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I need to prove that the congruence $f(x)=x^3+3x+9 \equiv 0 (\bmod 5^n)$ has only one solutions for every $n \geq 2$.

I checked with Hensel theorem that for $n=2$ there is one solution indeed. I want to use induction and to use Hensel theorem again, so I assumed that for $n$ there is one solution,r, but for using Hensel Lemma for $n+1$ how do i know that $f'(r)\not\equiv 0 (\bmod 5) $?

Thanks!

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    Sorry, it's $9$. :)2012-06-24

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Here $f'(x)=3x^2+3=3(x^2+1)$ vanishes modulo $5$, iff $x\equiv\pm2\pmod 5$. Modulo $25$ considerations dictate $r\equiv19\pmod{25}$, so you always have $r\equiv-1\pmod5$, and the assumptions of Hensel's lemma are satisfied.

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    @Jozef, when you apply Hensel lifting, the residue class modulo lower powers **never** changes. That's the whole point. The sequence of solutions $x_2,x_3,\ldots$ such that $x_k$ is a solution modulo $5^k$ is built to satisfy the congruences $x_k\equiv x_\ell\pmod{5^\ell}$ for all k>\ell\ge2. The present case is a bit exceptional in the sense that among the solutions modulo $5^1$ there is a double root $x\equiv3$ that cannot be lifted to a root modulo higher powers of $5$. This is illuminating: the derivative will vanish modulo $5$ at that point exactly because it is a double root.2012-06-24