# The congruence $f(x)=x^3+3x+9 \equiv 0 (\bmod 5^n)$ has only one solutions for every $n \geq 2$

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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I am wondering what $p$ might be. – Arthur Jun 24 '12 at 11:32
Sorry, it's $9$. :) – Jozef Jun 24 '12 at 11:45

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.
Thank you for the answer. I don't understand what led you to conclude that $r$ would always be that way, it keeps changes, why does it have to have remainder $-1$ modulo $5$? – Jozef Jun 24 '12 at 12:36
@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. – Jyrki Lahtonen Jun 24 '12 at 15:30