solve $x^2 -4x +13 \equiv 0 \pmod{81}$?

Question

How do I solve $x^2 -4x +13 \equiv o \pmod{81}$ ?

I know that this is the same as $x^2 -4x +13 \equiv x^2 + 2x + 1 \equiv (x +1)^2\equiv 0\pmod{3^4}$

but why is $x \equiv -1\pmod{3}$ the only solution of the congruence $f(x) \equiv 0\pmod{3}$ ? (why is $x \equiv 2\pmod{3}$ not a solution?)

How do I use hensel's lemma to find/ prove that there are no solutions to this problem?

Answer 1

We have $\left(x-2\right)^2\equiv -9\left(\text{mod}\,81\right)$. The left-hand side is a multiple of 9, so its square root is a multiple of 3, say $x-2=3y$. Then $y^2\equiv -1\left(\text{mod}\,9\right)$. But integers cannot even solve $y^2\equiv -1\left(\text{mod}\,3\right)$.

Answer 2

$$x^2-4 x+13\equiv 0\pmod {81}\iff81|(x^2-4 x+13)=(x-2)^2+9\iff$$ $$\iff 9|((x-2)^2/9+1\iff \exists y\;(\;[x-2=3 y]\land [9|y^2+1]\;)\implies$$ $$\implies \exists y\;(3|y^2+1)\iff \exists y\;(y^2\equiv 2\pmod 3\;).$$ Which can't happen because if $y\equiv 0\pmod 3$ then $y^2\equiv 0\pmod 3,$ and if $y\equiv \pm 1\pmod 3$ then $y^2\equiv 1\not \equiv 2\pmod 3.$