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.

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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.