Talk:Cipolla's algorithm: Difference between revisions
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We're supposed to solve '''x² ≡ n (mod p)''' but step 3 has us solving for ω given ω² in Fp². But if we could solve for ω given ω² in Fp² why do we need this algorithm? --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 04:34, 26 March 2016 (UTC) |
We're supposed to solve '''x² ≡ n (mod p)''' but step 3 has us solving for ω given ω² in Fp². But if we could solve for ω given ω² in Fp² why do we need this algorithm? --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 04:34, 26 March 2016 (UTC) |
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: Precision added to step 3 . The result is x + 0 * ω in Fp2 , that is x in Fp. The 'value' of ω is not needed.Same thing : we do'nt need the 'value' of i when dealing with complex numbers. Thx. --[[User:G.Brougnard|G.Brougnard]] ([[User talk:G.Brougnard|talk]]) 08:56, 26 March 2016 (UTC) |
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Revision as of 08:56, 26 March 2016
Something seems to be missing here...
We're supposed to solve x² ≡ n (mod p) but step 3 has us solving for ω given ω² in Fp². But if we could solve for ω given ω² in Fp² why do we need this algorithm? --Rdm (talk) 04:34, 26 March 2016 (UTC)
- Precision added to step 3 . The result is x + 0 * ω in Fp2 , that is x in Fp. The 'value' of ω is not needed.Same thing : we do'nt need the 'value' of i when dealing with complex numbers. Thx. --G.Brougnard (talk) 08:56, 26 March 2016 (UTC)