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Solve $3x^2+6x+5\equiv 0\pmod{89^2}$.

To do this, I first solved $3x^2+6x+5\equiv 0\pmod{89}$.

This has a solution since $3x^2+6x+5 = 3(x+1)^2 + 2$ and $3(x+1)^2 \equiv -2 \pmod{89}$ has a solution since $(3,89)=1$, we can check whether $(x+1)^2 = 3^{-1}(-2)\pmod{89}$ which is equivalent to calculating $\left(\dfrac{3^{-1}(-2)}{89}\right)$. The value is 1, so this is a quadratic residue. Since we have a number that when squared gives us this, we have a solution to the polynomial congruence mod 89. Since we have 1, we have 2 since solutions come in pairs $\{\pm x_0\}\pmod{89}$.

Here is my question: without finding the residue, how can I use Hensel's lemma to see whether I can lift the solutions to solutions modulo $89^2$?

I hope someone can help. Thank you.

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    Hensel's Lemma tells you how to go from a solution modulo $p^i$ to a solution modulo $p^{i+1}$; it is utterly silent on the question of how to find the first solution; but the value of $i$ is irrelevant. As far as solving polynomials of degree higher than $2$, the problems have to do with our inherent difficulty in solving them, rather than any particular issue with $n$th residues. Quadratic residues are used to solve quadratic polynomials because of the quadratic formula; you can use the Cardano formulas for cubic polynomials ($p\neq 2,3$); after degree 4, though, no ready formulas.2011-12-12

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