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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?

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  • $\begingroup$ aren't they the same? $\endgroup$ – Yimin Apr 13 '16 at 16:57
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    $\begingroup$ You have $x^2-4x+13 = (x-2)^2+9$. That pretty much explains why $x\equiv -1\equiv 2$ is the only solution modulo $3$. $\endgroup$ – Thomas Andrews Apr 13 '16 at 16:57
  • $\begingroup$ $x\equiv -1\mod3$ is the same as $x\equiv 2\mod3$. $\endgroup$ – Arthur Apr 13 '16 at 17:00
  • $\begingroup$ Ok so how do I use hensel's lemma to show that there is no solution? $\endgroup$ – Jamelia Jones Apr 13 '16 at 17:02
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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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$$x^2-4 x+13\equiv 0\pmod {81}\iff81|(x^2-4 x+13)=(x-2)^2+9\iff$$ $$\iff 9|((x-2)^2/9+1\iff \exists y\;(\;[x-2=3 y]\land [9|y^2+1]\;)\iff$$ $$\implies \exists y\;(3|y^2+1)\iff \exists y\;(y^2\equiv 2\pmod 3\;).$$ 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 $y^2\equiv 1\not \equiv 2\pmod 3.$

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