Question: I'm attempring to solve a congruence $x^2 \equiv 17\pmod{128}$ but not quite sure how to go about it. I see that $128 = 2^7$, but the Chinese Remainder Theorem doesn't apply to $\gcd > 1$. I found one solution quite easily by finding solutiong to $x^2 \equiv\pmod{32}$ which was $x \equiv 23\pmod{32}$ and it turned out that $23 + 128x\_0 \equiv 17\pmod{128}$. How do I find the other solution ans what's the proper way of doing it?

Answer: 
$x^2\equiv17\pmod{128}\implies x^2\equiv17\pmod{16}\equiv1$

$(4a+1)^2\equiv8a+1\pmod{16},$ 

we need $8a+1\equiv17\pmod{16}\implies a$ must be even $=2b$(say)

Again, $x^2\equiv17\pmod{128}\implies x^2\equiv17\pmod{32}$

Now $(8b+1)^2\equiv16b+1\pmod{32},$ we need $16b+1\equiv17\pmod{32}\implies b=2c+1$

$8b+1=8(2c+1)+1=16c+9$

Again, $x^2\equiv17\pmod{128}\implies x^2\equiv17\pmod{64}$

and $(16c+9)^2\equiv288c+81\pmod{64}\equiv32c+17$

For $(16c+9)^2\equiv17\pmod{64},c=2d$

$16c+9=16(2d)+9=32d+9$

Now $(32d+9)^2\equiv576d+81\equiv64d+81\pmod{128}$

We need $\displaystyle64d+81\equiv17\pmod{128}$
$\iff64d\equiv-64\pmod{128}\iff d\equiv-1\pmod2\implies d=2e-1$

$32d+9=32(2r-1)+9=64r-23\equiv-23\pmod{64}$

Similarly start with $4a-1$ to get $x\equiv23\pmod{64}$

That these are the two exclusive in-congruent solutions can be verified as follows:

$$x^2\equiv17\pmod{128}\equiv23^2$$

$$\iff128|(x-23)(x+23)$$

$$\iff32\mid\dfrac{x-23}2\cdot\dfrac{x+23}2$$

As $\dfrac{x+23}2-\dfrac{x-23}2=23$ is odd, so, they must be of opposite parity

If $32\mid\dfrac{x-23}2,x\equiv23\pmod{64}$

and $\cdots$
The [series](https://en.wikipedia.org/wiki/Binomial_theorem#Newton.27s_generalised_binomial_theorem) for the square root is
\begin{eqnarray}
\sqrt{1+16 t} =1 + 8 t - 32 t^2 + 256 t^3 - 2560 t^4 + 28672 t^5 - 344064 t^6+\\ + 
4325376 t^7 - 56229888 t^8 + 749731840 t^9 - 10196353024 t^{10}+ \cdots
\end{eqnarray}
The first three terms truncation $1 + 8 t - 32t^2$ has square $1 + 16 t - 512 t^3 + 1024 t^4$. Therefore, for $t=1$ we obtain
$$(1+8-32)^2 = 17 -512 + 1024= 17 + 512 \equiv 17 \!\!\!\!\mod 512$$

therefore, $23^2 \equiv 17 \!\!\!\!\mod 128$.

Let $x$ any other solution of $x^2 \equiv 17 \!\!\!\!\mod 128$. Then 
$y\colon =\frac{x}{23} \!\!\!\mod 128$ satisfies $y^2\equiv 1 \!\!\!\mod 128$
and for this we have $4$ solutions $\pm 1$, $\pm 65 \!\!\!\mod 128$. 

Therefore, the solutions are $\pm 23$, $\pm 23 \cdot 65 \!\!\!\mod 128$, that is 
$$23, 41, 87, 105$$