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?
Best Answer
I'll solve it without using Hensel's Lemma. Multiply both sides by $4$.
$$(2x+1)^2\equiv -135\equiv 27\pmod{81}$$
$\implies 27\mid (2x+1)^2\implies 9\mid 2x+1\implies 81\mid (2x+1)^2$, but $81\nmid 27$. No solutions.