How to solve $x^{2}\equiv-7\:\left(128\right)$? How to solve $x^{2}\equiv-7\:\left(128\right)$ ?
I know that $x^{2}\equiv a\:\left(2^{l}\right)$ is solvable iff $a\equiv1\:\left(8\right)$
which is the case here, but how can we find the solution?

The general method would be to first solve mod 2 and discover that $x \equiv 1 \pmod{2}$.  Then "lift" to mod $2^2$, which means that you observe that if $x \equiv 1 \pmod{2}$, then $x\equiv 1$ or $3 \pmod{4}$.  Both of these are solutions.  So lift them again to mod $2^3$ and get that $x \equiv 1, 3, 5,$ or $7 \pmod{8}$.  All of these are solutions.  So lift again to mod $2^4$.  
Now you get that $x \equiv 1,3, 5, 7, 9, 11, 13,$ or $15\pmod{16}$  but only 4 of these solve the congruence mod 16: 3, 5, 11, and 13. 
lift these to mod 32:  $3 \pmod{15}$ lifts to $3$ and $19 \pmod{32}$.  5 lifts to 5 and 21.  11 lifts to 11 and 27.  13 lifts to 13 and 29.  Of these 8, only 5, 11, 21 and 27 solve the congruence mod 32.
Lift two more times to get all four solutions mod 128.
