Proving a property about quadratic residues

Question

I am trying to determine which primes satisfy the equation $x^2+x\equiv y \pmod{p}$ for a given $y$.

For example, the $y$ which have solutions for $p = 11$ are the set $\{0,1,2,6,8,9\}$. However, instead I am interested in the reverse - which primes have, for instance, have a congruence which satisfies $y=3$?

As it turns out, the answer appears to be the set of primes for which $4y+1$ is a quadratic residue. This has held so far for $y=3,5,7,8,11,15$.

How can I prove this?

Answer 1

For $p=2$, $y$ must be even. Said this, let's assume that $p$ is odd.

You can use the quadratic formula to solve $x$: $$x=2^{-1}(-1\pm\sqrt{1+4y})$$ This belongs to $\Bbb Z_p$ if and only if $1+4y$ is a square modulo $p$.

Answer 2

Note that you can multiply by $4$ and add $1$ to obtain $$4x^2+4x+1=(2x+1)^2\equiv 4y+1$$ so $4y+1$ is a quadratic residue. You should be able to complete the argument from there.