What is the probability of having a quadratic residue?

Question

Let $a$ be a random integer and $p$ be a prime. What is the probability of $\left(\frac{a}{p}\right)=1$, as $p$ goes to infinity. And similarly, what about $\left(\frac{a}{p}\right)=-1$ or $0$?

Answer 1

$x^2\equiv y^2\iff x\equiv \pm y\,$ mod $p$, so $1^2,2^2,\ldots,\left(\frac{p-1}{2}\right)^2$ generate all the non-zero quadratic residues and there are $\frac{p-1}{2}$ non-zero quadratic residues and $\frac{p-1}{2}$ quadratic non-residues, so

$$\lim_{p\to\infty}P\left(\left(\frac{a}{p}\right)=1\right)=\lim_{p\to\infty}\frac{\frac{1}{2}(p-1)}{p}=\lim_{p\to\infty}\left(\frac{1}{2}-\frac{1}{2p}\right)=\frac{1}{2}$$

Same for $\left(\frac{a}{p}\right)=-1$.

$$\lim_{p\to\infty}P\left(\left(\frac{a}{p}\right)=0\right)=\lim_{p\to\infty}\frac{1}{p}=0$$