# $$\sqrt{-1}$$ isn't imaginary!

If $$x_1,x_2,\ldots, x_i$$ are the distinct solutions to the following congruence (with $$0\leq x_k<13$$ for each $$k$$, $$k \in \{1,2,\ldots ,i\}$$), $\large x^2 \equiv -1 \pmod{13}$

then enter the remainder when $$(x_1+x_2+\cdots+x_i)$$ is divided by 13.

×