Applying Hensel's lemma to solve $x^2 + 8 \equiv 0\pmod {121}$.

Question

When solving for $x^2 + 8 \equiv 0 \pmod {121}$, How can we apply Hensel's lemma to solve for its solutions? What I currently understand is that for a prime $p$ and $e \geq 2$, then $f(x) \equiv 0 \pmod {p^{e-1}}$ and comes in the form of $x_e = x_{e-1} - kp^{e-1}$.

Answer 1

Start with $f(x)=x^2+8\equiv0\bmod11$, one solution of which is $x_0\equiv5\bmod11$. Hensel's lemma to get the solution modulo the next prime power is essentially Newton's method: $$\begin{align} x_1&\equiv x_0-\frac{f(x_0)}{f'(x_0)}\bmod p^2\\ &\equiv 5-\frac{33}{10}\bmod11^2\\ &\equiv 5+33\cdot12\bmod11^2\\ &\equiv 38\bmod11^2 \end{align}$$ So the two solutions to the modulo-$11^2$ equation are $x\equiv\pm38\bmod11^2$.