Find out all solutions of the congruence $x^2 \equiv 9 \mod 256$. I need to find all the solutions of the congruence $x^2 \equiv 9 \mod 256$. 
I tried (apparently naively) to do this:
$x^2 \equiv 9 \mod 256$
   $\Leftrightarrow$      $x^2 -9 \equiv 0 \mod 256$  $\Leftrightarrow$ $256 | (x-3)(x+3)$ $\Leftrightarrow$
$256|(x-3)$ or  $256|(x+3)$  $\Leftrightarrow$ $x \equiv 3, -3 \mod 256$ 
but I found out that it's not fully correct because there are other solutions that I didn't succeed to find such as $x = 125$, so I want to know what is the general way to solve such questions? thank you!

If $(x-3)(x+3)\equiv0\pmod{2^{m+2}}$ for integer $m\ge1$
both $(x-3),(x+3)$ must be even
$\implies\dfrac{x-3}2\cdot\dfrac{x+3}2\equiv0\pmod{2^m}$
Now $\dfrac{x-3}2-\dfrac{x+3}2$ is odd  $\implies$ they have opposite parity
If $\dfrac{x-3}2$ is even, $\dfrac{x-3}2\equiv0\pmod{2^m},x\equiv3\pmod{2^{m+1}}$
Can you take it from here?
Here $m+2=8$ 
