Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$ Find the solutions of $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{13}$
I know that they are both soluble since $5\equiv 1\pmod{4}$ and $13\equiv 1\pmod{4}$
What is the method to solving this simultaneous equation.
Looking for a standard method to use with this type of problem.

Hint: $x^2 \equiv -1$ mod $5$ $\iff$ $x^2 \equiv 4$ mod $5$ $\iff$ $(x-2)(x+2) \equiv 0$ mod $5$, whence $x \equiv 2$ or $x \equiv -2$ mod $5$. 
$x^2 \equiv -1$ mod $13$ $\iff$ $x^2 \equiv 25$ mod $13$ $\iff$ $(x-5)(x+5) \equiv 0$ mod $13$, whence $x \equiv 5$ or $x \equiv -5$ mod $13$.
Now cross-solve these $4$ possibilities with the Chinese Remainder Theorem.

 Example: if $x \equiv 2$ mod $5$ and $x \equiv 5$ mod $13$, this should give you $x \equiv 57$ mod $65$

