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 $\ $ By CRT the square roots mod $5$ and $13$ lift to $4$ square roots mod $65$.  There is an obvious root $\,{\rm mod}\ 65\!:\ x^2\equiv -1\equiv 64\,$ if $\,x\equiv \pm8\equiv \pm (2,5)\pmod {5,13}$. The other pair $\pm (-2,5),\,$ arise by multiplying the first pair by $\,(-1,1)\equiv 14,\,$ i.e. $\,\pm 8\cdot 14\equiv \pm18\pmod{65}.\ $ QED
