$?
Also, this question was given in an assignment about Legendre and Jacobi Symbols.
Please do not spoil the answer, or use a formula that would make the problem trivial. I am trying to develop my intuition on the topic. If it is really necessary, please put a "warning". :-)
Thank you,
EDITED QUESTION: Apparently we also know $p=421.$
$$ x^2 + 12 x - 37 \equiv 0 \pmod p $$ add $73=73.$ $$ x^2 + 12 x + 36 \equiv 73 \pmod p $$ $$ (x+6)^2 \equiv 73 \pmod p $$ let $w = x+6$ $$ w^2 \equiv 73 \pmod p $$
This has solutions for some $p$ but not others.