# Finding $p$-primes which a congruential modulo $p$ has a solution.

I need to find the primes for which the congruential $x^2-4x-3=0$ (mod $p$). First I checked for $p=2$ and $x=1$ is a solution. Assuming $p>2$, my approach was that the discriminant $=28$ needs to be a quadratic residue modulo $p$.

Using Legendre's symbol, $(28/p)=(7/p)(4/p)=(7/p)$ (as $4$ is a quadratic residue). Now I need to find for which primes is $7$ a quadratic residue. I sense that I need to use the Quadratic Reciprocity to continue, but when I try I get stuck with splitting to separate cases.

I couldn't find a way or an algorithm online to solve such a question other than equations involving only $(-1/p)$, $(2/p)$, $(3/p)$.

Could someone show me how to continue from the spot I got stuck?

• Check link – Lozenges Jul 13 '18 at 9:10
• That doesn't really help me since 5 is 1 modulo 4, then it's simple to proceed by knowing which are the quadratic residues modulo 5. In this case 7 is 3 modulo 4 so that's where I'm stuck. – Tamir Shalev Jul 13 '18 at 9:40
• If $p=1(4)$ then $(p/7)(7/p)=1$ which implies $p=1,2,4(7)$. Similarly for $p=3(4)$ – Lozenges Jul 13 '18 at 9:50
• Alright so let me see if I get this right: If p=1(4), then (7/p)=(p/7), therefore (7/p)=1 iff p=1,2,4(7). If p=3(4), then (7/p)=-(p/7), therefore (7/p)=1 iff p=3,5,6(7). Is that correct? – Tamir Shalev Jul 13 '18 at 9:58
• Looks good. Let's not forget $p=7$ – Lozenges Jul 13 '18 at 10:03