Using Hensel's Lifting Lemma to Solve $x^2 + x + 34 \equiv 0 \pmod{81}$ As in the title, I'm trying to solve $$x^2 + x + 34 \equiv 0 \pmod{81}.$$
Let $f(x) = x^2 + x + 34$ throughout.
I'm using Hensel's lemma, but it's a bit dense and I'm not sure my interpretation is correct.  My method involves checking for solutions through mod 3, mod 9, mod 27 and mod 81, and checking the conditions on $f$ and $f'$ each time.  But should I check the conditions for mod 81 too?  Or just substitute my penultimate solutions into the original congruence?  More specifically, in the penultimate step (mod 27) I get the solutions $$x \equiv 4 \pmod{27}, \quad x \equiv 13 \pmod{27}, \quad x \equiv 22 \pmod{27}.$$
Now, is all that remains to substitute $x = 4,13,22$ into the original equation?  None of them are solutions and thus by Hensel's lemma there are no solutions?

You probably can't use Hensel's Lemma:
we need $f'(x) = 2x + 1 \not \equiv 0 \mod 3$ so we need $x \equiv 0, 2 \mod 3$ and that just isn't likely to solve $x^2 + x + 7 + 27 \equiv 0 \mod 3^k$
But  $x^2 + x + 34 \equiv x^2 - 80x + (27 + 7) \equiv$
$x^2 -80 x + 40^2 - 40^2 + (27+7) \equiv $
$(x-40)^2 -(27 + 13)^2 + (27 + 7) \equiv $
$(x-40)^2 - 26*27 - 13^2 + (27 + 7)\equiv $
$(x - 40)^2 - 25*27 - 169 +7\equiv $
$(x-40)^2 - 27 - 7+7 \equiv (x-40)^2 - 27 \mod 81$
$\implies (x-40)^2 \equiv 27 \mod 81$.
Which means $27 + 81k$ is a perfect square for some $k$.
So $27(3k + 1)$ is a perfect square.   But as $3\not |3k + 1$ that is impossible.
So there are no solutions.
.....
(Or in hindsight:  $x^2 + x + 34 \equiv 0 \mod 81 \iff$
$x^2 + x \equiv -27 - 7 \mod 81\iff$
$4x^2 + 4x \equiv -4*27 - 28 \equiv -5*27 - 1\equiv 27 -1 \mod 81\iff$
$4x^2 + 4x+1 = (2x + 1)^2 \equiv 27 \mod 81$
Same contradiction.)
