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: 
Outline: Let $a$ be a particular solution of $x^2\equiv 17\pmod{128}$. Let $x$ be any solution. Then there is a unique $y$ (modulo $128$) such that $x\equiv ay\pmod{128}$. 

We have $(ay)^2\equiv 17\pmod{128}$ if and only if $y^2\equiv 1\pmod{128}$. So we need to find all solutions $y$ of this congruence. 

That may have been done in your course already. If $n\ge 3$, then the congruence $y^2\equiv 1\pmod{2^n}$ has exactly $4$ solutions, namely $y\equiv \pm 1\pmod{2^n}$ and $y\equiv \pm (1+2^{n-1})\pmod{2^n}$.
We have to solve $x^2\equiv 17 \pmod{128}$

Now,$128=2^7$

So,we are to solve an equation of the form $x^2\equiv a\pmod {2^\alpha}$

We follow the ascending method:

First we find a root of $x^2\equiv 17 \pmod{16}$ which is $x^2\equiv 1\pmod{16}$

We can see that $x\_4=1$ is a solution to the given equation.

Now the ascending method says that if $x\_i$ is a solution of $x^2\equiv a \pmod{2^i}$,then a solution of $x^2\equiv a\pmod{2^{i+1}}$ is $x\_{i+1}=x\_i+2^{i-1}k$

Putting this solution in the equation,we can find a value of $k$.

Hence we can find $x\_{i+1}$.

Using this method it is easy to see that $x\_4=1,x\_5=9,x\_6=9,x\_7=41$.

Now,the square roots of $1$ modulo $128$ are $\pm1,2^{7-1}\pm1$ .So multiplying by $41$ we get all the solutions of $x^2\equiv 17 \pmod{128}$ as $41,87,105,23$.