I'm failing to understand the notes we've been given and have struggled to find something on the internet in the form of help. I'm currently stuck on a question for a class.
The specific question is solve $x^2 =-3\pmod{13^3}$.
As far as I can figure out I need to let $f(x) = (x^2)+3$ and then try to solve $f(x)= 0 \pmod{13^3}$. Beyond that I can't really understand what is going on.
All of the questions are of the form $f(x)= a \pmod{b^n}$. I've only been able to find help on questions where $b^n$ doesn't only have one prime factor and you split the question into two or more equations and solve, and as far as I have seen solving for $x^2 = -3\pmod{13^3}$ gives incorrect answers or leaves some out.
Any help would be much appreciated!
You must first solve the congruence $x^2\equiv -3\pmod{13}$. To do this note that $x_o\equiv 6\pmod{13}$ and $x_1\equiv 7\pmod{13}$ are solutions as $x_o^2=6^2=36\equiv -3\pmod{13}$ and $x_1^2=7^2=49\equiv -3\pmod{13}$. Now like you mentioned define the function $f$ such that $f(x)=x^2+3$ and $f'(x)=2x$. Note that $f'(x_o)=f'(6)=2*6=12\not\equiv 0 \pmod{13^2}$ and $f'(x_1)=f'(7)=2*7=14\not\equiv 0 \pmod{13^2}$. Thus by Hensel's Lemma a unique lift exists for both solutions and they are given by the formula $$x_k=x_{k-1}-f(x_{k-1})\overline{f'(x_{k-1})}$$ After you find these two solutions to the congruence $x^2\equiv -3\pmod{13^2}$ use Hensel's Lemma again to find the solutions to $x^2\equiv -3\pmod{13^3}$. Hope this helps!