<s>[INST] Solve polynomial congruence using Hensel's lemma Solve $x^4+2x+46 \equiv 0$ $(\mod 4375 )$ for x. 
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My attempt:
$x^4+2x+46 \equiv 0$ $(\mod 5^47 )$
breaks down to 
a Chinese Remainder Problem with the 2 following congruence's':
(1) $x^4+2x+46 \equiv 0$ $(\mod 7 )$ which by inspection is x $\equiv 2,6 (\mod 7)$
and
(2) $x^4+2x+46 \equiv 0$ $(\mod 5^4)$
It's (2) that's really difficult for me and I know I am supposed to use Hensel's Lemma, can any one help with this?
 [/INST] The Hensel lemma states that if there exist integer $x$ such that:
$$f(x) \equiv 0 \pmod {p^k} \quad \quad \text{and} \quad \quad f'(x) \equiv 0 \pmod p$$
then there exist integer $s$ such that:
$$f(s) \equiv 0 \pmod {p^{k+m}} \quad \quad \text{and} \quad \quad s \equiv r \pmod {p^k}$$
Now it's obvious that $s = r + tp^k$ for some $t \in \mathbb{Z}$.
Now let's move to the problem. Find a solution for the equation:
$$x^4 + 2x + 46 \equiv 0 \pmod 5$$
Since $x\equiv 0 \pmod 5$ obviously isn't solution, Fermat's Little Theorem come in use so we have:
$$x^4 + 2x + 46 \equiv 1 + 2x + 46 \equiv 2x + 47 \equiv 2x + 52 \equiv 0 \pmod 5$$
From this obviously $x \equiv 4 \pmod 5$ is a solution. So take $x_1 = 4$
Now find the derivative of the function. This is easy and we have:
$$f'(x) = 4x^3 + 2$$
Now we need to find a solution: $f(x_2) \equiv 0 \pmod {5^2}$ i.e. $f(x_1 + 5t) \equiv 0 \pmod {5^2}$. The last expression is equivalent to:
$$f(x_1) + 5tf'(x_1) \equiv \pmod {5^2}$$
Now substitute $x_1 = 4$ and we have:
$$f(4) + 5tf'(4) \equiv 0 \pmod {5^2}$$
$$310 + 5t \cdot 258 \equiv 0 \pmod {5^2}$$
Divide with $5$.
$$62 + 258t \equiv 2 + 3t \equiv \pmod 5$$
From this we get: $t \equiv 1 \pmod 5$. Choose $t=1$ and we have: $x_2 = x_1 + 5t = 9$
So now we have $f(9) \equiv 0 \pmod {5^2}$...
Did you get the concept. Can you continue on your own and lift the exponent first to $3$ and then to $4$? 
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