Talk:Cipolla's algorithm: Difference between revisions
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:::::: If the examples do not make sense, someone has to fix the Wikipedia page. Nevertheless I will publish to-morrow - it is late here :-) - a working solution. It only uses the rules of arithmetic. To make things work, you have to implement Fp2 arithmetic, like complex arithmetic is implemented. A number is a pair (x y) , etc. Remark: Indeed, -1 -3ω = -92 -16ω (mod 13) --[[User:G.Brougnard|G.Brougnard]] ([[User talk:G.Brougnard|talk]]) 02:20, 27 March 2016 (UTC)
::::::: Let ω = 2.64575 (in other words: <math>\sqrt{7}</math>) then -1 - 3ω =(mod 13) 4.06275 however, -92 - 16ω =(mod 13) 8.66798. And, ok, there's a slight precision issue because I've only shown the first six digits of those numbers. But neither that precision issue, nor the mod 13 issue, convinces me that 4.06275 equals 8.66798. --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 03:11, 27 March 2016 (UTC)
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