# Solving polynomials along with congruences

Say $g(x) = x^3 + x^2 − 5$. I need to show that for $i = 1,2,3,4,..$ There exists a unique $x_i \mod 7^i$ s.t

$g(x_i) ≡ 0 \mod 7^i$

-
If $7_i$ means $7^i$ then you may use Hensel's Lemma, on which there are already many answers. – Math Gems Feb 26 '13 at 1:48
Where to start is with $i=1$ (I'm assuming your $j$ was meant to be an $i$). Can you solve $x^3+x^2-5\equiv0\pmod7$? Then after that, it's Hensel time. – Gerry Myerson Feb 26 '13 at 2:26
I don't really understand hensel lifting. I can solve for $g(x_i) ≡ 0 \mod 7$ .. but after that do I just try and solve for $i=2,3,4,..$ until I find a pattern? – ker12 Feb 26 '13 at 2:39

Read Hensel's Lemma Statement here (which, in fact, is part of the proof) , and then you'll know what to do.

For example:

$$g(x)=x^3+x^2-5=0\pmod 7\Longrightarrow x^3+x^2+2=(x-2)(x^2+3x-1)\pmod 7$$

Note that the quadratic above is irreducible in $\,\Bbb F_7[x]=\left(\Bbb Z/7\Bbb Z\right)[x]\,$ (why?)

Now, checking the link given above, define (check the root $\,x=2\,$ is simple and thus $\,g'(2)\neq 0\,$ !)

$$t:=-\frac{g(2)}{7}g'(2)^{-1}=-\frac{14}{7}16=-32\Longrightarrow$$

$$s=2-32\cdot 7\pmod {7^2=49}=-26\pmod{49}=23\pmod{49}$$

and indeed:

$$g(23)=23^3+23^2+44=0\pmod{49}$$

Well, continue as above with higher powers of $\,7\,$...

-