The Diophantine equation $x^2 + py + a = 0$.

Question

Let $p$ be an odd prime. Show that the Diophantine equation $$x^2 + py + a = 0$$ with $\gcd(a, p) = 1$ has an integral solution if and only if $\left(\frac{-a}{p}\right) = 1$

Answer 1

Hint for $(\Rightarrow)$:

Suppose there is a solution $(x_{0},y_{0})$ to the equation. Then

$x_{0}^{2}=-py_{0}-a$

What is the reminder when the RHS is divided by $p$? Also, if $gcd(a,p)=1$, then what can be said about $-a$ being a quadratic residue $mod p$?

Hint for $(\Leftarrow)$:

Suppose $(-a/p)=1$ i.e. $x_{0}^{2}=-a+kp$ for some integers $x_{0}$ and $k$.

Can you then find an $y_{0}$ such that $(x_{0},y_{0})$ is a solution to the equation?