Quadratic congruences equivalent statements

Question

Prove that for every prime $p>3$,

$L= x^2-x+2 \equiv 0\mod p$ has a solution iff $D=x^2-x+16 \equiv 0\mod p$ has a solution.

This is not true, right? If $L \equiv 0$ mod$p$, then $D=L+14 \equiv 14\mod p$, and $14$ is not congruent to $0\mod p$, when $p \neq 7$.

Answer 1

The first congruence has solutions if and only if $\Delta_1=-7$ is a square$\bmod p$.

The second congruence has solutions if and only if $\Delta_2=-63$ is a square$\bmod p$.

Now it is obvious that $$-7 \,\text{ is a square}\bmod p\iff-63=3^2\cdot -7 \,\text{ is a square}\bmod p.$$

Answer 2

$$(2x-1)^2\equiv-7\pmod p$$

$$(2y-1)^2\equiv-63\equiv\{3(2x-1)\}^2$$

Answer 3

Hint:

The discriminant of $x^2-x+2$ is $-7$.

The discriminant of $x^2-x+16$ is $-63=3^2(-7)$.