The number of solutions to x2≡a(modp)x^2 \equiv a \pmod{p}x2≡a(modp) where ppp is an odd prime and gcd(a,p)=1\gcd(a, p) = 1gcd(a,p)=1 is:
Always 2
1 + (a/p)
1 - (a/p)
0