Hensel’s Lemma Number Theory Confusion

Question

I have been given an example, finding the solutions of the congruence $f(x) ≡ 0$ (mod $5^4$) for $f(x)=x^2+1$

This solution finds that for mod $5$ we have $x_0=2$ . So through the 'lifting' process, it starts with

$x_1= x_0 - f(x_0)/f'(x_0) = 2 - (5)/(4)=2-(5)/(-1)=7$(mod $5^2$) Now, the next step in the example states that $x_2= x_1 - f(x_1)/f'(x_1) = 7 - 50/14 = 7- 50/-1= 57$ (mod $5^3$)

and so on until $5^4$.

BUT, how does $5/4$ become $5/-1$ in mod ($5^2$) and how does $50/14$ become $50/-1$ in mod $5^3$, is this a mistake in the example or is there something I am missing because the same thing happens in the next step for $5^4$.

Thanks

Answer 1

I'm not sure of the pattern ur getting. But an explanation could be :

if $a$ is invertible and $a\equiv b \: mod (5)$ then $1/a \equiv 1/b \: mod (5)$. Therefore, $5/a=5(1/b+\alpha 5) \equiv 5/b \:mod(5^2)$, and $50/a=50(1/b+\alpha 5)\equiv 50/b \: mod (5^3)$...

Answer 2

If you want to compute $\dfrac{5}{4} \pmod{5^{2}}$, you should compute the inverse $-6 = -1 - 5$of $4$ modulo $5^{2}$. But it is enough to use the inverse $-1$ of $4$ modulo $5$, as $$ 5 \cdot (-6) = 5 (-1 - 5) = -5 - 5^{2} \equiv -5 = 5 (-1) \pmod{5^{2}}. $$

Similarly, if you want to compute $\dfrac{50}{14} \pmod{5^{3}}$, you should compute the inverse $9 = -1 + 10$ of $14$ modulo $5^{3}$. But it is enough to use the inverse $-1$ of $14$ modulo $5$, as $$ 50 \cdot 9 = 2 \cdot 5^{2} (-1 + 5 \cdot 2) \equiv -2 \cdot 5^{2} \equiv 50 (-1) \pmod{5^{3}}. $$