How to use the method of "Hensel lifting" to solve $x^2 + x -1 \equiv 0\pmod {11^4}$? I got $x^2 + x + 10 \equiv 0\pmod {11}$
I am confused about using the Hensel lifting method. Can someone just help me out with that please.
 A: First our function is:
$$f(x) = x^2 + x + 10$$
And it's derivative is:
$$f'(x) = 2x + 1$$
The Hensel's Lemma states that for:
$$f(x) \equiv 0 \pmod {p^k} \quad \quad \text{and} \quad \quad f'(x) \not\equiv 0 \pmod p$$
then there is an unique integer $s$ modulo $p^{k+m}$ satisfying these relations:
$$f(s) \equiv 0 \pmod {p^{k+m}} \quad \quad \text{and} \quad \quad r \equiv s \pmod {p^k}$$
From the last equation it's obvious that $s$ is of the form $r + tp^k$
Now play a little guess game and find a solution such that:
$$f(x) \equiv 0 \pmod p$$
Obviously $x_1 = 3$ will do the job, also $f'(x) \not\equiv 0 \pmod {11}$. 
Now we need to find solutions such that $f(x_2) \equiv 0 \pmod {11^2}$ i.e $f(x_1 + 11t) \equiv 0 \pmod {11^2}$. The last expresion is equivalent to:
$$f(x_1) + 11tf'(x_1) \equiv 0 \pmod {11^2}$$
Divide both sides by $p$ we have and substituting values we have:
$$\frac{22}{11} + \frac{77t}{11} \equiv 0 \pmod {11}$$
$$2 + 7t \equiv 0 \pmod {11} \implies t \equiv 6 \pmod {11}$$
Now we can take $t=6$ as the smallest integer to satisfy the relation and we have:
$$x_2 = 3 + 6\cdot 11 = 69$$
And indeed $f(69) \equiv 0 \pmod {121}$
Now you can continue on your own and raise it to fourth power. For additional information I found this link quite useful.
