# Let $f(x)=x^3+x^2-5$. Show that for $n=1, 2,3, ...$ there is a unique $x_n$ modulo $7^n$ such that $f(x_n)\equiv 0\pmod{7^n}$.

My gut feeling for solving this problem is to use strong induction.

Starting with the base case $$n=1$$ we can check each of the seven congruence classes and find that $$x_1=2$$ is the unique solution. Then assuming for $$1\le m\le k$$ there is a unique root modulo $$7^m$$. So we know from Hensel's Lemma we know that since there is a unique solution modulo $$x_k$$ to the congruence $$f(x)\equiv 0\pmod{7^k}$$ that $$7$$ does not divide $$f'(x_{k-1})$$. To show that when $$n = k+1$$ we have a unique solution it would be sufficient to show that $$7$$ does not divide $$f'(x_k)$$.

This is where I keep hitting a wall. I can't seem to figure out how to show that $$7$$ does not divide $$f'(x_k)$$. Using Hensel's Lemma Ive been able to find $$x_k$$ in terms of $$x_{k-1}$$ but its really ugly and involves inverses of $$f'(x_{k-1})$$.

Could I get a hint to try and steer me in the right direction, and if I'm going way off track what can I do to correct it?

• The sequence $x_1,x_2,\dotsc$ has the property that $x_{k+1},x_{k+2},x_{k+3},\dotsc$ are all congruent to $x_k$ mod $7^k$. In particular, $x_1,x_2,\dotsc$ are all congruent to $x_1\equiv 2\pmod 7$. But $2$ is not a root of $f'(x)=3x^2+2x$ mod $7$. Feb 29 '20 at 20:18

You have that $$f'(x)=3x^2+2x=x(3x+2)$$. Mod $$7$$, the only roots to that are $$\{0,4\}$$. Could $$x_k$$ have been one of these mod $$7$$?
If $$x_k\equiv0$$ mod $$7$$, then $$7$$ does not divide $$x_k^3+x_k^2-5$$, so $$x_k$$ was not a root mod $$7^k$$.
If $$x_k\equiv4$$ mod $$7$$, then ... (leaving as a hint. See answer history for the full answer.)
• See I was thinking that I needed to use that $x_k\equiv 0$ or $4\pmod{7}$ but I couldn't figure out where but that makes complete sense. Thank you. Feb 29 '20 at 20:18