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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?

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  • $\begingroup$ 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 $7$. $\endgroup$
    – W-t-P
    Feb 29 '20 at 20:18
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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.)

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  • $\begingroup$ 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. $\endgroup$ Feb 29 '20 at 20:18

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