let us suppose we have following congruence equation
$${y^2}\equiv {a} \pmod {p}.$$
where $p$ is prime, i now that if for instance $r$ is solution, then $p-r$ is also solution , for instance
$${y^2}\equiv {10} \pmod {13}.$$
i can check all possible starting values of $y$ , for instance
>> y=1:12;
>> y=y.^2;
>> r=y-10;
>> find(mod(r,13)==0)
ans =
6 7
which satisfies given theorem, but is that only solution method for prime $p$ ? i know when $p$ is composite , i can make prime factorization and solve system using Chinese remainder theorem , but what about when $p$ is prime?