<s>[INST] solve $x^2 -4x +13 \equiv 0 \pmod{81}$? 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?
 [/INST] 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 $y^2\equiv -1\left(\text{mod}\,3\right)$.
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