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I tried to solve it by using Chinese remainder theorem ($2023 = 7 \cdot 289$): $$ x^3 \equiv x^2 + x - 1 \pmod{2023} \Leftrightarrow \begin{cases} x^3 \equiv x^2 + x - 1 \pmod{7} \\ x^3 \equiv x^2 + x - 1 \pmod{289} \end{cases} $$ The first congruence is easy to solve just by putting number from $\{0,1,...,6\}$ in congruence, and the numbers are $1, 6$. But I don't really know how to solve the second without Hensel's Lemma.

How can I approach second congruence? Thank you in advance!

Donald
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    You might want to use $x^3-x^2-x+1=(x+1)(x-1)^2$ – Sil Sep 01 '24 at 17:10
  • Using the hint given by @Sil I could find $x=188$, but probably there are more solutions – Jacaré Sep 01 '24 at 17:22
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    Do you seek a method avoiding Hensel lifting? Why? – Bill Dubuque Sep 01 '24 at 17:25
  • Once you have that factorization, $x\equiv \pm1\pmod n$ for each of $n=289=17^2$ and $n=7.$ You can solve $n=289$ by solving for $n=17$ then using Hansel. – Thomas Andrews Sep 01 '24 at 17:30
  • $289=17^2$, $\gcd(x-1,x+1)\in{1,2}$, making it easy to see when you have a solution modulo $289$. The Chinese remainder theorem is your friend. – Jyrki Lahtonen Sep 01 '24 at 17:33
  • One of the roots modulo $17$ is a double root, so Hensel is a bit problematic. But not needed. – Jyrki Lahtonen Sep 01 '24 at 17:35
  • Same as here in the linked dupe, i.e. use Euclid's Lemma and $(x-1,x+1)$ coprime to $17\ \ $ – Bill Dubuque Sep 01 '24 at 17:37
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    Not the same @Bill. The double root needs a different (but common sense, and easy) treatment when lifting. You sometimes criticize others for careless choice of duplicates. Please strive to include all the key features rather than just picking the best match from your own old answers. – Jyrki Lahtonen Sep 01 '24 at 17:48
  • @JyrkiLahtonen Please be patient - it takes time to find dupe targets using multiple results - now you've destroyed the chance of that, sigh. I'll leave the work to you now. since you've already wasted my time once again. Your comment about "picking" is way off the mark and uncalled for. – Bill Dubuque Sep 01 '24 at 17:53
  • @Jyrki fyi: the reason that I close on the first dupe link (when there are a two or more needed) is that if one doesn't then often that leaves time for fgitw answerers to post more dupes. Time is of the essence to beat (low quality) fgitw answers. You can find many examples of this in my prior dupe closures. It works well (except when subverted in the process as here). – Bill Dubuque Sep 01 '24 at 18:03
  • FWIW here's how Hensel can be used when a root is not simple. Not very relevant here, if you ask me. – Jyrki Lahtonen Sep 01 '24 at 18:14

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