<s>[INST] Hensel Lifting and solving with mods I need to use the Hensel-Newton method (aka Hensel Lifting) to find all solutions of:
$$x^2 + x + 47 \equiv 0\:(\text{mod } 343)$$
Note: $$ 343 = 7^3 $$
I don't really understand Hensel Lifting so I am not sure where to begin.  Can someone help give me a better understanding of Hensel Lifting and how to start this problem?
 [/INST] Let's start modulo $7$.
By trial and error, we find $x^2+x+5\equiv 0\pmod 7$ has the solutions $x\equiv 1$, $x\equiv 5$.
Now if $x^2+x+47\equiv 0\pmod{49}$ then $x=x_0+7y$ with $x_0\in\{1,5\}$.
Hence $x^2+x+47=(x_0^2+14x_0y+49y^2)+(x_0+7y)+47\equiv (x_0^2+x_0+5)+7\cdot(2x_0y+y+6)\pmod{49}$, where we already know that the left hand side is a multiple of $7$. Hence we have to solve
$$ \frac{x_0^2+x_0+5}7+(2x_0+1)y+6\equiv0\pmod 7.$$
If $x_0=1$, this leads to $3y+7\equiv 0$, i.e. $y\equiv 0$. If $x_0=5$, we obtain $11y+11\equiv 0$, i.e. $y\equiv 6$.
Thus the only solutions modulo $49$ are $x\equiv 1\pmod{49}$, $x\equiv 47\equiv -2\pmod{49}$.
Now if $x^2+x+47\equiv 0\pmod{343}$, then $x=x_1+49z$ with $x_1\in \{1,-2\}$.
Hence 
$$ x^2+x+47 = x_1^2+2\cdot 7^2x_1z+7^4z^2+x_1+7^2z+47\\\equiv(x_1^2+x_1+47)+7^2(2x_1+1)z\pmod{343}.$$
Again, this is trivially a multiple of $49$ and we are lead to solve
$$\frac{x_1^2+x_1+47}{49}+(2x_1+1)z\equiv 0\pmod{7}.$$
With $x_1=1$, this is $3z+1\equiv 0$, i.e. $z=2$ and $x\equiv99\pmod{343}$.
With $x_1=-2$, this is $-3z+1\equiv0$, i.e. $z=-2$ and $x=-100\pmod{343}$.
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