Timeline for Method to solve modular quadratic polynomial
Current License: CC BY-SA 4.0
6 events
when toggle format | what | by | license | comment | |
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Mar 23, 2022 at 13:53 | comment | added | Turbo | Is it possible to avoid Hensel lifting? My equation is of form $$x^2-(a+b)x + ab = 0 \bmod q^2$$ where $a+b=0\bmod q$ and so I get $(x-r)^2=0\bmod q$. I know the answer is not $a,b$ for my purpose but one of the Hensel lifts. So there is no canonical uniqueness conditions for root modulo $q^2$? | |
Mar 16, 2022 at 21:46 | comment | added | Turbo | In my case $q|f'(root)$ and so there is no inverse of $f'(root)$ where $f(root)\equiv 0\bmod q$ holds and $disc\equiv0\bmod q$. Posted in math.stackexchange.com/questions/4405084/…. | |
Mar 16, 2022 at 18:44 | comment | added | Turbo | $f(x)$ for me is quadratic and the discriminant mod $q$ is $0$ while discriminant mod $q^2$ is not $0$. Are there any complications to applying the technique? Is there anything that would be different? | |
Mar 16, 2022 at 7:44 | comment | added | Peter Taylor | @Turbo, it is described in great detail in en.wikipedia.org/wiki/Hensel%27s_lemma | |
Mar 16, 2022 at 0:22 | comment | added | Turbo | Can you explain the Hensel lifting process for the two solutions? | |
Oct 5, 2021 at 8:24 | history | answered | Peter Taylor | CC BY-SA 4.0 |