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When solving for $x^2 + 8 \equiv 0 \pmod {121}$, How can we apply Hensel's lemma to solve for its solutions? What I currently understand is that for a prime $p$ and $e \geq 2$, then $f(x) \equiv 0 \pmod {p^{e-1}}$ and comes in the form of $x_e = x_{e-1} - kp^{e-1}$.

Christian E. Ramirez
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gus f
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    Here is an alternative method: This is solvable be subtracting 8 on both sides to get $x^2\equiv -8\equiv 113\pmod{11^2}$. This system can be solvable using theorem 2.2 here: https://www.uvm.edu/~cvincen1/files/teaching/spring2017-math255/quadraticequation.pdf to get 83 and 38 as the two solutions. – Captain Chicky Feb 25 '23 at 00:39
  • $x_1\equiv\pm5\pmod{11}.$ – Thomas Andrews Feb 25 '23 at 01:06

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Start with $f(x)=x^2+8\equiv0\bmod11$, one solution of which is $x_0\equiv5\bmod11$. Hensel's lemma to get the solution modulo the next prime power is essentially Newton's method: $$\begin{align} x_1&\equiv x_0-\frac{f(x_0)}{f'(x_0)}\bmod p^2\\ &\equiv 5-\frac{33}{10}\bmod11^2\\ &\equiv 5+33\cdot12\bmod11^2\\ &\equiv 38\bmod11^2 \end{align}$$ So the two solutions to the modulo-$11^2$ equation are $x\equiv\pm38\bmod11^2$.

Parcly Taxel
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  • How did you get $x_0 \equiv 5 mod 11$? Sorry I am a bit lost on that part – gus f Feb 25 '23 at 03:14
  • @gusf Just try all $11$ values to see if the equation holds $\bmod11$. – Parcly Taxel Feb 25 '23 at 03:15
  • @gusf Finding square roots modulo a prime is not easy. As Parcly says, brute force is best for small primes. For larger primes there is the Tonelli–Shanks algorithm: https://en.wikipedia.org/wiki/Tonelli–Shanks_algorithm but it's quite a bit of work. – B. Goddard Feb 25 '23 at 13:25
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 25 '23 at 22:12