I started learning number theory, specifically polynomial congruences, and need help with the following exercise. Here it is:
Does the congruence $x^2-6x-13 \equiv 0 \pmod{127}$ has solutions?
I tried to follow the method for solving general quadratic congruence but I didn't get really far. Here's what I've done so far:
Since $(4, 127) = 1$, we may complete the square by multiplying by $4$ without having to change the modulus in order to get the following equivalent congruence
$$(2x-6)^2 \equiv 36 - 4(-13) \pmod{127} \iff (2x-6)^2 \equiv 88 \pmod{127}.$$
If I'm heading in the right direction then I don't know how to continue from here. I suspect there is another method for solving this problem since I didn't make use of the fact that $127$ is prime. Also, the problem doesn't require to actually find the solutions but only determine if there are solutions. Possibly we can avoid computations and make use of some theorem/lemma to find if the congruence has solutions or not.